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# ch03 - 3.1 For a data set with an odd number of...

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3.1 For a data set with an odd number of observations, first we rank the data set in increasing (or decreasing) order and then find the value of the middle term. This value is the median. For a data set with an even number of observations, first we rank the data set in increasing (or decreasing) order and then find the average of the two middle terms. The average gives the median. 3.2 3.3 Suppose the 2002 sales (in millions of dollars) of five companies are: 10, 21, 14, 410, and 8. The mean for the data set is: Mean = (10 + 21 + 14 + 410 + 8) / 5 = \$92.60 million Now, if we drop the outlier (410), the mean is: Mean: (10 + 21 + 14 + 8) / 4 = \$13.25 million This shows how an outlier can affect the value of the mean. 3.5 The mode can assume more than one value for a data set. Examples 3–8 and 3–9 of the text present such cases. 3.3 3.7 For a symmetric histogram (with one peak), the values of the mean, median, and mode are all equal. Figure 3.2 of the text shows this case. For a histogram that is skewed to the right, the value of the mode is the smallest and the value of the mean is the largest. The median lies between the mode and the mean. Such a case is presented in Figure 3.3 of the text. For a histogram that is skewed to the left, the value of the mean is the smallest, the value of the mode is the largest, and the value of the median lies between the mean and the mode. Figure 3.4 of the text exhibits this case. 3.4 3.9 x = 5 + (–7) +2 + 0 + (–9) + 16 +10 + 7 = 24 ( N + 1) / 2 = (8 + 1) / 2 = 4.5 µ =( x ) / N = 24 / 8 = 3 Median = value of the 4.5 th term in ranked data = (2 + 5) / 2 = 3.50 33

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This data set has no mode. 3.5 3.11 x = ( x ) / n = 9907/ 12 = \$825.58 ( n +1) / 2 = (12 + 1) / 2 = 6.5 Median = value of the 6.5 th term in ranked data set = (769 + 798) / 2 = \$783.50 3.6 3.13 x = ( x ) / n = 16.682 / 12 = \$1.390 (n + 1) / 2 = (12 + 1) / 2 = 6.5 Median = (1.360 + 1.351) / 2 = \$1.356 This data set has no mode. 3.15 x = ( x ) / n = 85.81 /12 = \$7.15 ( n +1) / 2 = (12 + 1) / 2 = 6.5 Median = (6.99 + 7.03) / 2 = \$7.01
3.17 μ = ( x ) / N = 35,629 / 6 = \$5938.17 thousand ( n +1) / 2 = (6 +1) / 2 = 3.5 Median = (750 + 8500) / 2 = \$4625 thousand This data set has no mode because no value appears more than once. 3.19 x = ( x ) / n = 64 / 10 = 6.40 hours ( n +1) / 2 = (10 +1) / 2 = 5.5 Median = (7 + 7) / 2 =7 hours Mode = 0 and 7 hours 3.21 x = ( x ) / n = 294 / 10 = 29.4 computer terminals ( n + 1) / 2 = (10 +1) / 2 = 5.5 Median = (28 + 29) / 2 = 28.5 computer terminals Mode = 23 computer terminals 3.23 a. x = ( x ) / n = 257 /13 = 19.77 newspapers ( n + 1) / 2 = (13 + 1) / 2 = 7 Median = 12 newspapers b. Yes, 92 is an outlier. When we drop this value, Mean = 165 / 12 = 13.75 newspapers ( n + 1) / 2 = (12 + 1) / 2 = 6.5 Median = (11+12) / 2 = 11.5 newspapers As we observe, the mean is affected more by the outlier. c. The median is a better measure because it is not sensitive to outliers. 3.26

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3.27 Total money spent by 10 persons = x = n x = 10(85.50) = \$855 3.27 3.29 Sum of the ages of six persons = 6 x 46 = 276 years, so the age of sixth person = 276 – (57 + 39 + 44 + 51 + 37) = 48 years. 3.28 3.31 For Data Set I: Mean = 123 / 5 = 24.60 For Data Set II: Mean = 158 / 5 = 31.60 The mean of the second data set is greater than the mean of the first data set by 7. 3.33 The ranked data are: 19 23 26 31 38 39 47 49 53 67 By dropping 19 and 67, we obtain: x = 23 + 26 + 31 + 38 + 39 + 47 + 49 + 53 = 306 10% Trimmed Mean = ( x ) / n = 306/ 8 = 38.25 years 3.35 From the given information: x 1 = 73, x 2 = 67, x 3 = 85, w 1 = w 2 = 1, w 3 =2 Weighted mean = 5 . 77 4 310 4 ) 85 )( 2 ( ) 67 )( 1 ( ) 73 )( 1 ( = = + + = w xw 3.37 Suppose the monthly income of five families are: \$1445 \$2310 \$967 \$3195 \$24,500 Then, Range = Largest value – Smallest value = 24,500 – 967 = \$23,533 Now, if we drop the outlier (\$24,500) and calculate the range, then:
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