Solution-HW2

Solution-HW2 - HOMEWORK SOLUTION-2 SPRING 2007 Chapter 7 24...

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Unformatted text preview: HOMEWORK SOLUTION -2 SPRING 2007 Chapter 7 24. a. Normal distribution, ( ) 4260 E x = (2 pt) / / . 900 50 127 28 x n σ σ = = = b. Within \$250 (4pt) P (4010 ≤ x ≤ 4510) 4510 4260 . 1 96 . 127 28 z- = = Area = .4750 2(.4750) = .95 c. Within \$100 (4pts) P (4160 ≤ x ≤ 4360) 4360 4260 . 79 . 127 28 z- = = Area = .2852 2(.2852) = .5704 ( Total 10 pts) 29. μ = 1.46 σ = .15 a. n = 30 . 03 . 1 10 / . / 15 30 x z n μ σ- = = = P (1.43 ≤ x ≤ 1.49) = P (-1.10 ≤ z ≤ 1.10) = 2(.3643) = .7286 (4pts) b. n = 50 . 03 . 1 41 / . / 15 50 x z n μ σ- = = = P (1.43 ≤ x ≤ 1.49) = P (-1.41 ≤ z ≤ 1.41) = 2(.4207) = .8414 (4pts) c. n = 100 . 03 . 2 00 / . / 15 100 x z n μ σ- = = = P (1.43 ≤ x ≤ 1.49) = P (-2 ≤ z ≤ 2) = 2(.4772) = .9544 (4pts) d. A sample size of 100 is necessary. (1pt) (Total 13 pts) 37. a. Normal distribution (1 pt) E ( p ) = .50 (1 pt) ( ) (. )( . ) 1 50 1 50 . 0206 589 p p p n σ-- = = = (2 pts) b....
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Solution-HW2 - HOMEWORK SOLUTION-2 SPRING 2007 Chapter 7 24...

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