Chpt02 - Chapter 2 Numerical Summary Measures Section 2.1...

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Chapter 2 Numerical Summary Measures Section 2.1 1. (a) The sum of the n = 11 data points is 514.90, so x = 514.90/11 = 46.81. (b) The sample size (n = 11) is odd, so there will be a middle value. Sorting from smallest to largest: 4.4 16.4 22.2 30.0 33.1 36.6 40.4 66.7 73.7 81.5 109.9. The sixth value, 36.6 is the middle, or median, value. The mean differs from the median because the largest sample observations are much further from the median than are the smallest values. (c) Deleting the smallest (x = 4.4) and largest (x = 109.9) values, the sum of the remaining 9 observations is 400.6. The trimmed mean tr x is 400.6/9 = 44.51. The trimming percentage is 100(1/11) 9.1%. tr x lies between the mean and median. = (100.4/8) = 12.55. 2. (a) The sample mean is x The sample size (n = 8) is even. Therefore, the sample median is the average of the (n/2) and (n/2) + 1 values. By sorting the 8 values in order, from smallest to largest: 8.0 8.9 11.0 12.0 13.0 14.5 15.0 18.0, the forth and fifth values are 12 and 13. The sample median is (12.0 + 13.0)/2 = 12.5. The 12.5% trimmed mean requires that we first trim (.125)(n) or 1 value from the ends of the ordered data set. Then we average the remaining 6 values. The 12.5% trimmed mean ) 5 . 12 ( tr x is 74.4/6 = 12.4. All three measures of center are similar, indicating little skewness to the data set. (b) The smallest value (8.0) could be increased to any number below 12.0 (a change of less than 4.0) without affecting the value of the sample median. (c) The values obtained in part (a) can be used directly. For example, the sample mean of 12.55 psi could be re-expressed as (12.55 psi) x ksi 70 . 5 psi 2 . 2 ksi 1 = .
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Chapter 2 Numerical Summary Measures 2 3. (a) A stem-and leaf display of this data appears below: Stem-and-leaf of C1 N = 26 Leaf Unit = 1.0 2 32 55 4 33 49 4 34 8 35 6699 13 36 34469 13 37 03345 8 38 9 7 39 2347 3 40 23 1 41 1 42 4 The display is reasonably symmetric, so the mean and median will be close. (b) The sample mean is x x ~ = (369+370)/2 = 369.50. = 9638/26 = 370.7. The sample median is (c) The largest value (currently 424) could be increased by any amount. Doing so will not change the fact that the middle two observations are 369 and 370, and hence, the median will not change. However, the value x = 424 can not be changed to a number less than 370 (a change of 424-370 = 54) since that will lower the values(s) of the two middle observations. (d) Expressed in minutes, the mean is (370.7 sec)/(60 sec) = 6.18 min; the median is 6.16 min. 4. The three quantities of interest are: . Their relationship is as follows: and , , 1 1 + + n n n x x x [] + + = + + = + + = + = + + = + + = + 1 1 1 1 1 1 1 1 1 1 1 1 1 1 n x x n x x n n x x n x n x n n n n n i n i n i i n For the strength observations, 34 . 43 10 = x 11 = x and . Therefore, 5 . 81 81 . 46 1 10 5 . 81 ) 34 . 43 )( 10 ( 11 = + + = x x 5. The sum of the n = 10, observations is 120.10. so the mean is 12.01. The sorted data is: 6.5 7.9 9.2 10.0 10.7 12.0 12.5 14.5 14.9 21.9, so the middle two values are 10.7 and 12.0. Therefore, the median is (10.7+12.0)/2 = 11.35. Trimming off the largest and smallest values (i.e, using a 100(1/10)% = 10% trimming factor), the trimmed mean is (120.10-6.5-21.9)/8 = 11.45. Because the trimmed mean and median are close, either of these values (11.35 or 11.46) would be a good representative value for the data.
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Chpt02 - Chapter 2 Numerical Summary Measures Section 2.1...

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