.1 CHAPTER
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9 Frequency Distributions; Measures of Central Tendency 9.2 Measures of Variation 9.3 The Normal Distribution 9.4 Normal Approximation to the Binomial Distribution Review Exercises Extended Application: Statistics in the Law The Castaneda Decision
Statistics
To understand the economics of large-scale farming, analysts look at historical data on the farming
industry. In an exercise in Section 1 you will calculate basic descriptive statistics for U.S. wheat prices and production levels over a recent decade. Later sections in this chapter develop more sophisticated techniques for extracting useful information from this kind of data.
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Statistics is a branch of mathematics that deals with the collection and summarization of data. Methods of statistical analysis make it possible for us to draw conclusions about a population based on data from a sample of the population. Statistical models have become increasingly useful in manufacturing, government, agriculture, medicine, and the social sciences, and in all types of research. In this chapter we give a brief introduction to some of the key topics from statistical theory.
9.1
FREQUENCY DISTRIBUTIONS; MEASURES OF CENTRAL TENDENCY
THINK ABOUT IT
How can the results of a survey of business executives on the number of college courses in management needed by a business major best be organized to provide useful information?
Frequency distributions can provide an answer to this question. Often, a researcher wishes to learn something about a characteristic of a population, but because the population is very large or mobile, it is not possible to examine all of its elements. Instead, a limited sample drawn from the population is studied to determine the characteristics of the population. For these inferences to be correct, the sample chosen must be a random sample. Random samples are representative of the population because they are chosen so that every element of the population is equally likely to be selected. A hand dealt from a well-shuffled deck of cards is a random sample. A random sample can be difficult to obtain in real life. For example, suppose you want to take a random sample of voters in your congressional district to see which candidate they prefer in the next election. If you do a telephone survey, you have a random sample of people who are at home to answer the telephone, underrepresenting those who work a lot of hours and are rarely home to answer the phone, or those who have an unlisted number, or those who cannot afford a telephone, or those who refuse to answer telephone surveys. Such people may have a different opinion than those you interview. A famous example of an inaccurate poll was made by the Literary Digest in 1936. Their survey indicated that Alfred Landon would win the presidential election; in fact, Franklin Roosevelt won with 62% of the popular vote. The Digests major error was mailing their surveys to a sample of those listed in telephone directories. During the Depression, many poor people did not have telephones, and the poor voted overwhelmingly for Roosevelt. Modern pollsters use sophisticated techniques to ensure that their sample is as random as possible. Once a sample has been chosen and all data of interest are collected, the data must be organized so that conclusions may be more easily drawn. One method of organization is to group the data into intervals; equal intervals are usually chosen. EXAMPLE 1 Business Executives A survey asked a random sample of 30 business executives for their recommendations as to the number of college units in management that a business major should have. The results are shown on the next page. Group the data into intervals and nd the frequency of each interval.
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8 34 21 3 25 22 16 0 9 14 8 15 20 12 28 19 93 15 12 17 16 23 19 12 14 29 13 24 18
Solution The highest number in the list is 34 and the lowest is 0; one convenient
way to group the data is in intervals of size 5, starting with 04 and ending with 3034. This gives an interval for each number in the list and results in seven equal intervals of a convenient size. Too many intervals of smaller size would not simplify the data enough, while too few intervals of larger size would conceal information that the data might provide. A rule of thumb is to use from six to fteen intervals. First tally the number of college units falling into each interval. Then total the tallies in each interval as in the following table. This table is an example of a grouped frequency distribution.
College Units 04 5 9 1014 1519 2024 2529 30 34
Tally 111 1111 1111 1 1111 111 1111 111 1
Frequency 3 4 6 8 5 3 1 Total: 30
The frequency distribution in Example 1 shows information about the data that might not have been noticed before. For example, the interval with the largest number of recommended units is 1519, and 19 executives (more than half) recommended between 10 and 24 units, inclusive. Also, the frequency in each interval increases rather evenly (up to 8) and then decreases at about the same pace. However, some information has been lost; for example, we no longer know how many executives recommended 12 units. The information in a grouped frequency distribution can be displayed in a histogram similar to the histograms for probability distributions in the previous chapter. The intervals determine the widths of the bars; if equal intervals are used, all the bars have the same width. The heights of the bars are determined by the frequencies.
N OT E In this section, the heights of the histogram bars give the frequencies. The histograms in the previous chapter were for probability distributions, and so the heights gave the probabilities. I
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A frequency polygon is another form of graph that illustrates a grouped frequency distribution. The polygon is formed by joining consecutive midpoints of the tops of the histogram bars with straight line segments. The midpoints of the
Finite Mathematics, Eighth Edition, by Margaret L. Lial, Raymond N. Greenwell, and Nathan P. Ritchey. Published by Addison Wesley. Copyright 2005 by Pearson Education, Inc.
9 .1 Frequency Distri but ions; Measures of Central Tendency
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8
6 Frequency
6 5
4 3 2
4 3
1
4.5 9.5 14.5 19.5 24.5 29.5 34.5 Number of units
FIGURE 1
rst and last bars are joined to endpoints on the horizontal axis where the next midpoint would appear. EXAMPLE 2 Frequency Distributions A grouped frequency distribution of college units was found in Example 1. Draw a histogram and a frequency polygon for this distribution.
Solution First draw a histogram, shown in red in Figure 1. To get a frequency polygon, connect consecutive midpoints of the tops of the bars. The frequency polygon is shown in blue.
10
Many graphing calculators have the capability of drawing a histogram. Figure 2 shows the data of Example 1 drawn on a TI-83/84 Plus.
0 0
35
FIGURE 2
Mean The average value of a probability distribution is the expected value of the distribution. Three measures of central tendency, or averages, are used with frequency distributions: the mean, the median, and the mode. The most important of these is the mean, which is similar to the expected value of a probability distribution. The arithmetic mean (the mean) of a set of numbers is the sum of the numbers, divided by the total number of numbers. Recall from Section 1.3 that we can write the sum of n numbers x1, x2, x3, * , xn in a compact way using summation notation:
x1 1 x2 1 x3 1 ) 1 xn 5 g x. The symbol x (read x-bar) is used to represent the mean of a sample.
MEAN
The mean of the n numbers x1, x2, x3, * , xn is x5 Sx . n
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EXAMPLE
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Bankruptcy The number of bankruptcy petitions (in thousands) led in the United States in the scal years 1996 2001 are given in the table.* Find the mean number of bankruptcy petitions led annually during this period.
Solution Let x1 5 1179, x2 5 1404, and so on. Here, n 5 6, since there are six
Year 1996 1997 1998 1999 2000 2001
Petitions Filed 1179 1404 1443 1319 1253 1492
numbers. x5 1179 1 1404 1 1443 1 1319 1 1253 1 1492 6 8090 x5 < 1348 6
The mean number of bankruptcy petitions led during the given years is about 1,348,000. Notice that this average is greater than three of the six values and less than the other three. Such symmetry does not always occur, however. The average for the years 1992 through 1997 (not included here) is greater than four of the six values because the value for 1997 was much greater than for the previous ve years. One large value can have a signicant effect on the mean, as we shall see later. As another example, the mean response for the number of college units in management that a business major should have, based on the sample of 30 business executives described in Example 1, is x 5 1 3 1 25 1 22 1 ) 1 18 2 / 30 5 478 / 30 5 15.93. EXAMPLE 4 M ean for Frequency Distributions Find the mean for the data shown in the following frequency distribution.
Value 30 32 33 37 42
Frequency 6 9 7 12 6 Total: 40
Value 3 Frequency 30 . 6 5 180 32 . 9 5 288 33 . 7 5 231 37 . 12 5 444 42 . 6 5 252 Total: 1395
Solution The value 30 appears six times, 32 nine times, and so on. To nd the
mean, rst multiply 30 by 6, 32 by 9, and so on. A new column, Value 3 Frequency, has been added to the frequency distribution. Adding the products from this column gives a total of 1395. The total from the frequency column is 40. The mean is x5 1395 5 34.875. 40
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*The U.S. Trustee Programs Efforts to Prevent Bankruptcy Fraud and Abuse, March 2003, Department of Justice, www.usdoj.gov.
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The mean of grouped data is found in a similar way. For grouped data, intervals are used, rather than single values. To calculate the mean, it is assumed that all these values are located at the midpoint of the interval. The letter x is used to represent the midpoints and f represents the frequencies, as shown in the next example. EXAMPLE 5 Business Executives Listed below is the grouped frequency distribution for the 30 business executives described in Example 1. Find the mean from the grouped frequency distribution. Interval 04 5 9 10 14 15 19 20 24 25 29 30 34 Midpoint, x 2 7 12 17 22 27 32 Frequency, f 3 4 6 8 5 3 1 Total: 30 Product, xf 6 28 72 136 110 81 32 Total: 465
Solution A column for the midpoint of each interval has been added. The numbers in this column are found by adding the endpoints of each interval and dividing by 2. For the interval 0 4, the midpoint is 1 0 1 4 2 / 2 5 2. The numbers in the product column on the right are found by multiplying each frequency by its corresponding midpoint. Finally, we divide the total of the product column by the total of the frequency column to get
x5
465 5 15.5. 30
Notice that this mean is slightly different from the earlier mean of 15.93. The reason for this difference is that we have acted as if each piece of data is at the midpoint, which is not true here, and is not true in most cases. Information is always lost when the data are grouped. It is more accurate to use the original data, rather than the grouped frequency, when calculating the mean, but the original data might not be available. Furthermore, the mean based upon the grouped data is typically not too far from the mean based upon the original data, and there may be situations in which the extra accuracy is not worth the extra effort.
N OT E 1. The midpoint of the intervals in a grouped frequency distribution may be values that the data cannot take on. For example, if we grouped the data for the 30 business executives into the intervals 0 5, 6 11, 1217, 18 23, 24 29, and 30 35, the midpoints would be 2.5, 8.5, 14.5, 20.5, 26.5, and 32.5, even though all the data are whole numbers. 2. If we used different intervals in Example 5, the mean would come out to be a slightly different number. Verify that with the intervals 0 5, 6 11, 1217, 18 23, 24 29, and 30 35, the mean in Example 5 is 16.1. I
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The formula for the mean of a grouped frequency distribution is given below.
MEAN OF A GROUPED DISTRIBUTION
The mean of a distribution, where x represents the midpoints, f the frequencies, and n 5 g f, is x5 Sxf . n
The mean of a random sample is a random variable, and for this reason it is sometimes called the sample mean. The sample mean is a random variable because it assigns a number to the experiment of taking a random sample. If a different random sample were taken, the mean would probably have a different value, with some values more probable than others. If another set of 30 business executives were selected in Example 1, the mean number of college units in management recommended for a business major might be 13.22 or 17.69. It is unlikely that the mean would be as small as 1.21 or as large as 32.75, although these values are remotely possible. We saw in Section 8.5 how to calculate the expected value of a random variable when we know its probability distribution. The expected value is sometimes called the population mean, denoted by the Greek letter m. In other words, E 1 x 2 5 m.
Furthermore, it can be shown that the expected value of x is also equal to m; that is, E 1 x 2 5 m.
For instance, consider again the 30 business executives in Example 1. We found that x 5 15.93, but the value of m, the average for all possible business executives, is unknown. If a good estimate of m were needed, the best guess (based on this data) is 15.93.
Median Asked by a reporter to give the average height of the players on his team, a Little League coach lined up his 15 players by increasing height. He picked the player in the middle and pronounced that player to be of average height. This kind of average, called the median, is dened as the middle entry in
Odd Number of Entries 8 7 Median 5 4 3 1
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Even Number of Entries 2 3 4 f 7 9 12
Median
5 4 1 7 5 5.5 2
Finite Mathematics, Eighth Edition, by Margaret L. Lial, Raymond N. Greenwell, and Nathan P. Ritchey. Published by Addison Wesley. Copyright 2005 by Pearson Education, Inc.
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a set of data arranged in either increasing or decreasing order. If there is an even number of entries, the median is dened to be the mean of the two center entries.
EXAMPLE
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Median Find the median for each list of numbers. (a) 11, 12, 17, 20, 23, 28, 29 Solution The median is the middle number; in this case, 20. (Note that the numbers are already arranged in numerical order.) In this list, three numbers are smaller than 20 and three are larger. (b) 15, 13, 7, 11, 19, 30, 39, 5, 10 Solution First arrange the numbers in numerical order, from smallest to largest. 5, 7, 10, 11, 13, 15, 19, 30, 39 The middle number, or median, can now be determined; it is 13. (c) 47, 59, 32, 81, 74, 153
Solution Write the numbers in numerical order.
32, 47, 59, 74, 81, 153 There are six numbers here; the median is the mean of the two middle numbers. Median 5 59 1 74 133 1 5 5 66 2 2 2
Both the mean and the median are examples of a statistic, which is simply a number that gives information about a sample. In some situations, the median gives a truer representation or typical element of the data than the mean. For example, suppose in an office there are 10 salespersons, 4 secretaries, the sales manager, and Carter Fenton, who owns the business. Their annual salaries are as follows: secretaries, $15,000 each; salespersons, $25,000 each; manager, $35,000; and owner, $200,000. The mean salary is x5 1 15,000 2 4 1 1 25,000 2 10 1 35,000 1 200,000 5 $34,062.50. 16
However, since 14 people earn less than $34,062.50 and only 2 earn more, this does not seem very representative. The median salary is found by ranking the salaries by size: $15,000, $15,000, $15,000, $15,000, $25,000, $25,000, * , $200,000. Since there are 16 salaries (an even number) in the list, the mean of the eighth and ninth entries will give the value of the median. The eighth and ninth entries are both $25,000, so the median is $25,000. In this example, the median gives a truer representative element than the mean.
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Mode Sues scores on ten class quizzes include one 7, two 8s, six 9s, and
one 10. She claims that her average grade on quizzes is 9, because most of her scores are 9s. This kind of average, found by selecting the most frequent entry, is called the mode.
EXAMPLE
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Mode Find the mode for each list of numbers. (a) 57, 38, 55, 55, 80, 87, 98 Solution The number 55 occurs more often than any other, so it is the mode. It is not necessary to place the numbers in numerical order when looking for the mode. (b) 182, 185, 183, 185, 187, 187, 189 Solution Both 185 and 187 occur twice. This list has two modes. (c) 10,708; 11,519; 10,972; 17,546; 13,905; 12,182 Solution No number occurs more than once. This list has no mode.
The mode has the advantages of being easily found and not being inuenced by data that are very large or very small compared to the rest of the data. It is often used in samples where the data to be averaged are not numerical. A major disadvantage of the mode is that we cannot always locate exactly one mode for a set of values. There can be more than one mode, in the case of ties, or there can be no mode if all entries occur with the same frequency. The mean is the most commonly used measure of central tendency. Its advantages are that it is easy to compute, it takes all the data into consideration, and it is reliablethat is, repeated samples are likely to give very similar means. A disadvantage of the mean is that it is inuenced by extreme values, as illustrated in the previous salary example. The median can be easy to compute and is inuenced very little by extremes. Like the mode, the median can be found in situations where the data are not numerical. For example, in a taste test, people are asked to rank ve soft drinks from the one they like best to the one they like least. The combined rankings then produce an ordered sample, from which the median can be identied. A disadvantage of the median is the need to rank the data in order; this can be difficult when the number of items is large.
EXAMPLE
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Seed Storage Seeds that are dried, placed in an airtight container, and stored in a cool, dry place remain ready to be planted for a long time. The table on the next page gives the amount of time that each type of seed can be stored and still remain viable for planting.*
*The Handy Science Answer Book, 2nd ed., The Carnegie Library of Pittsburgh, 1997, p. 247.
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9 .1 Frequency Distri but ions; Measures of Central Tendency
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Vegetable Beans Cabbage Carrots Cauliower Corn Cucumbers Melons Peppers Pumpkin Tomatoes
Years 3 4 1 4 2 5 4 2 4 3
Find the mean, median, and mode of the information in the table.
Solution
Method 1: Calculating by Hand
The mean amount of time that the seeds can be stored is x5 3141114121514121413 5 3.2 years. 10
After the numbers are arranged in order from smallest to largest, the middle number, or median, is found; it is 3.5. The number 4 occurs more often than any other, so it is the mode.
Method 2: Graphing Calculator
Most scientic calculators have some statistical capability and can calculate the mean of a set of data; graphing calculators can often calculate the median as well. For example, Figure 3 shows the mean and the median for the data above calculated on a TI-83/84 Plus, where the data was stored in the list L1. This calculator does not include a command for nding the mode.
FIGURE 3
Method 3: Spreadsheet
Using Microsoft Excel, place the data in cells A1 through A10. To nd the mean of this data, type =AVERAGE(A1:A10) in cell A11, or any other unused cell, and then press Enter. The result of 3.2 will appear in cell A11. To nd the median of this data, type =MEDIAN(A1:A10) in cell A12, or any other unused cell, and press Enter. The result of 3.5 will appear in cell A12. To nd the mode of this data, type =MODE(A1:A10) in cell A13, or any other unused cell, and press Enter. The result of 4 will appear in cell A13.
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9 .1 E X E R C I S E S
For Exercises 14, do the following: a. Group the data as indicated. b. Prepare a frequency distribution with a column for intervals and frequencies. c. Construct a histogram. d. Construct a frequency polygon. 1. Use six intervals, starting with 0 24. 74 103 149 42 4 79 60 74 59 55 98 61 67 83 71 71 46 63 66 69 42 75 62 71 77 133 27 132 50 140 78 65 87 57 78 91 82 73 94 48 4 139 64 95 12 87 65 62 81 63 66 65 49 45 51 127 118 141 56 88 69 56 84 93 63 60 68 51 73 54 20 138 130 65 119 50 88 76 93 48 70 39 76 95 57 30 121 76 104 64 63 94 82 54 89 64 77 94 72 69 51 56 67 88 81 70 81 54 66 87
2. Use seven intervals, starting with 30 39.
3. Repeat Exercise 1 using eight intervals, starting with 0 19. 4. Repeat Exercise 2 using six intervals, starting with 39 48. 5. How does a frequency polygon differ from a histogram? 6. Discuss the advantages and disadvantages of the mean as a measure of central tendency. Find the mean for each list of numbers. Round to the nearest tenth. 7. 8, 10, 16, 21, 25 9. 21,900; 22,850; 24,930; 29,710; 28,340; 40,000 11. 9.4, 11.3, 10.5, 7.4, 9.1, 8.4, 9.7, 5.2, 1.1, 4.7 Find the mean for the following. Round to the nearest tenth. 13. Value 3 5 9 12 Frequency 4 2 1 3 14. Value 9 12 15 18 Frequency 3 5 1 1 8. 44, 41, 25, 36, 67, 51 10. 38,500; 39,720; 42,183; 21,982; 43,250 12. 30.1, 42.8, 91.6, 51.2, 88.3, 21.9, 43.7, 51.2
Find the median for each list of numbers. 15. 12, 18, 32, 51, 58, 92, 106 17. 100, 114, 125, 135, 150, 172
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16. 596, 604, 612, 683, 719 18. 1072, 1068, 1093, 1042, 1056, 1005, 1009 20. .6, .4, .9, 1.2, .3, 4.1, 2.2, .4, .7, .1
19. 28.4, 9.1, 3.4, 27.6, 59.8, 32.1, 47.6, 29.8
Finite Mathematics, Eighth Edition, by Margaret L. Lial, Raymond N. Greenwell, and Nathan P. Ritchey. Published by Addison Wesley. Copyright 2005 by Pearson Education, Inc.
9 .1 Frequency Distri but ions; Measures of Central Tendency
Use a graphing calculator or spreadsheet to calculate the mean and median for the data in the indicated exercises. 21. Exercise 1 Find the mode or modes for each list of numbers. 23. 25. 27. 29. 4, 9, 8, 6, 9, 2, 1, 3 24. 21, 32, 46, 32, 49, 32, 49 74, 68, 68, 68, 75, 75, 74, 74, 70 26. 158, 162, 165, 162, 165, 157, 163 6.8, 6.3, 6.3, 6.9, 6.7, 6.4, 6.1, 6.0 28. 12.75, 18.32, 19.41, 12.75, 18.30, 19.45, 18.33 When is the median the most appropriate measure of central tendency? 22. Exercise 2
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30. Under what circumstances would the mode be an appropriate measure of central tendency? For grouped data, the modal class is the interval containing the most data values. Give the mean and modal class for each collection of grouped data. 31. Use the distribution in Exercise 1. 32. Use the distribution in Exercise 2. 33. To predict the outcome of the next congressional election, you take a survey of your friends. Is this a random sample of the voters in your congressional district? Explain why or why not.
Applications
BUSINESS AND ECONOMICS
Wheat Production U.S. wheat prices and production gures for a recent decade are given in the following table.* Price ($ per bushel) 3.24 3.26 3.45 4.55 4.30 3.38 2.65 2.48 2.62 2.80 Production (millions of bushels) 2467 2396 2321 2183 2285 2481 2547 2299 2232 1958
36. Salaries The total pay (in thousands of dollars) for the 15 highest paid executives in 2001 is given in the following table. Person, Company Lawrence J. Ellison, Oracle Jozef Straus, Uniphase Howard Solomon, Forest Laboratories Gregory P. Dougherty, Uniphase Charles B. Wang, Computer Associates Richard S. Fuld, Jr., Lehman Brothers Donald R. Scifres, Uniphase Jeffrey O. Henley, Oracle M. Zita Cobb, Uniphase Robert J. Herbold, Microsoft Tony L. White, Applera David M. Rickey, Applied Micro Circuits John F. Gifford, Maxim Jeffrey S. Raikes, Microsoft Paul F. Folino, Emulex Total Pay 706,077 150,817 148,476 121,375 119,097 105,183 94,356 86,534 69,496 68,213 61,877 59,504 58,011 57,962 56,239
Year 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001
Find the mean and median for the following. 34. Price per bushel of wheat 35. Wheat production
*The World Almanac and Book of Facts 2003, pp. 137138. http://www.businessweek.com/magazine/content/02_12/b3775069.htm.
a. Find the mean total pay for this group of people. b. Find the median total pay for this group of people.
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37. Household Income The total household income for fulltime African American workers making under $100,000 in 2000 is given in the following table.*
Airline America West United TWA American Continental Delta Northwest U.S. Airways American Eagle Alaska Southwest
Complaints 729 2448 528 1964 952 2021 1065 1049 204 174 281
Complaints per 100,000 Passenger Boardings 3.72 3.24 2.54 2.51 2.23 2.16 1.97 1.87 1.70 1.27 .38
Income Range Under $5,000 $5,000 $9,999 $10,000 $14,999 $15,000 $24,999 $25,000 $34,999 $35,000 $49,999 $50,000 $74,999 $75,000 $99,999
Midpoint Salary $2500 $7500 $12,500 $20,000 $30,000 $42,500 $62,500 $87,500
Frequency (in thousands) 814 1389 1268 2203 1722 2243 2030 868
Use this table to estimate the mean household income for full-time African American workers in 2000. 38. Household Income The total household income for fulltime white American workers making under $100,000 in 2000 is given in the following table.
a. By considering the numbers in the column labeled Complaints, calculate the mean and median number of complaints per airline. b. Explain why the average found in part a is not meaningful. c. There were a total of 539,805,753 passenger boardings in 2001. Divide the total number of complaints by the total number of passenger boardings to nd the mean number of complaints per 100,000 passenger boardings. d. Find the mean of the numbers in the column labeled Complaints per 100,000 passenger boardings. Explain why this number is different from the answer to part c, and which is more meaningful.
Income Range Under $5,000 $5,000 $9,999 $10,000 $14,999 $15,000 $24,999 $25,000 $34,999 $35,000 $49,999 $50,000 $74,999 $75,000 $99,999
Midpoint Salary $2500 $7500 $12,500 $20,000 $30,000 $42,500 $62,500 $87,500
Frequency (in thousands) 2037 4870 5844 11,511 11,157 13,636 17,178 9740
LIFE SCIENCES
40. Pandas The size of the home ranges (in square kilometers) of several pandas were surveyed over a years time, with the following results.
Home Range .1.5 .6 1.0 1.11.5 1.6 2.0 2.12.5 2.6 3.0 3.1 3.5
Frequency 11 12 7 6 2 1 1
a. Use this table to estimate the mean household income for full-time white American workers in 2000. b. Compare this estimate with the estimate found in Exercise 37. Does this provide some evidence that full-time white American workers have higher household earnings than full-time African American workers? 39. Airlines The number of consumer complaints against the top U.S. airlines during 2001 is given in the following table.
Sketch a histogram and frequency polygon for the data.
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*Time Almanac 2003, Time Inc., p. 638. Ibid. Time Almanac 2003, Time Inc., p. 216.
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9 .1 Frequency Distri but ions; Measures of Central Tendency
41. Blood Types The number of recognized blood types varies by species, as indicated by the following table.* Find the mean, median, and mode of this data. Animal Pig Cow Chicken Horse Human Sheep Dog Rhesus monkey Mink Rabbit Mouse Rat Cat Number of Blood Types 16 12 11 9 8 7 7 6 5 5 4 4 2 Year 1970 1971 1972 1973 1974 1975 1976 1977 1978 1979 1980 1981 Days 5 11 11 8 11 3 8 11 5 7 12 12 Year 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 Days 11 20 7 4 8 14 21 10 6 21 4 25 Year 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003
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d. Compare the histogram in Exercise 42 with the following histogram. What seems to be true of the U.S. population?
09
1019
2029
3039
4049
5059
6069
GENERAL INTEREST
44. Temperature The following table gives the number of days in June and July of recent years in which the temperature reached 90 degrees or higher in New Yorks Central Park.
70+
14 13 12 11 10
Percent of population
Days 16 14 0 10 5 24 3 4 13 11
SOCIAL SCIENCES
42. Population The following histogram shows estimates of the percent of the U.S. population in each age group in the year 2000. What percent of the population is estimated to be in each age group? a. 10 19 b. 60 69 c. What age range has the largest percent of the population?
20 15 10 5 0 Percent of population
a. Prepare a frequency distribution with a column for intervals and frequencies. Use six intervals, starting with 0 4.
1019 2029 3039 4049 5059 6069
b. Sketch a histogram and a frequency polygon, using the intervals in part a. c. Find the mean for the original data. d. Find the mean using the grouped data from part a. e. Explain why your answers to parts c and d are different. f. Find the median and the mode for the original data. 45. Temperature The table on the following page gives the average monthly temperatures in degrees Fahrenheit for a certain area.
09
43. Population The following histogram shows estimates of the percent of the U.S. population in each age group in the year 2025. What percent of the population is estimated to be in each age group then? a. 20 29 b. 701 c. What age group will have the smallest percent of the population?
*The Handy Science Answer Book, Carnegie Library of Pittsburgh, 1997, p. 264. U.S. Census Bureau, Jan. 13, 2000. U.S. Census Bureau, Jan. 13, 2000. The New York Times, July 31, 1996, p. B4, and www.accuweather.com.
70+
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48. Personal Wealth Washington Post writer John Schwartz pointed out that if Microsoft Corp. cofounder Bill Gates, who, at the time, was reportedly worth $10 billion, lived in a town with 10,000 totally penniless people, the average personal wealth in the town would make it seem as if everyone were a millionaire. a. Verify Schwartzs statement. b. What would be the median personal wealth in this town? c. What would be the mode for the personal wealth in this town? d. In this example, which average is most representative: the mean, the median, or the mode? 49. Baseball Salaries According to CBS Sportsline, the salary in 2003 for each player on the Chicago Cubs baseball team is given in the following table.
Month January February March April May June July August September October November December
Maximum 39 39 44 50 60 69 79 78 70 51 47 40
Minimum 16 18 21 26 32 37 43 42 37 31 24 20
Name Find the mean and median for the following. a. The maximum temperature b. The minimum temperature 46. Olympics The number of nations participating in the winter Olympic games, from 1968 to 2002, is given below.* Find the following measures for the data. a. Mean b. Median c. Mode Year 1968 1972 1976 1980 1984 1988 1992 1994 1998 2002 Nations Participating 37 35 37 37 49 57 64 67 72 77 Sammy Sosa Moises Alou Eric Karros Kerry Wood Mark Grudzielanek Antonio Alfonseca Matt Clement Aramis Ramirez Shawn Estes Damian Miller Mike Remlinger Dave Veres Mark Guthrie Mark Prior Lenny Harris Ramon Martinez Paul Bako Troy OLeary Tom Goodwin Kyle Farnsworth Mark Bellhorn Joe Borowski Corey Patterson Juan Cruz Carlos Zambrano Will Ohman Scott Chiasson Hee Seop Choi
Salary (in U.S. dollars) 16,875,000 9,500,000 8,375,000 6,190,000 5,500,000 4,000,000 4,000,000 3,000,000 3,000,000 2,700,000 2,633,333 2,000,000 1,600,000 1,450,000 800,000 800,000 750,000 750,000 635,000 600,000 465,000 410,000 365,000 340,000 340,000 305,000 305,000 305,000
47. Personal Wealth When Russian billionaire Roman Abramovich became governor of the Russian province Chukotka (in the Bering Straits, opposite Alaska), it instantly became the fourth most prosperous region in Russia, even though its 80,000 other residents are poor. Mr. Abramovich was then worth $5.7 billion. Suppose each of the 80,000 other residents of Chukotka was worth $100. a. Calculate the average worth of a citizen of Chukotka. b. What does this example tell you about the use of the mean to describe an average?
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*The New York Times Almanac 2003, p. 901. http://discover.npr.org/features/feature.jhtml?wfId=1318509. Schwartz, J., Mean Statistics: When Is Average Best? The Washington Post, Jan. 11, 1995, p. H7. http://cbs.sportsline.com/mlb/teams/salaries/CHC.
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a. Find the mean, median, and mode of the salaries. b. Which average best describes this data? c. Why is there such a difference between the mean and the median?
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50. SAT I: Reasoning Test Given the following sequence of numbers* 1, a, a2, a3, * , an, where n is a positive even integer, with the additional assumption that a is a positive number, the median is best described as a. greater than an/2; b. smaller than an/2; c. equal to an/2. d. The relationship cannot be determined from the information given.
9.2 MEASURES OF VARIATION
THINK ABOUT IT
How can we tell when a manufacturing process is out of control?
To answer this question, we need to understand measures of variation, which tell us how much the numbers in a sample vary from the mean. The mean gives a measure of central tendency of a list of numbers, but tells nothing about the spread of the numbers in the list. For example, look at the following three samples.
I II III
3 4 10
5 4 1
6 4 0
3 4 0
3 4 9
Each of these three samples has a mean of 4, and yet they are quite different; the amount of dispersion or variation within the samples is different. Therefore, in addition to a measure of central tendency, another kind of measure is needed that describes how much the numbers vary.
*Permission to reprint SAT materials does not constitute review or endorsement by Educational Testing Service or the College Board of this publication as a whole or of any other questions or testing information it may contain. This problem appeared, minus the additional assumption, on an SAT in 1996. Colin Rizzio, a high school student at the time, became an instant celebrity when he noticed that the additional assumption was needed to complete the problem. The New York Times, Feb. 7, 1997, p. A1.
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The largest number in sample I is 6, while the smallest is 3, a difference of 3. In sample II this difference is 0; in sample III, it is 10. The difference between the largest and smallest number in a sample is called the range, one example of a measure of variation. The range of sample I is 3, of sample II, 0, and of sample III, 10. The range has the advantage of being very easy to compute, and gives a rough estimate of the variation among the data in the sample. It depends only on the two extremes, however, and tells nothing about how the other data are distributed between the extremes.
EXAMPLE
1
Range Find the range for each list of numbers. (a) 12, 27, 6, 19, 38, 9, 42, 15
Solution The highest number here is 42; the lowest is 6. The range is the difference between these numbers, or 42 2 6 5 36.
(b) 74, 112, 59, 88, 200, 73, 92, 175
Solution
Range 5 200 2 59 5 141
The most useful measure of variation is the standard deviation. Before dening it, however, we must nd the deviations from the mean, the differences found by subtracting the mean from each number in a sample. EXAMPLE 2
Deviations from the Mean Find the deviations from the mean for the numbers 32, 41, 47, 53, 57.
Solution Adding these numbers and dividing by 5 gives a mean of 46. To nd the deviations from the mean, subtract 46 from each number in the list. For example, the rst deviation from the mean is 32 2 46 5 214; the last is 57 2 46 5 11. To check your work, nd the sum of these deviations. It should always equal 0. (The answer is always 0 because the positive and negative numbers cancel each other.)
Deviation Number 32 41 47 53 57
From Mean 214 25 1 7 11
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To nd a measure of variation, we might be tempted to use the mean of the deviations. As mentioned above, however, this number is always 0, no matter how widely the data are dispersed. One way to solve this problem is to use absolute value and nd the mean of the absolute values of the deviations from the mean. Absolute value is awkward to work with algebraically, and there is an alternative approach that provides better theoretical results. In this method, the way to get a list of positive numbers is to square each deviation and then nd the mean. When nding the mean of the squared deviations, most statisti-
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cians prefer to divide by n 2 1, rather than n. We will give the reason later in this section. For the data above, this gives 1 214 2 2 1 1 25 2 2 1 12 1 72 1 112 196 1 25 1 1 1 49 1 121 5 521 4 5 98. This number, 98, is called the variance of the distribution. Since it is found by averaging a list of squares, the variance of a sample is represented by s 2. For a sample of n numbers x1, x2, x3, * , xn, with mean x, the variance is s2 5 g 1 x 2 x 22 . n21
The following shortcut formula for the variance can be derived algebraically from the formula above. This is left as an exercise.
VARIANCE
The variance of a sample of n numbers x1, x2, x3, * , xn, with mean x, is s2 5 Sx 2 2 nx 2 . n21
To nd the variance, we squared the deviations from the mean, so the variance is in squared units. To return to the same units as the data, we use the square root of the variance, called the standard deviation.
STANDARD DEVIATION
The standard deviation of the n numbers x1, x2, x3, * , xn, with mean x, is s5 Sx 2 2 nx 2 . n21
As its name indicates, the standard deviation is the most commonly used measure of variation. The standard deviation is a measure of the variation from the mean. The size of the standard deviation tells us something about how spread out the data are from the mean. EXAMPLE 3 Standard Deviation Find the standard deviation of the numbers 7, 9, 18, 22, 27, 29, 32, 40.
Solution
Method 1: Calculating by Hand
The mean of the numbers is 7 1 9 1 18 1 22 1 27 1 29 1 32 1 40 5 23. 8
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Arrange the work in columns, as shown in the table.
Number 7 9 18 22 27 29 32 40
Square of the Number 49 81 324 484 729 841 1024 1600 Total: 5132
The total of the second column gives g x2 5 5132. Now nd the variance. The variance is s2 5 g x 2 2 nx2 n21 5132 2 8 1 23 2 2 5 821 < 128.6, rounded, and the standard deviation is "128.57 < 11.3
Method 2: Graphing Calculator
The data are entered into the L5 list on a TI-83/84 Plus calculator. Figure 4 shows how the variance and standard deviation are then calculated. Figure 5 shows an alternative method, going through the STAT menu, which calculates the mean, the standard deviation using both n 2 1 and n in the denominator, and other statistics.
FIGURE 4
FIGURE 5
Method 3: Spreadsheet
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The data are entered in cells A1 through A8. Then, in cell A9, type =VAR(A1:A8) and press Enter. The standard deviation can be calculated by either taking the square root of cell A9 or by typing =STDEV(A1:A8) in cell A10 and pressing Enter.
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C A U T I O N Be careful to divide by n 2 1, not n, when calculating the standard deviation of a sample. Many calculators are equipped with statistical keys that compute the variance and standard deviation. Some of these calculators use n 2 1 and others use n for these computations; some may have keys for both. Check your calculators instruction book before using a statistical calculator for the exercises. I
One way to interpret the standard deviation uses the fact that, for many populations, most of the data are within three standard deviations of the mean. (See Section 9.3.) This implies that, in Example 3, most of the population from which this sample is taken are between x 2 3s 5 23 2 3 1 11.3 2 5 210.9 and x 1 3s 5 23 1 3 1 11.3 2 5 56.9. This has important implications for quality control. If the sample in Example 3 represents measurements of a product that the manufacturer wants to be between 5 and 45, the standard deviation is too large, even though all the numbers are within these bounds. We saw in the previous section that the mean of a random sample is a random variable. It should not surprise you, then, to learn that the variance and standard deviation are also random variables. We will refer to the variance and standard deviation of a random sample as the sample variance and sample standard deviation. Recall from the previous section that the sample mean x is not the same as the population mean m, which is dened by m 5 E 1 x 2 , but that x gives a good approximation to m because E 1 x 2 5 m. Similarly, there is a population variance, denoted s 2, dened by s 2 5 E 1 x 2 m 2 2, which measures the amount of variation in a population. The population standard deviation is simply s, the square root of the population variance s 2. (The Greek letter s is the lowercase version of sigma. You have already seen g , the uppercase version.) In more advanced courses in statistics, it is shown that E 1 s2 2 5 s 2. The reason many statisticians prefer n 2 1 in the denominator of the standard deviation formula is that it makes E 1 s2 2 5 s 2 true; this is not true if n is used in the denominator. It may surprise you, then, that E 1 s 2 5 s is false, whether n or n 2 1 is used. If n is large, the difference between E 1 s 2 and s is slight, so, in practice, the sample standard deviation s gives a good estimate of the population standard deviation s. For data in a grouped frequency distribution, a slightly different formula for the standard deviation is used.
FOR REVIEW
Recall from Section 8.5 that a random variable is a function that assigns a real number to each outcome of an experiment. When the experiment consists of drawing a random sample, the standard deviation and the variance are two real numbers assigned to each outcome. Every time the experiment is performed, the standard deviation and variance will most likely have different values, with some values more probable than others.
STANDARD DEVIATION FOR A GROUPED DISTRIBUTION
The standard deviation for a distribution with mean x, where x is an interval midpoint with frequency f, and n 5 g f, is s5 g fx 2 2 nx 2 . n21
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The formula indicates that the product fx2 is to be found for each interval. Then these products are summed, n times the sum of the mean is subtracted, and the difference is divided by one less than the total frequency; that is, by n 2 1. The square root of this result is s, the standard deviation.
C A U T I O N In calculating the standard deviation for either a grouped or ungrouped distribution, using a rounded value for the mean may produce an inaccurate value. I
EXAMPLE
4
Standard Deviation for Grouped Data Find s for the grouped data of Example 5, Section 9.1.
Solution Begin by adding columns for x 2 (the midpoint of the interval) and fx 2.
Recall from Example 5 of Section 9.1 that x 5 15.5.
Interval 04 5 9 10 14 15 19 20 24 25 29 30 34
x 2 7 12 17 22 27 32
x2 4 49 144 289 484 729 1024
f 3 4 6 8 5 3 1 Total: 30
fx2 12 196 864 2312 2420 2187 1024 Total: 9015
Use the formula above with n 5 30 to nd s. s5 5 g fx2 2 nx2 n21
< 7.89
9015 2 30 1 15.5 2 2 30 2 1
Verify that the standard deviation of the original, ungrouped data in Example 1 of Section 9.1 is 7.92.
EXAMPLE
5
Quality Assurance Statistical process control is a method of determining when a manufacturing process is out of control, producing defective items. The procedure involves taking samples of a measurement on a product over a production run and calculating the mean and standard deviation of each sample. These results are used to determine when the manufacturing process is out of control. For example, three sample measurements from a manufacturing process on each of four days are given
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Day Sample Number Measurements
1 1 23 0 2 21 / 3 2.5 2 0 5 2 7/3 2.5 3 4 3 2 3 1 1 5 4 3 4 1
2 2 22 0 1 21 / 3 1.5 3 4 3 4 11/3 .6 1 3 22 0 1/3 2.5
x s
3 2 21 0 1 0 1
3 0 0 22 22 / 3 1.2
1 4 3 3 10/3 .6
4 2 22 0 21 21 1
3 1 3 0 4/3 1.5
in the table above. The mean x and standard deviation s are calculated for each sample. Next, the mean of the 12 sample means, X, and the mean of the 12 sample standard deviations, s, are found (using the formula for x). Here, these measures are X 5 1.3 and s 5 1.41.
The control limits for the sample means are given by X 6 k1s, where k1 is a constant found from a manual.* For samples of size 3, k1 5 1.954, so the control limits for the sample means are 1.3 6 1 1.954 2 1 1.41 2 . The upper control limit is 4.06, and the lower control limit is 21.46. Similarly, the control limits for the sample standard deviations are given by k2 . s and k3 . s, where k2 and k3 also are values given in the same manual. Here, k2 5 2.568 and k3 5 0, with the upper and lower control limits for the sample standard deviations equal to 2.568 1 1.41 2 and 0 1 1.41 2 , or 3.62 and 0. As long as the sample means are between 21.46 and 4.06 and the sample standard deviations are between 0 and 3.62, the process is in control.
9.2 EXERCISES
1. How are the variance and the standard deviation related? 2. Why cant we use the sum of the deviations from the mean as a measure of dispersion of a distribution? Find the range and standard deviation for each set of numbers. 3. 42, 38, 29, 74, 82, 71, 35 5. 241, 248, 251, 257, 252, 287 7. 3, 7, 4, 12, 15, 18, 19, 27, 24, 11 4. 122, 132, 141, 158, 162, 169, 180 6. 51, 58, 62, 64, 67, 71, 74, 78, 82, 93 8. 15, 42, 53, 7, 9, 12, 28, 47, 63, 14
*For example, see Statistical Process Control by Leonard A. Doty, Industrial Press, Inc., 1991, p. 317.
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Find the standard deviation for the following grouped data. 9. (From Exercise 1, Section 9.1) Interval 0 24 25 49 50 74 75 99 100 124 125 149 Frequency 4 3 6 3 5 9 10. (From Exercise 2, Section 9.1) Interval 30 39 40 49 50 59 60 69 70 79 80 89 90 99 Frequency 1 6 13 22 17 13 8
Chebyshevs theorem states that for any set of numbers, the fraction that will lie within k standard deviations of the mean 1 for k . 1 2 is at least 12 1 . k2
For example, at least 1 2 1 / 22 5 3 / 4 of any set of numbers lie within 2 standard deviations of the mean. Similarly, for any probability distribution, the probability that a number will lie within k standard deviations of the mean is at least 1 2 1 / k2. For example, if the mean is 100 and the standard deviation is 10, the probability that a number will lie within 2 standard deviations of 100, or between 80 and 120, is at least 3 / 4. Use Chebyshevs theorem to nd the fraction of all the numbers of a data set that must lie within the following numbers of standard deviations from the mean. 11. 3 12. 4 13. 5
In a certain distribution of numbers, the mean is 50 with a standard deviation of 6. Use Chebyshevs theorem to tell the probability that a number lies in each interval. 14. Between 38 and 62 16. Less than 38 or more than 62 18. Discuss what the standard deviation tells us about a distribution. 19. Explain the difference between the sample mean and standard deviation, and the population mean and standard deviation. 20. Derive the shortcut formula for the variance s2 5 from the formula s2 5 g 1 x 2 x 22 n21 g x2 2 nx2 n21 15. Between 32 and 68 17. Less than 32 or more than 68
and the following summation formulas, in which c is a constant: g cx 5 c g x, g c 5 nc,
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and
(Hint: Multiply out 1 x 2 x 2 2.)
g 1 x 6 y 2 5 g x 6 g y.
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Applications
BUSINESS AND ECONOMICS
21. Battery Life Forever Power Company analysts conducted tests on the life of its batteries and those of a competitor (Brand X). They found that their batteries had a mean life (in hours) of 26.2, with a standard deviation of 4.1. Their results for a sample of 10 Brand X batteries were as follows: 15, 18, 19, 23, 25, 25, 28, 30, 34, 38. a. Find the mean and standard deviation for the sample of Brand X batteries. b. Which batteries have a more uniform life in hours? c. Which batteries have the highest average life in hours? 22. Sales Promotion The Quaker Oats Company conducted a survey to determine whether a proposed premium, to be included in boxes of cereal, was appealing enough to generate new sales.* Four cities were used as test markets, where the cereal was distributed with the premium, and four cities as control markets, where the cereal was distributed without the premium. The eight cities were chosen on the basis of their similarity in terms of population, per capita income, and total cereal purchase volume. The results were as follows. Percent Change in Average Market Share per Month 118 115 17 110 11 28 25 0
e. Find the difference between the means of parts a and b. This difference represents the estimate of the percent change in sales due to the premium. f. The two standard deviations from parts c and d were used to calculate an error of 67.95 for the estimate in part e. With this amount of error, what are the smallest and largest estimates of the increase in sales? On the basis of the interval estimate of part f, the company decided to mass-produce the premium and distribute it nationally. 23. Process Control The following table gives 10 samples of three measurements, made during a production run. Sample Number 1 2 22 1 2 3 21 4 3 22 0 1 4 23 1 2 5 21 2 4 6 3 2 2 7 0 1 2 8 21 2 3 9 2 3 2 10 0 0 2
Use the information in Example 5 to nd the following. a. Find the mean x for each sample of three measurements. b. Find the standard deviation s for each sample of three measurements. c. Find the mean X of the sample means. d. Find the mean s of the sample standard deviations. e. Using k1 5 1.954, nd the upper and lower control limits for the sample means. f. Using k2 5 2.568 and k3 5 0, nd the upper and lower control limits for the sample standard deviations. 24. Process Control Given the following measurements from later samples on the process in Exercise 23, decide whether the process is out of control. (Hint: Use the results of Exercise 23e and f.) Sample Number 1 3 25 2 2 24 2 1 3 2 0 1 4 5 1 24 5 4 21 22 6 0 1 26
City 1 Test Cities 2 3 4 1 Control Cities 2 3 4
a. Find the mean of the change in market share for the four test cities. b. Find the mean of the change in market share for the four control cities. c. Find the standard deviation of the change in market share for the test cities. d. Find the standard deviation of the change in market share for the control cities.
*This example was supplied by Jeffery S. Berman, Senior Analyst, Marketing Information, Quaker Oats Company.
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cise 41 of the previous section, the mean was found to be 7.38. Animal Pig Cow Chicken Horse Human Year 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 Number Unemployed 9.61 8.94 8.00 7.40 7.24 6.74 6.21 5.88 5.66 6.74 a. Find the variance and the standard deviation of these data. b. How many of these animals have blood types that are within 1 standard deviation of the mean? 29. Tumor Growth The amount of time that it takes for various slow-growing tumors to double in size are listed in the following table. Doubling Time (days) 84 91 128 131 143 153 164 Sheep Dog Rhesus monkey Mink Rabbit Mouse Rat Cat Number of Blood Types 16 12 11 9 8 7 7 6 5 5 4 4 2
25. Washer Thickness An assembly-line machine turns out washers with the following thicknesses (in millimeters). 1.20 2.05 1.64 1.01 1.46 2.19 1.25 1.90 2.25 2.20 2.03 2.08 2.58 2.13 1.96 2.19 1.86 1.83 1.29 1.65 1.17 1.15 2.27 2.24
Find the mean and standard deviation of these thicknesses. 26. Unemployment The number of unemployed workers in the United States in recent years (in millions) is given below.*
a. Find the mean number unemployed (in millions) in this period. Which year has unemployment closest to the mean? b. Find the standard deviation for the data. c. In how many of these years is unemployment within 1 standard deviation of the mean? d. In how many of these years is unemployment within 3 standard deviations of the mean?
Type of Cancer Breast cancer Rectal cancer Synovioma Skin cancer Lip cancer Testicular cancer Esophageal cancer
LIFE SCIENCES
27. Blood pH A medical laboratory tested 21 samples of human blood for acidity on the pH scale, with the following results. 7.1 7.4 7.5 7.5 7.5 7.4 7.3 7.3 7.4 7.4 7.2 7.1 7.6 7.4 7.3 7.2 7.3 7.4 7.3 7.5 7.4
a. Find the mean and standard deviation of these data. b. How many of these cancers have doubling times that are within 2 standard deviations of the mean? c. If a person had a nonspecied tumor that was doubling every 200 days, discuss whether this particular tumor is growing at a rate that would be expected.
a. Find the mean and standard deviation. b. What percent of the data is within 2 standard deviations of the mean? 28. Blood Types The number of recognized blood types between species is given in the following table. In Exer-
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*The World Almanac and Book of Facts 2003, p. 141. The Handy Science Answer Book, Carnegie Library of Pittsburgh, 1997, p. 264. Collins, Vincent, R. Kenneth Lodffer, and Harold Tivey, Observations on Growth Rates of Human Tumors, American Journal of Roentgen, Vol. 76, No. 5, Nov. 1956, pp. 988 1000.
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GENERAL INTEREST
30. Governors In 2002, eleven state governors earned at least $124,000 annually (not counting expense allowances) as listed below. (Salaries are given in thousands of dollars and are rounded to the nearest $1000.)* State California Georgia Illinois Massachusetts Michigan New Mexico New York Pennsylvania Vermont Virginia Washington Salary 175 127 150 135 177 150 179 142 125 125 142
b. Find the standard deviation for the data. c. What percent of the governors have salaries within 1 standard deviation of the mean? d. What percent of the governors have salaries within 3 standard deviations of the mean? 31. Baseball Salaries The table in Exercise 49 in the previous section listed the total salary in 2003 for each player on the Chicago Cubs baseball team. a. Calculate the standard deviation of these data. b. What percent of the 2003 Chicago Cubs players have salaries that are beyond 3 standard deviations from the mean? c. What does your answer to part b suggest? 32. Cookies Marie Revak and Jihan Williams performed an experiment to determine whether Oreo Double Stuf cookies contain twice as much lling as traditional Oreo cookies. The following table gives the results in grams of the amount of lling inside 49 traditional cookies and 52 Double Stuf cookies. a. Find the mean, maximum, minimum, and standard deviation of the weights for traditional Oreo cookies. b. Find the mean, maximum, minimum, and standard deviation of the weights for Oreo Double Stuf cookies. Double Stuf 4.7 6.5 5.5 5.6 5.1 5.3 5.4 5.4 3.5 5.5 6.5 5.9 5.4 4.9 5.6 5.7 5.3 6.9 Double Stuf 6.5 6.3 4.8 3.3 6.4 5.0 5.3 5.5 5.0 6.0 5.7 6.3 6.0 6.3 6.1 6.0 5.8 Double Stuf 5.8 5.9 6.2 5.9 6.5 6.5 6.1 5.8 6.0 6.2 6.2 6.0 6.8 6.2 5.4 6.6 6.2
a. Find the mean salary of these governors. Which states have the governor with the salary closest to the mean? Traditional 2.9 2.8 2.6 3.5 3.0 2.4 2.7 2.4 2.5 2.2 2.6 2.6 2.9 2.6 2.6 3.1 2.9 Traditional 2.4 2.8 3.8 3.1 2.9 3.0 2.1 3.8 3.0 3.0 2.8 2.9 2.7 3.2 2.8 3.1 Traditional 2.7 2.8 2.6 2.6 3.0 2.8 3.5 3.3 3.3 2.8 3.1 2.6 3.5 3.5 3.1 3.1
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*The World Almanac and Book of Facts 2003, p. 68. Data from Revak, Marie, and Jihan Williams, The Double Stuf Dilemma, Mathematics Teacher, Vol. 92, No. 8, Nov. 1999, pp. 674 675.
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c. What percent of the data of traditional Oreo cookies is within 2 standard deviations of the Double Stuf Oreo mean? (Hint: Use the mean and standard deviation for the Double Stuf data.) d. What percent of the data of traditional Oreo cookies, when multiplied by 2, is within 2 standard deviations of the Double Stuf Oreo mean? (Hint: Use the mean and standard deviation for the Double Stuf data.) e. Is there evidence that Double Stuf Oreos have twice as much lling as the traditional Oreo cookie? Explain.
9.3 THE NORMAL DISTRIBUTION
THINK ABOUT IT
What is the probability that a salesperson drives between 1200 miles and 1600 miles per month?
FOR REVIEW
Empirical probabilities, discussed in Section 7.4, are derived from grouped data by dividing the frequency or amount for each group by the total for all the groups. This gives one example of a probability distribution, discussed further in Sections 7.4 and 8.5.
This question can be answered by using the normal probability distribution introduced in this section. Suppose a bank is interested in improving its services to customers. The manager decides to begin by nding the amount of time tellers spend on each transaction, rounded to the nearest minute. The times for 75 different transactions are recorded, with the results shown in the following table. The frequencies listed in the second column are divided by 75 to nd the empirical probabilities.
Time 1 2 3 4 5 6 7 8 9 10
Frequency 3 5 9 12 15 11 10 6 3 1
Probability 3 / 75 5 .04 5 / 75 < .07 9 / 75 5 .12 12 / 75 5 .16 15 / 75 5 .20 11 / 75 < .15 10 / 75 < .13 6 / 75 5 .08 3 / 75 5 .04 1 / 75 < .01
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Figure 6(a) shows a histogram and frequency polygon for the data. The heights of the bars are the empirical probabilities, rather than the frequencies. The transaction times are given to the nearest minute. Theoretically at least, they could have been timed to the nearest tenth of a minute, or hundredth of a minute,
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.20 Probability .15 .10 .05 1 2 3 4 5 6 7 8 9 10 Minutes Probability
.20 .15 .10 .05 1 2 3 4 5 6 7 8 9 10 Minutes
(a) FIGURE 6
(b)
or even more precisely. In each case, a histogram and frequency polygon could be drawn. If the times are measured with smaller and smaller units, there are more bars in the histogram and the frequency polygon begins to look more and more like the curve in Figure 6(b) instead of a polygon. Actually, it is possible for the transaction times to take on any real number value greater than 0. A distribution in which the outcomes can take any real number value within some interval is a continuous distribution. The graph of a continuous distribution is a curve. The distribution of heights (in inches) of college women is another example of a continuous distribution, since these heights include innitely many possible measurements, such as 53, 58.5, 66.3, 72.666, * , and so on. Figure 7 shows the continuous distribution of heights of college women. Here the most frequent heights occur near the center of the interval shown. Another continuous curve, which approximates the distribution of yearly incomes in the United States, is given in Figure 8. The graph shows that the most frequent incomes are grouped near the low end of the interval. This kind of distribution, where the peak is not at the center, is called skewed.
Frequency
0
60
63
66
69
72
Frequency 0 Heights of female college freshmen (in inches) 100,000 Income in the United States (in dollars)
FIGURE 7
FIGURE 8
Many natural and social phenomena produce continuous probability distributions whose graphs can be approximated very well by bell-shaped curves, such as those shown in Figure 9. Such distributions are called normal distributions and their graphs are called normal curves. Examples of distributions that are approximately normal are the heights of college women and the errors made in lling 1-lb cereal boxes. We use the Greek letters m (mu) to denote the mean, and s (sigma) to denote the standard deviation, of a normal distribution.
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Three normal distributions
FIGURE 9
There are many normal distributions. Some of the corresponding normal curves are tall and thin and others are short and wide, as shown in Figure 9. But every normal curve has the following properties. 1. Its peak occurs directly above the mean m. 2. The curve is symmetric about the vertical line through the mean (that is, if you fold the page along this line, the left half of the graph will t exactly on the right half). 3. The curve never touches the x-axisit extends indenitely in both directions. 4. The area under the curve (and above the horizontal axis) is always 1. (This agrees with the fact that the sum of the probabilities in any distribution is 1.) It can be shown that a normal distribution is completely determined by its mean m and standard deviation s.* A small standard deviation leads to a tall, narrow curve like the one in the center of Figure 9. A large standard deviation produces a at, wide curve, like the one on the right in Figure 9. Since the area under a normal curve is 1, parts of this area can be used to determine certain probabilities. For instance, Figure 10(a) is the probability distribution of the annual rainfall in a certain region. Calculus can be used to show that the probability that the annual rainfall will be between 25 in. and 35 in. is the area under the curve from 25 to 35. The general case, shown in Figure can 10(b), be stated as follows. The area of the shaded region under the normal curve from a to b is the probability that an observed data value will be between a and b. To use normal curves effectively, we must be able to calculate areas under portions of these curves. These calculations have already been done for the normal curve with mean m 5 0 and standard deviation s 5 1 (which is called the standard normal curve) and are available in a table in the Appendix. The following examples demonstrate how to use the table to nd such areas. Later we shall see how the standard normal curve may be used to nd areas under any normal curve.
*As is shown in more advanced courses, its graph is the graph of the function 1 2 2 f1 x 2 5 e21x 2m2 / 12s 2, s"2p where e < 2.71828 is a real number.
20 30 40 50 60 70 80
(a)
a
b
x
(b) FIGURE 10
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EXAMPLE
1
Standard Normal Curve The horizontal axis of the standard normal curve is usually labeled z. Find the following areas under the standard normal curve. (a) The area to the left of z 5 1.25 Solution Look up 1.25 in the normal curve table. (Find 1.2 in the left-hand column and .05 at the top, then locate the intersection of the corresponding row and column.) The specied area is .8944, so the shaded area shown in Figure 11 is .8944. This area represents 89.44% of the total area under the normal curve, and so the probability that z # 1.25 is P 1 z # 1.25 2 5 .8944.
Method 1: Using a Table
Shaded area = P(z 1.25) = .8944
z = 1.25
FIGURE 11
(b) The area to the right of z 5 1.25 Solution From part (a), the area to the left of z 5 1.25 is .8944. The total area under the normal curve is 1, so the area to the right of z 5 1.25 is 1 2 .8944 5 .1056. See Figure 12, where the shaded area represents 10.56% of the total area under the normal curve, and the probability that z $ 1.25 is .1056.
Shaded area = P(z 1.25) = 1 .8944 = .1056
z = 1.25
FIGURE 12
(c) Between z 5 21.02 and z 5 .92 Solution To nd this area, which is shaded in Figure 13, start with the area to the left of z 5 .92 and subtract the area to the left of z 5 21.02. See the two shaded regions in Figure 14 on the next page. The result is
z = 1.02 z = .92
P 1 21.02 # z # .92 2 5 .8212 2 .1539 5 .6673.
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FIGURE 13
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z = .92
z = 1.02
FIGURE 14
Method 2: Graphing Calculator
FIGURE 15
Because of convenience and accuracy, graphing calculators and computers have made normal curve tables less important. Figure 15 shows how parts (a) and (b) of this example can be done on a TI-83/84 Plus using the normalcdf command in the DISTR menu. In Figure 15, 21E99 stands for 21 . 1099. The area between 21 . 1099 and 1.25 is essentially the same as the area to the left of 1.25. Similarly, the area between 1.25 and 1 . 1099 is essentially the same as the area to the right of 1.25. Verify the results of part (c) with a graphing calculator. Many statistical software packages are widely used today. All of these packages are set up in a way that is similar to a spreadsheet, and they all can be used to generate normal curve values. In addition, most spreadsheets can also perform a wide range of statistical calculations. For example, Microsoft Excel can be used to generate the answers to parts (a), (b), and (c) of this example. In any cell, type =NORMDIST(1.25,0,1,1) and press Enter. The value of .894350161 is returned. The rst three input values represent the z value, mean, and standard deviation. The fourth value is always either a 0 or 1. For applications in this text, we will always place a 1 in this position to indicate that we want the area to the left of the rst input value. Similarly, by typing =1-NORMDIST(1.25,0,1,1) and pressing Enter, we nd that the area to the right of z 5 1.25 is .105649839.
Method 3: Spreadsheet
N OT E Notice in Example 1 that P 1 z # 1.25 2 5 P 1 z , 1.25 2 . The area under the curve is the same, whether we include the endpoint or not. Notice also that P 1 z 5 1.25 2 5 0, because no area is included. I C A U T I O N When calculating normal probability, it is wise to draw a normal curve with the mean and the z-scores every time. This will avoid confusion as to whether you should add or subtract probabilities. I
EXAMPLE
2
Normal Probabilities Find a value of z satisfying the following conditions. (a) 12.1% of the area is to the left of z. Solution Use the table backwards. Look in the body of the table for an area of .1210, and nd the corresponding value of z using the left column and the top column of the table. You should nd that z 5 21.17 corresponds to an area of .1210.
Method 1: Using a Table
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(b) 20% of the area is to the right of z.
Solution If 20% of the area is to the right, 80% is to the left. Find the value of
z corresponding to an area of .8000. The closest value is z 5 .84.
Method 2: Graphing Calculator
Figure 16 illustrates how a TI-83/84 Plus can be used to nd z values for the particular probabilities given in parts (a) and (b) of this example. The command invNorm is found in the DISTR menu.
FIGURE 16
Method 3: Spreadsheet
Microsoft Excel can also be used to generate the answers to parts (a) and (b) of this example. In any cell, type =NORMINV(.121,0,1) and press Enter. The value of 21.170002407 is returned. Similarly, by typing =NORMINV(.8,0,1) and pressing Enter, we nd that the corresponding z value is .8416212335.
The key to nding areas under any normal curve is to express each number x on the horizontal axis in terms of standard deviation above or below the mean. The z-score for x is the number of standard deviations that x lies from the mean (positive if x is above the mean, negative if x is below the mean). EXAMPLE 3 z-Scores If a normal distribution has mean 50 and standard deviation 4, nd the following z-scores. (a) The z-score for x 5 46 Solution Since 46 is 4 units below 50 and the standard deviation is 4, 46 is 1 standard deviation below the mean. So, its z-score is 21. (b) The z-score for x 5 60
Solution The z-score is 2.5 because 60 is 10 units above the mean (since 60 2 50 5 10), and 10 units is 2.5 standard deviations (since 10 / 4 5 2.5).
In Example 3(b), we found the z-score by taking the difference between 60 and the mean and dividing this difference by the standard deviation. The same procedure works in the general case. If a normal distribution has mean m and standard deviation s, then the z-score for the number x is z5 x2m . s
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The importance of z-scores lies in the following fact.
AREA UNDER A NORMAL CURVE
The area under a normal curve between x 5 a and x 5 b is the same as the area under the standard normal curve between the z-score for a and the z-score for b.
Therefore, by converting to z-scores and using the table for the standard normal curve, we can nd areas under any normal curve. Since these areas are probabilities, we can now handle a variety of applications. EXAMPLE 4 Sales Dixie Office Supplies nds that its sales force drives an average of 1200 miles per month per person, with a standard deviation of 150 miles. Assume that the number of miles driven by a salesperson is closely approximated by a normal distribution. (a) Find the probability that a salesperson drives between 1200 miles and 1600 miles per month. Solution Here m 5 1200 and s 5 150, and we must nd the area under the normal distribution curve between x1 5 1200 and x2 5 1600. We begin by nding the z-score for x1 5 1200. z1 5 x1 2 m 1200 2 1200 0 5 5 50 s 150 150
The z-score for x2 5 1600 is z2 5 x2 2 m 400 1600 2 1200 5 5 2.67. 5 s 150 150
From the table, the area to the left of z2 5 2.67 is .9962, the area to the left of z1 5 0 is .5000, and .9962 2 .5000 5 .4962. Therefore, the probability that a salesperson drives between 1200 miles and 1600 miles per month is .4962. See Figure 17. (b) Find the probability that a salesperson drives between 1000 miles and 1500 miles per month.
Solution As shown in Figure 18, z-scores for both x1 5 1000 and x2 5 1500 are needed.
= 1200 x = 1600
FIGURE 17
= 1200
x = 1000
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x = 1500
FIGURE 18
For x1 5 1000, 1000 2 1200 z1 5 150 2200 5 150 z1 < 21.33.
For x2 5 1500, 1500 2 1200 z2 5 150 300 5 150 z2 5 2.00.
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From the table, z1 5 21.33 leads to an area of .0918, while z2 5 2.00 corresponds to .9772. A total of .9772 2 .0918 5 .8854, or 88.54%, of the drivers travel between 1000 and 1500 miles per month. The probability that a driver travels between 1000 miles and 1500 miles per month is .8854.
.003
800 0
1600
FIGURE 19
Example 4 can also be done using a graphing calculator or computer, as described before, putting 1200 and 150 in place of 0 and 1 for the mean and standard deviation. On the TI-83/84 Plus, we could also use the common ShadeNorm(1000,1500,1200,150) for Example 4(b), with the results shown in Figure 19. The answer of .886039 is more accurate than the value of .8854 found using the normal curve table, which required rounding the z-scores to two decimal places.
N OT E The answers given to the exercises in this text are found using the normal curve table. If you use a graphing calculator or computer program, your answers will differ slightly. I
As mentioned above, z-scores are the number of standard deviations from the mean, so z 5 1 corresponds to 1 standard deviation above the mean, and so on. Looking up z 5 1.00 and z 5 21.00 in the table shows that .8413 2 .1587 5 .6826, or 68.3% of the area under a normal curve lies within 1 standard deviation of the mean. Also, .9772 2 .0228 5 .9544, or 95.4% of the area lies within 2 standard deviations of the mean. These results, summarized in Figure 20, can be used to get a quick estimate of results when working with normal curves.
3
2
+
+2
+3
68.3% of area 95.4% of area 99.7% of area
FIGURE 20
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9.3 EXERCISES
1. The peak in a normal curve occurs directly above ___________. 2. The total area under a normal curve (above the horizontal axis) is __________. 3. How are z-scores found for normal distributions where m 2 0 or s 2 1? 4. How is the standard normal curve used to nd probabilities for normal distributions? Find the percent of the area under a normal curve between the mean and the given number of standard deviations from the mean. 5. 2.50 6. .81 7. 21.71 8. 22.04
Find the percent of the total area under the standard normal curve between each pair of z-scores. 9. z 5 1.41 and z 5 2.83 11. z 5 22.48 and z 5 2.05 13. z 5 23.11 and z 5 1.44 Find a z-score satisfying the following conditions. 15. 5% of the total area is to the left of z. 17. 15% of the total area is to the right of z. 16. 1% of the total area is to the left of z. 18. 25% of the total area is to the right of z. 10. z 5 .64 and z 5 2.11 12. z 5 21.74 and z 5 21.02 14. z 5 22.94 and z 5 .43
19. For any normal distribution, what is the value of P 1 x # m 2 ? P 1 x $ m 2 ?
20. Compare the probability that a number will lie within 2 standard deviations of the mean of a probability distribution using Chebyshevs theorem and using the normal distribution. (See Exercises 1117, Section 9.2.) Explain what you observe. 21. Repeat Exercise 20 using 3 standard deviations.
Applications
In all of the following applications, assume the distributions are normal. In each case, you should consider whether this is reasonable. Quality Control A box of oatmeal must contain 16 oz. The machine that lls the oatmeal boxes is set so that, on the average, a box contains 16.5 oz. The boxes lled by the machine have weights that can be closely approximated by a normal curve. What fraction of the boxes lled by the machine are underweight if the standard deviation is as follows? 29. .5 oz 31. .2 oz 30. .3 oz 32. .1 oz
BUSINESS AND ECONOMICS
Life of Light Bulbs A certain type of light bulb has an average life of 500 hours, with a standard deviation of 100 hours. The length of life of the bulb can be closely approximated by a normal curve. An amusement park buys and installs 10,000 such bulbs. Find the total number that can be expected to last for each period of time. 22. 24. 25. 26. 27. 28.
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At least 500 hours 23. Less than 500 hours Between 650 and 780 hours Between 290 and 540 hours Less than 740 hours More than 300 hours Find the shortest and longest lengths of life for the middle 80% of the bulbs.
Quality Control The chickens at Colonel Thompsons Ranch have a mean weight of 1850 g, with a standard deviation of 150 g. The weights of the chickens are closely approximated by a normal curve. Find the percent of all chickens having weights in the following ranges. 33. 35. 37. 38. More than 1700 g 34. Less than 1800 g Between 1750 and 1900 g 36. Between 1600 and 2000 g More than 2100 g or less than 1550 g Find the smallest and largest weights for the middle 95% of the chickens.
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9 .3 The Normal Distri but ion
39. Quality Control A machine produces bolts with an average diameter of .25 in. and a standard deviation of .02 in. What is the probability that a bolt will be produced with a diameter greater than .3 in.? 40. Quality Control A machine that lls quart milk cartons is set up to average 32.2 oz per carton, with a standard deviation of 1.2 oz. What is the probability that a lled carton will contain less than 32 oz of milk? 41. Grocery Bills At the Discount Market, the average weekly grocery bill is $52.25, with a standard deviation of $19.50. What are the largest and smallest amounts spent by the middle 50% of this markets customers? 42. Grading Eggs To be graded extra large, an egg must weigh at least 2.2 oz. If the average weight for an egg is 1.5 oz, with a standard deviation of .4 oz, how many eggs in a sample of ve dozen would you expect to grade extra large?
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fares are set arbitrarily and often are articially low. According to traffic engineers, the ideal limit should be the 85th percentile speed. This means the speed at or below which 85 percent of the traffic moves. Assuming speeds are normally distributed, nd the 85th percentile speed for roads with the following conditions. 49. The mean speed is 50 mph with a standard deviation of 10 mph. 50. The mean speed is 30 mph with a standard deviation of 5 mph. Education The grading system known as grading on the curve is based on the assumption that grades are often distributed according to the normal curve, and that a certain percent of a class should receive each grade, regardless of the performance of the class as a whole. The following is how one professor might grade on the curve.
LIFE SCIENCES
Vitamin Requirements In nutrition, the Recommended Daily Allowance of vitamins is a number set by the government as a guide to an individuals daily vitamin intake. Actually, vitamin needs vary drastically from person to person, but the needs are very closely approximated by a normal curve. To calculate the Recommended Daily Allowance, the government rst nds the average need for vitamins among people in the population, and the standard deviation. The Recommended Daily Allowance is then dened as the mean plus 2.5 times the standard deviation. 43. What percent of the population will receive adequate amounts of vitamins under this plan? Find the Recommended Daily Allowance for each vitamin in Exercises 44 46. 44. Mean 5 1800 units; standard deviation 5 140 units 45. Mean 5 159 units; standard deviation 5 12 units 46. Mean 5 1200 units; standard deviation 5 92 units 47. Blood Clotting The mean clotting time of blood is 7.45 seconds, with a standard deviation of 3.6 seconds. What is the probability that an individuals blood clotting time will be less than 7 seconds or greater than 8 seconds? 48. Fish The average size of the sh caught by anglers in Lake Amotan is 12.3 in., with a standard deviation of 4.1 in. Find the probability of catching a sh longer than 18 in. in Lake Amotan. Grade A B C D F Total Points m 1 1 1 / 2 2 s to m 1 1 3 / 2 2 s m 2 1 1 / 2 2 s to m 1 1 1 / 2 2 s m 2 1 3 / 2 2 s to m 2 1 1 / 2 2 s Below m 2 1 3 / 2 2 s Greater than m 1 1 3 / 2 2 s
What percent of the students receive the following grades? 51. A 52. B 53. C
54. Do you think this system would be more likely to be fair in a large freshman class in psychology or in a graduate seminar of ve students? Why? Education A teacher gives a test to a large group of students. The results are closely approximated by a normal curve. The mean is 74, with a standard deviation of 6. The teacher wishes to give As to the top 8% of the students and Fs to the bottom 8%. A grade of B is given to the next 15%, with Ds given similarly. All other students get Cs. Find the bottom cutoff (rounded to the nearest whole number) for the following grades. 55. A 56. B 57. C 58. D
SOCIAL SCIENCES
Speed Limits New studies by Federal Highway Administration traffic engineers suggest that speed limits on many thorough-
59. Standardized Tests David Rogosa, a professor of educational statistics at Stanford University, has calculated the accuracy of tests used in California to abolish social promotion. Dr. Rogosa has claimed that a fourth grader whose true reading score is exactly at reading level (50th percentile
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lead level of 130.5 or higher. From this can we conclude that Andrew Jackson had lead poisoning? (Note: These standards may not be valid for this experiment.) 62. Mercury Poisoning Historians and biographers have also collected evidence that suggests that President Andrew Jackson suffered from mercury poisoning. Recently, researchers measured the amount of mercury in samples of Jacksons hair from 1815. The results of this experiment showed that Jackson had a mean mercury level of 6.0 ppm. a. If levels of mercury in hair samples from that time period follow a normal distribution with mean 6.9 and standard deviation 4.6,** nd the probability that a randomly selected person from that time period would have a mercury level of 6.0 ppm or higher. b. Discuss whether this provides evidence that Jackson suffered from mercury poisoning during this time period. c. Todays accepted normal mercury levels follow a normal distribution with approximate mean .6 ppm and standard deviation .3 ppm.|| By present standards, is it likely that a randomly selected person from today would have a mercury level of 6.0 ppm or higher? d. Discuss whether we can conclude that Andrew Jackson suffered from mercury poisoning. 63. Barbie The popularity and voluptuous shape of Barbie dolls have generated much discussion about the inuence these dolls may have on young children, particularly with regard to normal body shape. In fact, many people have speculated as to what Barbies measurements would be if they were scaled to a common human height. Researchers have done this and have compared Barbies measurements to the average 18 35 year old woman, labeled Reference, and with the average model. The table on the next page illustrates some of the results of their research, where each measurement is in centimeters. Assume that the distributions of measurements for the models and for the reference group follow a normal distribution with the given mean and standard deviation.
half of all the students read worse and half read better than this student) has a 58 percent chance of either scoring above the 55th percentile or below the 45th percentile on any one test.* Assume that the results of a given test are normally distributed with mean .50 and standard deviation .09. a. Verify that Dr. Rogosas claim is true. b. Find the probability that this student will either score above the 60th percentile or below the 40th percentile. c. Using the results of parts a and b, discuss problems with the use of standardized testing to prevent social promotion.
GENERAL INTEREST
60. Christopher Columbus Before Christopher Columbus crossed the ocean, he measured the heights of the men on his three ships and found that they were normally distributed with a mean of 69.60 in. and a standard deviation of 3.20 in. What is the probability that a member of his crew had a height less than 66.27 in.? (The answer has another connection with Christopher Columbus!) 61. Lead Poisoning Historians and biographers have collected evidence that suggests that President Andrew Jackson suffered from lead poisoning. Recently, researchers measured the amount of lead in samples of Jacksons hair in 1815. The results of this experiment showed that Jackson had a mean lead level of 130.5 ppm. a. If levels of lead in hair samples from that time period follow a normal distribution with mean 93 and standard deviation 16, nd the probability that a randomly selected person from this time period would have a lead level of 130.5 ppm or higher. Does this provide evidence that Jackson suffered from lead poisoning during this time period? b. Todays typical lead levels follow a normal distribution with approximate mean 10 ppm and standard deviation 5 ppm.|| By these standards, calculate the probability that a randomly selected person from today would have a
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*Rothstein, R., How Tests Can Drop the Ball, The New York Times, Sept. 13, 2000, p. B11. Deppisch, Lidwig, Jose Centeno, David Gemmel, and Norca Torres, Andrew Jacksons Exposure to Mercury and Lead, JAMA, Vol. 282, No. 6, Aug. 11, 1999, pp. 569 571. Weiss, D., B. Whitten, and D. Leddy, Lead Content of Human Hair (18711971), Science, Vol. 178, 1972, pp. 69 70. Although this provides evidence that Andrew Jackson had elevated lead levels, the authors of the paper concluded that Andrew Jackson did not die from lead poisoning. || Iyengar,V. and J. Woittiez, Trace Elements in Human Clinical Specimens, Clinical Chemistry, Vol. 34, 1988, pp. 474 481. **Suzuki, T., T. Hongo, M. Morita, and R. Yamamoto, Elemental Contamination of Japanese Womens Hair from Historical Samples, Sci. Total Environ., Vol. 39, 1984, pp. 8191. Norton, Kevin, Timothy Olds, Scott Olive, and Stephen Dank, Sex Roles, Vol. 34, Nos. 3/4, Feb. 1996, pp. 287294.
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Measurement
Football Mean s.d. 2.3 1.8 3.5 1.9 3.6
Reference Mean 53.7 34.2 91.2 28.8 80.9 s.d. 2.9 1.9 4.8 2.2 9.8
Ken
Head Neck Chest Upper Arm Waist
52.1 34.6 92.3 29.9 75.1
53.0 32.1 75.0 27.1 56.5
Measurement
Models Mean s.d. 2.4 1.0 3.0 .6 3.5
Reference Mean 55.3 32.7 90.3 16.1 69.8 s.d. 2.0 1.4 5.5 .8 4.7
Barbie
a. Find the probability of Kens head size or larger occurring for the reference group and for the football players. b. Find the probability of Kens neck size or smaller occurring for the reference group and for the football players. c. Find the probability of Kens chest size or smaller occurring for the reference group and for the football players. d. Find the probability of Kens upper arm size or smaller occurring for the reference group and for the football players. e. Find the probability of Kens waist size or smaller occurring for the reference group and for the football players. f. Compare the above values and discuss whether Kens measurements are representative of either the reference group or football players. Then compare these results with the results of Exercise 63. Any surprises?
Head Neck Chest (bust) Wrist Waist
50.0 31.0 87.4 15.0 65.7
55.0 23.9 82.3 10.6 40.7
a. Find the probability of Barbies head size or larger occurring for the reference group and for the models. b. Find the probability of Barbies neck size or smaller occurring for the reference group and for the models. c. Find the probability of Barbies bust size or larger occurring for the reference group and for the models. d. Find the probability of Barbies wrist size or smaller occurring for the reference group and for the models. e. Find the probability of Barbies waist size or smaller occurring for the reference group and for the models. f. Compare the above values and discuss whether Barbie represents either the reference group or models. Any surprises? 64. Ken The same researchers from Exercise 63 wondered how the famous Ken doll measured up to average males and with Australian football players. The table in the next column illustrates some of the results of their research, where each measurement is in centimeters.* Assume that the distributions of measurements for the football players and for the reference group follow a normal distribution with the given mean and standard deviation.
*Norton, Kevin, Timothy Olds, Scott Olive, and Stephen Dank, Sex Roles, Vol. 34, Nos. 3/4, Feb. 1996, pp. 287294.
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9.4 NORMAL APPROXIMATION TO THE BINOMIAL DISTRIBUTION
THINK ABOUT IT
What is the probability that at least 40 out of 100 drivers exceed the speed limit by at least 20 mph in Atlanta?
This is a binomial probability problem with a large number of trials (100). In this section we will see how the normal curve can be used to approximate the binomial distribution and answer this question. As we saw in Section 8.4 on Binomial Probability, many practical experiments have only two possible outcomes, sometimes referred to as success or failure. Such experiments are called Bernoulli trials or Bernoulli processes. Examples of Bernoulli trials include ipping a coin (with heads being a success, for instance, and tails a failure) or testing a computer chip coming off the assembly line to see whether or not it is defective. A binomial experiment consists of repeated independent Bernoulli trials, such as ipping a coin 10 times or taking a random sample of 20 computer chips from the assembly line. In Section 8.5 on Probability Distributions and Expected Value, we found the probability distribution for several binomial experiments, such as sampling ve people with bachelors degrees in education and counting how many are women. The probability distribution for a binomial experiment is known as a binomial distribution. As another example, it was reported in 2003 that 40% of drivers in Atlanta exceed the speed limit by at least 20 mph.* Suppose a state trooper wants to verify this statistic and records the speed of 10 randomly selected drivers. The trooper nds that 5 out of 10, or 50%, exceed the speed limit by at least 20 mph. How likely is this if the 40% gure is accurate? We can answer this question with the binomial probability formula
FOR REVIEW
Recall from Chapter 8 that the n symbol a b is dened as r n! . For example, r! 1 n 2 r 2 ! 10 10! a b5 5 5! 5! 10 . 9 . 8 . 7 . 6 5 . . . . 5 252. 54321
n a b . px . 1 1 2 p 2 n2x, x where n is the size of the sample (10 in this case), x is the number of speeders (5 in this example), and p is the probability that a driver is a speeder (.40). This gives P1 x 5 5 2 5 a 10 . b .405 . 1 1 2 .40 2 5 5 5 252 1 .01024 2 1 .07776 2 < .2007.
The probability is about 20%, so this result is not unusual. Suppose that the state trooper takes a larger random sample of 100 drivers. What is the probability that 50 or more drivers speed if the 40% gure is accurate? Calculating P 1 x 5 50 2 1 P 1 x 5 51 2 1 ) 1 P 1 x 5 100 2 is a formidable task. One solution is provided by graphing calculators or computers. On the TI-83/84 Plus, for example, we can rst calculate the probability that 49 or fewer drivers exceed the speed limit using the DISTR menu command binomcdf(100, .40,49). Subtracting the answer from 1 gives a probability of .0271. But this high-tech method fails as n becomes larger; the command binomcdf(1000000,
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*www.laseratlanta.com/pr08223_speedmotorist.htm.
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9 .4 Normal Approximat ion to the Binomial Distri but ion
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.40,50000) gives an error message. On the other hand, there is a low-tech method that works regardless of the size of n. It has further interest because it connects two different distributions: the normal and the binomial. The normal distribution is continuous, since the random variable can take on any real number. The binomial distribution is discrete, because the random variable can only take on integer values between 0 and n. Nevertheless, the normal distribution can be used to give a good approximation to binomial probability. In order to use the normal approximation, we rst need to know the mean and standard deviation of the binomial distribution. Recall from Section 8.5 that for the binomial distribution, E 1 x 2 5 np. In Section 9.1, we referred to E 1 x 2 as m, and that notation will be used here. It is shown in more advanced courses in statistics that the standard deviation of the binomial distribution is given by s 5 "np 1 1 2 p 2 .
MEAN AND STANDARD DEVIATION FOR THE BINOMIAL DISTRIBUTION
For the binomial distribution, the mean and standard deviation are given by m 5 np and s 5 "np 1 1 2 p 2 ,
where n is the number of trials and p is the probability of success on a single trial.
EXAMPLE
1
Coin Flip Suppose a fair coin is ipped 15 times. (a) Find the mean and standard deviation for the number of heads. Solution Using n 5 15 and p 5 1 / 2, the mean is 1 m 5 np 5 15 a b 5 7.5. 2 The standard deviation is s 5 "np 1 1 2 p 2 5 5 1 1 15 a b a 1 2 b 2 2 11 15 a b a b 5 "3.75 < 1.94. 22
We expect, on average, to get 7.5 heads out of 15 tosses. Most of the time, the number of heads will be within 3 standard deviations of the mean, or between 7.5 2 3 1 1.94 2 5 1.68 and 7.5 1 3 1 1.94 2 5 13.32. (b) Find the probability distribution for the number of heads, and draw a histogram of the probabilities.
Solution The probability distribution is found by putting n 5 15 and p 5 1 / 2 into the formula for binomial probability. For example, the probability of 9 heads is given by
P1 x 5 9 2 5 a
15 1 9 16 b a b a 1 2 b < .15274. 92 2
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x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
P(x) .00003 .00046 .00320 .01389 .04166 .09164 .15274 .19638 .19638 .15274 .09164 .04166 .01389 .00320 .00046 .00003
.18 .15 Probability .12
.09
.06 .03
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15
Number of heads
FIGURE 21
Probabilities for the other values of x between 0 and 15, as well as a histogram of the probabilities, are shown in the table and in Figure 21.
In Figure 21, we have superimposed the normal curve with m 5 7.5 and s 5 1.94 over the histogram of the distribution. Notice how well the normal distribution ts the binomial distribution. This approximation was rst discovered in 1733 by Abraham De Moivre (16671754) for the case p 5 1 / 2. The result was generalized by the French mathematician Pierre-Simon Laplace (17491827) in a book published in 1812.* As n becomes larger and larger, a histogram for the binomial distribution looks more and more like a normal curve. Histograms of the binomial distribution with p 5 .3, using n 5 8 and n 5 50 are shown in Figures 22(a) and (b), respectively on the next page. The probability of getting exactly 9 heads in 15 tosses, or .15274, is the same as the area of the bar in blue in Figure 21. As the graph suggests, the area in blue is approximately equal to the area under the normal curve from x 5 8.5 to x 5 9.5. The normal curve is higher than the top of the bar in the left half but lower in the right half. To nd the area under the normal curve from x 5 8.5 to x 5 9.5, rst nd z-scores, as in the previous section. Use the mean and the standard deviation for the distribution, which we have already calculated, to get z-scores for x1 5 8.5 and x2 5 9.5.
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*Laplaces generalization, known as the Central Limit theorem, states that the distribution of the sample mean from any distribution approaches the normal distribution as the sample size increases. For more details, see any statistics book, such as Elementary Statistics (9th ed.) by Mario F. Triola, AddisonWesley, 2004.
Finite Mathematics, Eighth Edition, by Margaret L. Lial, Raymond N. Greenwell, and Nathan P. Ritchey. Published by Addison Wesley. Copyright 2005 by Pearson Education, Inc.
9 .4 Normal Approximat ion to the Binomial Distri but ion
.3 .25 .2 .15 .06 .1 .04 .05 0 2 0 2 4 n=8 6 8 10 .02 0 10 .14 .12 .1 .08
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0
10
20 30 n = 50
40
50
60
(a) FIGURE 22
(b)
For x1 5 8.5, z1 5 5 8.5 2 7.5 1.94 1.00 1.94
For x2 5 9.5, z2 5 5 9.5 2 7.5 1.94 2.00 1.94
z1 < .52.
z2 < 1.03.
From the table in the Appendix, z1 5 .52 gives an area of .6985, and z2 5 1.03 gives .8485. The difference between these two numbers is the desired result. P 1 z # 1.03 2 2 P 1 z # .52 2 5 .8485 2 .6985 5 .1500 This answer (.1500) is not far from the more accurate answer of .15274 found above.
C A U T I O N The normal curve approximation to a binomial distribution is quite accurate provided that n is large and p is not close to 0 or 1. As a rule of thumb, the normal curve approximation can be used as long as both np and n 1 1 2 p 2 are at least 5. I
EXAMPLE
2
Speeding Consider the random sample discussed earlier of 100 drivers in Atlanta, where 40% of the drivers exceed the speed limit by at least 20 mph. (a) Use the normal distribution to approximate the probability that at least 50 drivers exceed the speed limit. Solution First nd the mean and the standard deviation using n 5 100 and p 5 .40.
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= 40
x = 49.5
FIGURE 23
m 5 100 1 .40 2 5 40
s 5 "100 1 .40 2 1 1 2 .40 2 5 "24 < 4.899 5 "100 1 .40 2 1 .60 2
As the graph in Figure 23 shows, we need to nd the area to the right of x 5 49.5 (since we want 50 or more speeders). The z-score corresponding to x 5 49.5 is z5 49.5 2 40 < 1.94. 4.899
From the table, z 5 1.94 leads to an area of .9738, so P 1 z . 1.94 2 5 1 2 .9738 5 .0262. This value is close to the value of .0271 found earlier with the help of a graphing calculator. Either method tells us there is roughly a 3% chance of nding 50 or more speeders out of a random sample of 100. If the trooper found 50 or more speeders in his sample, he might suspect that either his sample is not truly random, or that the 40% gure for the percent of drivers who speed is too low. (b) Find the probability of nding between 42 and 48 speeders in a random sample of 100.
Solution As Figure 24 on the next page shows, we need to nd the area between x1 5 41.5 and x2 5 48.5.
If x1 5 41.5, then z1 5 If x2 5 48.5, then z2 5
41.5 2 40 < .31. 4.899 48.5 2 40 < 1.74. 4.899
ISBN: 0-536-10718-1
Use the table to nd that z1 5 .31 gives an area of .6217, and z2 5 1.74 yields .9591. The nal answer is the difference of these numbers, or
Finite Mathematics, Eighth Edition, by Margaret L. Lial, Raymond N. Greenwell, and Nathan P. Ritchey. Published by Addison Wesley. Copyright 2005 by Pearson Education, Inc.
9 .4 Normal Approximat ion to the Binomial Distri but ion
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= 40 x1 = 41.5
x2 = 48.5
FIGURE 24
P 1 .31 # z # 1.74 2 5 P 1 z # 1.74 2 2 P 1 z # .31 2 5 .9591 2 .6217 5 .3374. The probability of nding between 42 and 48 speeders is about .3374.
9.4 EXERCISES
1. What must be known to nd the mean and standard deviation of a binomial distribution? 2. What is the rule of thumb for using the normal distribution to approximate a binomial distribution? Suppose 16 coins are tossed. Find the probability of getting the following results (a) using the binomial probability formula, and (b) using the normal curve approximation. 3. Exactly 8 heads 5. More than 12 tails 4. Exactly 7 heads 6. Fewer than 5 tails
For the remaining exercises in this section, use the normal curve approximation to the binomial distribution. Suppose 1000 coins are tossed. Find the probability of getting the following results. 7. Exactly 500 heads 8. Exactly 510 heads 9. 480 heads or more 10. Fewer than 470 tails
A die is tossed 120 times. Find the probability of getting the following results. 11. Exactly twenty 5s 13. More than eighteen 3s 12. Exactly twenty-four 6s 14. Fewer than twenty-two 6s
15. A reader asked Mr. Statistics (a feature in Fortune magazine) about the game of 26 once played in the bars of Chicago.* The player chooses a number between 1 and 6, and then rolls a cup full of 10 dice 13 times. Out of the 130 numbers rolled, if the number chosen appears at least 26 times, the player wins. Calculate the probability of winning.
*Seligman, Daniel and Patty De Llosa, Ask Mr. Statistics, Fortune, May 1, 1995, p. 141.
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Applications
BUSINESS AND ECONOMICS
16. Quality Control Two percent of the quartz heaters produced in a certain plant are defective. Suppose the plant produced 10,000 such heaters last month. Find the probabilities that among these heaters, the following numbers were defective. a. Fewer than 170 b. More than 222 17. Quality Control The probability that a certain machine turns out a defective item is .05. Find the probabilities that in a run of 75 items, the following results are obtained. a. Exactly 5 defectives b. No defectives c. At least 1 defective 18. Survey Results A company is taking a survey to nd out whether people like its product. Their last survey indicated that 70% of the population like the product. Based on that, of a sample of 58 people, nd the probabilities of the following. a. All 58 like the product. b. From 28 to 30 (inclusive) like the product. 19. Minimum Wage A recent study of minimum wage earners found that 55.2% of them are 16 to 24 years old.* Suppose a random sample of 600 minimum wage earners is selected. What is the probability that more than 350 of them are 16 to 24 years old?
a. Find the probability that an animal consumes exactly 80 units of food. b. Suppose the animal must consume at least 70 units of food to survive. What is the probability that this happens? 22. Coconuts A 4-year review of trauma admissions to the Provincial Hospital, Alotau, Milne Bay Providence, reveals that 2.5% of such admissions were due to being struck by falling coconuts. a. Suppose 20 patients are admitted to the hospital during a certain time period. What is the probability that no more than 1 of these patients are there because they were struck by falling coconuts? Do not use the normal distribution here. b. Suppose 2000 patients are admitted to the hospital during a longer time period. What is the approximate probability that no more than 70 of these patients are there because they were struck by falling coconuts? 23. Kidney Dialysis A study found that 23.6% of people on kidney dialysis in the United States die, more than in any other industrial country.|| A hospital surveyed 40 of its patients who had been on dialysis that year and found that 15 died so far. What is the probability that 15 or more people out of 40 would die if this hospitals dialysis patients are typical of those in the United States? 24. Drug Effectiveness A new drug cures 80% of the patients to whom it is administered. It is given to 25 patients. Find the probabilities that among these patients, the following results occur. a. Exactly 20 are cured. b. All are cured. c. No one is cured. d. Twelve or fewer are cured. 25. Flu Inoculations A u vaccine has a probability of 80% of preventing a person who is inoculated from getting the u. A county health office inoculates 134 people. Find the probabilities of the following. a. Exactly 10 of the people inoculated get the u. b. No more than 10 of the people inoculated get the u. c. None of the people inoculated get the u.
LIFE SCIENCES
20. Nest Predation For certain bird species, with appropriate assumptions, the number of nests escaping predation has a binomial distribution. Suppose the probability of success (that is, a nest escaping predation) is .3. Find the probability that at least half of 26 nests escape predation. 21. Food Consumption Under certain appropriate assumptions, the probability of a competing young animal eating x units of food is binomially distributed, with n equal to the maximum number of food units the animal can acquire, and p equal to the probability per time unit that an animal eats a unit of food. Suppose n 5 120 and p 5 .6.
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*The World Almanac and Book of Facts 2003, p. 146. Wilbur, H. M., American Naturalist, Vol. 111. deJong, G., American Naturalist, Vol. 110. Barss, Peter, Injuries Due to Falling Coconuts, The Journal of Trauma, Vol. 24, No. 11, 1984, pp. 990 991. || The New York Times, Dec. 4, 1995, p. A1.
Finite Mathematics, Eighth Edition, by Margaret L. Lial, Raymond N. Greenwell, and Nathan P. Ritchey. Published by Addison Wesley. Copyright 2005 by Pearson Education, Inc.
9 .4 Normal Approximat ion to the Binomial Distri but ion
26. Blood Types The blood types B2 and AB2 are the rarest of the eight human blood types, representing 1.5% and .6% of the population, respectively.* a. If the blood types of a random sample of 1000 blood donors are recorded, what is the probability that 10 or more of the samples are AB2? b. If the blood types of a random sample of 1000 blood donors are recorded, what is the probability that 20 to 40 inclusive of the samples are B2? c. If a particular city had a blood drive in which 500 people gave blood and 3% of the donations were B2, would we have reason to believe that this town has a higher than normal number of donors who are B2? (Hint: Calculate the probability of 15 or more donors being B2 for a random sample of 500 and then discuss the probability obtained.) 27. Motorcycles In 2002, there were 59 motorcyclist deaths per 100,000 registered motorcycles. Suppose that in 2003 the probability that a motorcyclist died in an accident remained the same as it was in 2002 and that total number of registered motorcycles remained at its 2002 value of 5,359,000. What would be the probability that the number of motorcyclist deaths in 2003 is between 3000 and 3200?
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so close, even if the electorate is evenly divided? Assume that the number of votes for Bush is binomially distributed with n 5 5,825,043 (the sum of the votes for the two candidates) and p 5 .5. a. Using the binomial probability feature on a graphing calculator, try to calculate P 1 2,912,253 # X # 2,912,790 2 . What happens? b. Use the normal approximation to calculate the probability in part a.
GENERAL INTEREST
31. Homework Only 1 out of 12 American parents requires that children do their homework before watching TV.|| If your neighborhood is typical, what is the probability that out of 51 parents, 5 or fewer require their children to do homework before watching TV? 32. True-False Test A professor gives a test with 100 true-false questions. If 60 or more correct is necessary to pass, what is the probability that a student will pass by random guessing? 33. Hole in One In the 1989 U.S. Open, four golfers each made a hole in one on the same par-3 hole on the same day. Sports Illustrated writer R. Reilly stated the probability of a hole in one for a given golf pro on a given par-3 hole to be 1 / 3709.# a. For a specic par-3 hole, use the binomial distribution to nd the probability that 4 or more of the 156 golf pros in the tournament eld shoot a hole in one.** b. For a specic par-3 hole, use the normal approximation to the binomial distribution to nd the probability that 4 or more of the 156 golf pros in the tournament eld shoot a hole in one. Why must we be very cautious when using this approximation for this application? c. If the probability of a hole in one remains constant and is 1 / 3709 for any par-3 hole, nd the probability that in 20,000 attempts by golf pros, there will be 4 or more hole in ones. Discuss whether this assumption is reasonable.
SOCIAL SCIENCES
28. Straw Votes In one state, 55% of the voters expect to vote for Nola Akala. Suppose 1400 people are asked the name of the person for whom they expect to vote. Find the probability that at least 750 people will say that they expect to vote for Akala. 29. Weapons and Youth A poll of 2000 teenagers found that 1 in 8 reported carrying a weapon for protection. In a typical high school with 1200 students, what is the probability that more than 120 students, but fewer than 180, carry a weapon? 30. Election 2000 The Florida recount in the 2000 presidential election gave George W. Bush 2,912,790 votes and Al Gore 2,912,253 votes. What is the likelihood of the vote being
*The Handy Science Answer Book, The Carnegie Library of Pittsburgh, 1997, p. 332. http://www.hwysafety.org/safety_facts/fatality_facts/motorcyl.htm The New York Times, Jan. 12, 1996, p. A6. The World Almanac and Book of Facts 2003, p. 577. || Harpers Index, Harpers, Sept. 1996, p. 15. # Reilly, R., King of the Hill, Sports Illustrated, June 1989, pp. 20 25. **Litwiller, Bonnie and David Duncan, The Probability of a Hole in One, School Science and Mathematics, Vol. 91, No. 1, Jan. 1991, p. 30.
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C H A P T E R S U M M A RY
In this chapter we used the concepts of the previous two chapters to introduce the eld of statistics. Measures of central tendency, such as mean, median, and mode, were dened. Information with regard to the spread of a group of numbers was obtained by calculating the variance and standard deviation. The normal distribution, perhaps the most important and widely used probability distribution, was dened and used to study a wide range of problems. The normal approximation of the binomial distribution was then developed, as were several important applications. The concepts that were introduced in this chapter serve only to familiarize a student to the wonderful world of statistics. Students are urged to further their study in this important area.
KEY TERMS
9.1 random sample grouped frequency distribution frequency polygon (arithmetic) mean sample mean population mean median statistic mode 9.2 range deviations from the mean variance standard deviation sample variance sample standard deviation population variance population standard deviation 9.3 continuous distribution skewed distribution normal distribution normal curve standard normal curve z-score 9.4 binomial distribution
CHAPTER 9 REVIEW EXERCISES
1. Discuss some reasons for organizing data into a grouped frequency distribution. 2. What is the rule of thumb for an appropriate interval in a grouped frequency distribution? In Exercises 3 and 4, (a) write a frequency distribution; (b) draw a histogram; (c) draw a frequency polygon. 3. The following numbers give the sales (in dollars) for the lunch hour at a local hamburger stand for the last 20 Fridays. Use intervals 450 474, 475 499, and so on. 480 516 451 535 501 492 478 558 512 488 473 547 509 461 515 475 458 492 566 471 4. The number of units carried in one semester by students in a business mathematics class was as follows. Use intervals 9 10, 1112, 13 14, 15 16. 10 12 9 12 16 15 12 12 13 14 15 10 13 12 16 14 15 15 11 15 13 13
Find the mean for the following. 5. 41, 60, 67, 68, 72, 74, 78, 83, 90, 97 6. 105, 108, 110, 115, 106, 110, 104, 113, 117 7. Interval 10 19 20 29 30 39 40 49 50 59 Frequency 6 12 14 10 8 8. Interval 40 44 45 49 50 54 55 59 60 64 65 69 Frequency 2 5 7 10 4 1
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9. What do the mean, median, and mode of a distribution have in common? How do they differ? Describe each in a sentence or two.
Finite Mathematics, Eighth Edition, by Margaret L. Lial, Raymond N. Greenwell, and Nathan P. Ritchey. Published by Addison Wesley. Copyright 2005 by Pearson Education, Inc.
C hapter 9 Review Exercises
Find the median and the mode (or modes) for each list of numbers. 10. 32, 35, 36, 44, 46, 46, 59 Find the modal class for the indicated distributions. 12. Exercise 7 14. What is meant by the range of a distribution? 15. How are the variance and the standard deviation of a distribution related? What is measured by the standard deviation? Find the range and standard deviation for each distribution. 16. 14, 17, 18, 19, 32 Find the standard deviation for the following. 18. Exercise 7 20. Describe the characteristics of a normal distribution. 21. What is meant by a skewed distribution? Find the following areas under the standard normal curve. 22. Between z 5 0 and z 5 1.27 24. Between z 5 21.88 and z 5 2.10 23. To the left of z 5 .41 25. Between z 5 1.53 and z 5 2.82 19. Exercise 8 17. 26, 43, 51, 29, 37, 56, 29, 82, 74, 93 13. Exercise 8 11. 38, 36, 42, 44, 38, 36, 48, 35
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26. Find a z-score such that 8% of the area under the curve is to the right of z. 27. Why is the normal distribution not a good approximation of a binomial distribution that has a value of p close to 0 or 1? 28. Suppose a card is drawn at random from an ordinary deck 1,000,000 times with replacement. a. What is the probability that between 249,500 and 251,000 hearts (inclusive) are drawn? b. Why must the normal approximation to the binomial distribution be used to solve part a? 29. Suppose four coins are ipped and the number of heads counted. This experiment is repeated 20 times. The data might look something like the following. (You may wish to try this yourself and use your own results rather than these.) Number of Heads 0 1 2 3 4 Frequency 1 5 7 5 2
a. Calculate the sample mean x and sample standard deviation s. b. Calculate the population mean m and population standard deviation s for this binomial population. c. Compare your answer to parts a and b. What do you expect to happen? 30. Much of our work in Chapters 8 and 9 is interrelated. Note the similarities in the following parallel treatments of a frequency distribution and a probability distribution.
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Probability Distribution A binomial distribution has n 5 10 and p 5 .5. Complete the following table. x 0 1 2 3 4 5 6 7 8 9 10 P(x) .001 .010 .044 .117 x ? P1 x 2
Frequency Distribution Complete the table below for the following data. (Recall that x is the midpoint of the interval.) 14, 7, 1, 11, 2, 3, 11, 6, 10, 13, 11, 11, 16, 12, 9, 11, 9, 10, 7, 12, 9, 6, 4, 5, 9, 16, 12, 12, 11, 10, 14, 9, 13, 10, 15, 11, 11, 1, 12, 12, 6, 7, 8, 2, 9, 12, 10, 15, 9, 3
Interval 1 3 46 79 10 12 13 15 16 18
x 2
Tally 1111 1
f 6
x?f 12
a. Find the mean (or expected value) for each distribution. b. Find the standard deviation for each distribution. c. Use the normal approximation of the binomial probability distribution to nd the interval that contains 95.44% of that distribution. d. Why cant we use the normal distribution to answer probability questions about the frequency distribution?
Applications
BUSINESS AND ECONOMICS
31. Stock Returns The annual returns of two stocks for 3 years are given below. Stock Stock I Stock II 1998 11% 9% 1999 21% 5% 2000 14% 10%
of the cartons vary normally, with a standard deviation of .1 oz, what percent of the cartons contain less than a quart (32 oz)? 33. Quality Control About 6% of the frankfurters produced by a certain machine are overstuffed and thus defective. Find the following probabilities for a sample of 500 frankfurters. a. Twenty-ve or fewer are overstuffed. b. Exactly 30 are overstuffed. c. More than 40 are overstuffed. 34. Bankruptcy The probability that a small business will go bankrupt in its rst year is .21. For 50 such small businesses, nd the following probabilities rst by using the binomial probability formula, and then by using the normal approximation. a. Exactly 8 go bankrupt. b. No more than 2 go bankrupt.
a. Find the mean and standard deviation for each stock over the 3-year period. b. If you are looking for security (hence, less variability) with an average 8% return, which of these stocks should you choose? 32. Quality Control A machine that lls quart orange juice cartons is set to ll them with 32.1 oz. If the actual contents
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C hapter 9 Review Exercises
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LIFE SCIENCES
35. Rat Diets The weight gains of 2 groups of 10 rats fed different experimental diets were as follows. Diet A B 1 2 0 1 3 1 Weight Gains 7 2 1 3 1 2 5 1 4 0 1 1 4 0
a. Find the mean, median, and mode of the data. b. Find the standard deviation of the data. c. What percent of the data is within 1 standard deviation of the mean? d. What percent of the data is within 3 standard deviations of the mean? Commuting Times The average resident of a certain East Coast suburb spends 42 minutes per day commuting, with a standard deviation of 12 minutes. Assume a normal distribution. Find the percent of all residents of this suburb who have the following commuting times. 41. At least 50 minutes per day
Compute the mean and standard deviation for each group. a. Which diet produced the greatest mean gain? b. Which diet produced the most consistent gain? Chemical Effectiveness White ies are devastating California crops. An area infested with white ies is to be sprayed with a chemical which is known to be 98% effective for each application. Assume a sample of 1000 ies is checked. 36. Find the approximate probability that exactly 980 of the ies are killed in one application. 37. Find the approximate probability that no more than 986 of the ies are killed in one application. 38. Find the approximate probability that at least 975 of the ies are killed in one application. 39. Find the approximate probability that between 973 and 993 (inclusive) of the ies are killed in one application.
42. No more than 35 minutes per day 43. Between 32 and 40 minutes per day 44. Between 38 and 60 minutes per day 45. I.Q. Scores On standard IQ tests, the mean is 100, with a standard deviation of 15. The results are very close to tting a normal curve. Suppose an IQ test is given to a very large group of people. Find the percent of those people whose IQ scores are as follows. a. More than 130 b. Less than 85 c. Between 85 and 115 46. Broadway A survey was given to 313 performers appearing in 23 Broadway companies. The percentage of performers injured during practice or a performance was 55.5%. If a random sample of 500 Broadway performers is taken, use the normal approximation to the binomial distribution to nd the approximate probability that more than 300 performers have been injured. 47. Broadway In the survey described in Exercise 46, the demographics of the Broadway performers were recorded as shown on the following page. Assume that all of these demographics follow a normal distribution, an assumption that always must be veried prior to using it in real situations. a. Find the probability that a female dancer is 35 years old or older. b. Find the probability that a male dancer is 35 years old or older. c. Compare your answers to parts a and b.
SOCIAL SCIENCES
40. Homicide The number of homicides in Vermont from 1992 through 2001 is given in the following table.* Homicide Victims 21 15 5 13 11 9 12 17 12 11
Year 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001
*http://170.222.24.9/cjs/crime_01/homicide_01.html. Evans, Randolph, Richard Evans, Scott Carvajal, and Susan Perry, A Survey of Injuries Among Broadway Performers, American Journal of Public Health, Vol. 86, No. 1, Jan. 1996, pp. 77 80.
ISBN: 0-536-10718-1
Finite Mathematics, Eighth Edition, by Margaret L. Lial, Raymond N. Greenwell, and Nathan P. Ritchey. Published by Addison Wesley. Copyright 2005 by Pearson Education, Inc.
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d. Find the probability that a female performer is 1.4 m tall or taller. e. Find the probability that a female performer has a career duration that is more than 1.5 standard deviations from the mean. f. Would a female who has more than 6 injuries during her career be considered a rare event? Explain. Dancers Age (female) Dancers Age (male) Height (in m) (female) Duration as Professional in yr (female) Total No. of Injuries as Performer (female)
Mean 28.0 32.2 1.64 11.0 3.0
Standard Deviation 5.5 8.4 .08 8.9 2.2
ISBN: 0-536-10718-1
Finite Mathematics, Eighth Edition, by Margaret L. Lial, Raymond N. Greenwell, and Nathan P. Ritchey. Published by Addison Wesley. Copyright 2005 by Pearson Education, Inc.
Extended Applicat ion
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EXTENDED APPLICATION: Statistics in the LawThe Castaneda Decision
Statistical evidence is now routinely presented in both criminal and civil cases. In this application well look at a famous case that established use of the binomial distribution and measurement by standard deviation as an accepted procedure.* Defendants who are convicted in criminal cases sometimes appeal their conviction on the grounds that the jury that indicted or convicted them was drawn from a pool of jurors that does not represent the population of the district in which they live. These appeals almost always cite the Supreme Courts decision in Castaneda v. Partida [430 U.S. 482], a case that dealt with the selection of grand juries in the state of Texas. The decision summarizes the facts this way: After respondent, a Mexican-American, had been convicted of a crime in a Texas District Court and had exhausted his state remedies on his claim of discrimination in the selection of the grand jury that had indicted him, he led a habeas corpus petition in the Federal District Court, alleging a denial of due process and equal protection under the Fourteenth Amendment, because of gross underrepresentation of Mexican-Americans on the county grand juries. The case went to the Appeals Court, which noted that the county population was 79% Mexican-American, but, over an 11-year period, only 39% of those summoned for grand jury service were Mexican-American, and concluded that together with other testimony about the selection process, the proof offered by respondent was sufficient to demonstrate a prima facie case of intentional discrimination in grand jury selection. . . . The state appealed to the Supreme Court, and the Supreme Court needed to decide whether the underrepresentation of Mexican-Americans on grand juries was indeed too extreme to be an effect of chance. To do so, they invoked the binomial distribution. Here is the argument: Given that 79.1% of the population is Mexican-American, the expected number of Mexican-Americans among the 870 persons summoned to serve as grand jurors over the 11-year period is approximately 688. The observed number is 339. Of course, in any given drawing some uctuation from the expected number is predicted. The important point, however, is that the statistical model shows that the results of a random drawing are likely to fall in the vicinity of the expected value. . . . The measure of the predicted uctuations from the expected value is the standard deviation, dened for the binomial distribution as the square root of the product of the total number in the sample (here 870) times the probability of selecting a Mexican-American (.791) times the probability of selecting a non-Mexican-American (.209) . . . . Thus, in this case the standard deviation is approximately 12. As a general rule for such large samples, if the difference between the expected value and the observed number is greater than two or three standard deviations, then the hypothesis that the jury drawing was random would be suspect to a social scientist. The 11-year data here reect a difference between the expected and observed number of Mexican-Americans of approximately 29 standard deviations. A detailed calculation reveals that the likelihood that such a substantial departure from the expected value would occur by chance is less than 1 in 10140. The Court decided that the statistical evidence supported the conclusion that jurors were not randomly selected, and that it was up to the state to show that its selection process did not discriminate against Mexican-Americans. The Court concluded: The proof offered by respondent was sufficient to demonstrate a prima facie case of discrimination in grand jury selection. Since the State failed to rebut the presumption of purposeful discrimination by competent testimony, despite two opportunities to do so, we affirm the Court of Appeals holding of a denial of equal protection of the law in the grand jury selection process in respondents case.
Exercises
1. Check the Courts calculation of 29 standard deviations as
the difference between the expected number of MexicanAmericans and the number actually chosen.
2. Where do you think the Courts gure of 1 in 10140 came
from?
1 3. The Castaneda decision also presents data from a 2 2-year
period during which the State District Judge supervised
*The Castaneda case and many other interesting applications of statistics in law are discussed in Finkelstein and Levin, Statistics for Lawyers, New York, Springer-Verlag, 1990. U.S. Supreme Court decisions are online at http://www.ndlaw.com/casecode/supreme.html, and most states now have important state court decisions online.
ISBN: 0-536-10718-1
Finite Mathematics, Eighth Edition, by Margaret L. Lial, Raymond N. Greenwell, and Nathan P. Ritchey. Published by Addison Wesley. Copyright 2005 by Pearson Education, Inc.
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there were 294 employees; 206 were men and 88 were women. a. How far from the expected number of women in management was the actual number, assuming that gender had nothing to do with promotion? Measure the difference in standard deviations. b. Does this look like evidence of purposeful discrimination?
the selection process. During this period, 220 persons were called to serve as grand jurors, and only 100 of these were Mexican-American. a. Considering the 220 jurors as a random selection from a large population, what is the expected number of Mexican-Americans, using the 79.1% population gure? b. If we model the drawing of jurors as a sequence of 220 independent Bernoulli trials, what is the standard deviation of the number of Mexican-Americans? c. About how many standard deviations is the actual number of Mexican-Americans drawn (100) from the expected number that you calculated in part a? d. What does the normal distribution table at the back of the book tell you about this result?
Directions for Group Project
Suppose that you and three other students are serving as interns at a prestigious law rm. One of the partners is interested in the use of probability in court cases and would like the four of you to prepare a brief on the Castaneda decision. She insists that you describe the case and highlight the mathematics used in your brief. Be sure to use the results from the case along with the results of Exercises 1 3 in preparing your brief. Also, make recommendations of other types of cases where probability may be used in law. Presentation software, such as Microsoft Powerpoint, should be used to present your brief to the partners of the rm.
4. The following information is from an appeal brought by
Hy-Vee stores before the Iowa Supreme Court, appealing a ruling by the Iowa Civil Rights Commission in favor of a female employee of one of their grocery stores. In 1985, there were 112 managerial positions in the ten Hy-Vee stores located in Cedar Rapids. Only 6 of these managers were women. During that same year
ISBN: 0-536-10718-1
Finite Mathematics, Eighth Edition, by Margaret L. Lial, Raymond N. Greenwell, and Nathan P. Ritchey. Published by Addison Wesley. Copyright 2005 by Pearson Education, Inc.