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5.121. Calculate the first four moments about the mean for the distribution of Problem 5.97. 5.122. Find (a) m1, (b) m2, (c) m3, (d) m4, (e) x, (f) s, (g) x2 (h) x3, (i) x4, (j) (x # Problem 5.100. 1)3 for the distribution of
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5.123. Find the coefficient of (a) skewness, (b) kurtosis for the distribution Problem 5.120. 5.124. Find the coefficient of (a) skewness, (b) kurtosis for the distribution of Problem 5.97. See Problem 5.121. 5.125. The second moments about the mean of two distributions are 9 and 16, while the third moments about the mean are 8.1 and 12.8, respectively. Which distribution is more skewed to the left 5.126. The fourth moments about the mean of the two distributions of Problem 5.125 are 230 and 780, respectively. Which distribution more nearly approximates the normal distribution from the viewpoint of (a) peakedness, (b) skewness
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CHAPTER 5 Sampling Theory
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5.127. A population of 7 numbers has a mean of 40 and a standard deviation of 3. If samples of size 5 are drawn from this population and the variance of each sample is computed, find the mean of the sampling distribution of variances if sampling is (a) with replacement, (b) without replacement. 5.128. Certain tubes produced by a company have a mean lifetime of 900 hours and a standard deviation of 80 hours. The company sends out 1000 lots of 100 tubes each. In how many lots can we expect (a) the mean lifetimes to exceed 910 hours, (b) the standard deviations of the lifetimes to exceed 95 hours What assumptions must be made 5.129. In Problem 5.128 if the median lifetime is 900 hours, in how many lots can we expect the median lifetimes to exceed 910 hours Compare your answer with Problem 5.128(a) and explain the results. 5.130. On a citywide examination the grades were normally distributed with mean 72 and standard deviation 8. (a) Find the minimum grade of the top 20% of the students. (b) Find the probability that in a random sample of 100 students, the minimum grade of the top 20% will be less than 76. 5.131. (a) Prove that the variance of the set of n numbers a, a d, a 2d, c, a (n 1) d (i.e., an arithmetic 1 progression with first term a and common difference d) is given by 12(n2 1)d2. [Hint: Use 1 2 c (n 1) 1 n(n 1), 12 22 32 c (n 1)2 1 n (n 1)(2n 3 1).] 2 6 (b) Use (a) in Problem 5.103. 5.132. Prove that the first four moments about the mean of the arithmetic progression a, a a (n 1)d are m1 0, m2 1 2 (n 12 1)d2, 24 m3 34 0, c m4 (n 1 (n2 240 1)4
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Compare with Problem 5.114. [Hint: 14
ANSWERS TO SUPPLEMENTARY PROBLEMS
5.49. (a) 9.0 (b) 4.47 (c) 9.0 (d) 3.16 5.51. (a) mX 5.52. (a) mX 22.40 oz, sX 22.40 oz, sX 5.50. (a) 9.0 (b) 4.47 (c) 9.0 (d) 2.58 22.40 oz, sX is slightly less than 0.008 oz 22.40 oz, sX 0.0057 oz
0.008 oz (b) mX 0.008 oz (b) mX
5.53. (a) 237 (b) 2 (c) none (d) 24
5.54. (a) 0.4972 (b) 0.1587 (c) 0.0918 (d) 0.9544 5.56. 0.0026
5.55. (a) 0.8164 (b) 0.0228 (c) 0.0038 (d) 1.0000 5.57. (a) 0.0029 (b) 0.9596 (c) 0.1446 5.59. (a) 0.0179 (b) 0.8664 (c) 0.1841 5.62. (a) 19 (b) 125
5.58. (a) 2 (b) 996 (c) 218 5.60. (a) 6 (b) 9 (c) 2 (d) 12 5.64. (a) 0.0028 (b) 0.9172
5.63. (a) 0.0077 (b) 0.8869
CHAPTER 5 Sampling Theory
5.66. 0.0482 5.67. 0.0188 5.68. 0.0410
5.65. (a) 0.2150 (b) 0.0064 (c) 0.4504 5.70. (a) 118.79 lb (b) 0.74 lb 5.73. (a) 40/3 (b) 28.10
5.71. 0.0228
5.72. (a) 10.00 (b) 11.49 5.75. (a) 0.36 (b) 0.49
5.74. (a) 0.50 (b) 0.17 (c) 0.28
5.80. (a) between 0.01 and 0.05 (b) greater than 0.05 5.82. (a) 799 (b) 1000 (c) 949.5 (d) 1099.5, 1199.5
5.81. (a) greater than 0.05 (b) greater than 0.05 0.155 or 15.5% (i) 19.0% ( j) 78.0%
(e) 100 (hours) (g) 62>400 (f ) 76 (h) 29.5%
5.85. (a) 24% (b) 11% (c) 46% 5.86. (a) 0.003 inch (b) 0.3195, 0.3225, 0.3255, . . . ,0.3375 inch (c) 0.320 0.322, 0.323 0.325, 0.326 0.328, . . . ,0.335 0.337 5.91. 86 5.92. 0.50 s 5.93. 8.25 5.94. (a) 82 (b) 79 5.95. 78 5.96. 80%, 20%
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