M316 Chapter 3 - M316 Chapter 3 Dr. Berg The Normal...

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Unformatted text preview: M316 Chapter 3 Dr. Berg The Normal Distributions Density curves are used to model the distribution of a real number valued variable. The most important of the real valued distributions are the normal distributions. This is partly because we can approximate many whole number valued distributions using a normal distribution. Density Curves For a whole number valued distribution, we can make a histogram of the proportion of observations that fall into each interval giving us what is known as a probability distribution for that variable. For real number valued variables we use a density curve. Definition A density curve satisfies: a) It is never negative (does not go below the xaxis). b) The entire area between the curve and the xaxis is one. c) The area under the curve that lies between two values a and b on the horizontal axis gives the probability that a randomly chosen individual has a value between a and b. The simplest density curve is for a uniform distribution. Such a curve is constant over an interval of real numbers [a, b]. Since the area under this curve is 1 one, the height over the interval [a, b] is f (x) = (and zero otherwise). b-a 1 M316 Chapter 3 Dr. Berg Example Suppose it is known that a call came into a switchboard sometime between 8 am and 10 am. Then the probability that the call came in between times a and b is 1 1 given by the area under the curve f (x) = = (on the interval [8,10] and zero 10 - 8 2 otherwise) that lies between a and b. Thus, the probability that it came in between 1 1 8:30 and 9:00 is p = (9 - 8.5) = . 2 4 Density functions are often used to approximate histograms. Here is an example: Generally, the approximation is good for a large number of observations and narrow intervals. Describing Density Curves Our measures of center and spread apply to density curves as well as actual sets of observations. The median and quartiles are relatively easy. Since the area 2 M316 Chapter 3 Dr. Berg under a density curve between two values represents the proportion of observations that fall in that interval, the median of a density curve is the equal areas point, the point with half the area to the left and half to the right. Similarly, the first quartile has a quarter of the area to the left, and the third quartile has a quarter of the area to the right. Density curves are described by a function, and calculating areas under curves is the subject of integral calculus. Before Newton and Liebnitz, these areas were approximated using rectangles (the opposite of approximating a histogram with a curve). We will use tables for this. The mean and median are defined in a similar manner as for a histogram. The median divides the area into two equal parts instead of the observations. The mean is a "balance point". Normal Distributions A normal distribution is described by a symmetric curve called a normal density curve. Each normal density curve is completely determined by two numbers, the mean and the standard deviation . The curve is given by - 1 2 f (x) = e 2 . 2 The axis of symmetry is located at the mean and the curve changes curvature either side of the mean exactly one standard deviation to the left of the mean and one standard deviation to the right. Normal distributions occur in the real world as good approximations to heights of people, SAT scores, IQ scores, repeated measurements of the same quantity, the number of heads in a large number of tosses of a fair coin, and many other things. ( x- ) 2 3 M316 Chapter 3 Dr. Berg Here is a graph of several normal curves with different values of the mean and standard deviation. The 689599.7 Rule All normal curves satisfy share a special ruleofthumb. In a normal distribution with mean and standard deviation : a) Approximately 68% of the observations fall within of . B) Approximately 95% of the observations fall within 2 of . c) Approximately 99.7% of the observations fall within 3 of . 4 M316 Chapter 3 Dr. Berg Example The heights of young women aged 20 to 29 approximately satisfy a normal distribution with mean 64 inches and standard deviation 2.7 inches. We conclude that approximately 68% of these young women are between 64 - 2.7 = 61.3 inches and 64 + 2.7 = 66.7 inches in height. Also, 95% are between 64 - 2(2.7) = 58.6 inches and 64 + 2(2.7) = 69.4 inches in height. And so on. The Standard Normal Distribution We use tables to find areas under normal curves. Fortunately we don't have to make a new table for every possible normal distribution. We only need a table for the standard normal distribution. Definition The standard normal distribution N(0,1) has mean 0 and standard deviation 1. If a variable x has the normal distribution N(, ) with mean and standard x - deviation , then the standard variable z = has the standard normal distribution. Example 72 - 64 The zscore for a young woman 6 feet tall is z = 2.963 (almost 3 2.7 standard deviations above average). Finding Normal Proportions Table A in our textbook gives the proportion of area to the left of the zscore for the standard normal distribution. Some tables do it differently. Our textbook refers to this as the cumulative proportion. Others call this a cumulative distribution function; it gives the probability that a randomly chosen individual will have a (standardized) measurement for that variable that falls to the left of the z score. Using the Standard Normal Table 1) 2) z. 3) To use Table A to find a normal proportion: State the problem in terms of the observed variable x. A picture might help. Standardize x to restate the problem in terms of a standard normal variable Use the table and the fact that the total area is 1 to find the required area. 5 M316 Chapter 3 Dr. Berg Example Find the proportion of young women who are between 5 feet 6 inches and 6 feet tall, or, between 66 and 72 inches. Indeed, the distribution is N(64, 2.7), so 66 - 64 72 - 64 P(66 x 72) = P z = P(0.74 z 2.96) 2.7 2.7 = 0.9985 - 0.7704 = 0.2281, so about 22.8% fall into this height range. Exercise a) Find the proportion of young women over 5 feet 8 inches tall. b) Find the proportion of young women between 5 feet 7 inches and 4 feet 2 inches tall. Finding a Value Given a Proportion We can also use table A to find values that place subjects above or below a certain percentage of the population. Example Find how tall a young woman would be were she in the top 10% of the population. Indeed, we look on table A for the score nearest .9 and convert this into a height. On table A 0.8997 corresponds to 1.28 standard deviations above the mean. Thus, the height is 64 + (2.7)(1.28) = 67.456 inches. Exercise (3.14) Scores on the Wechsler Adult Intelligence Scale are approximately N(100, 15). a) What scores fall in the lowest 25% of the distribution? b) How high a score is needed to be in the highest 5%? 6 ...
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This note was uploaded on 09/14/2009 for the course CH 310 N taught by Professor Blocknack during the Fall '08 term at University of Texas at Austin.

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