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hw6solu_STAT515Fall2011

# hw6solu_STAT515Fall2011 - Solution to Homework 6(STAT515...

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Solution to Homework 6 (STAT515, Fall 2011) 5.24. ( 8 points ) a. 0.0721; c. 0.2434; f. 0.9233; g. 0.9901. 5.28. ( 8 points ) a. - 1 . 75; b. 1.96; e. 0 . 83; f. 2.50. 5.30. ( 6 points ) a. 0.3830; c. 0.1525; f. 0.9545. 5.32. ( 8 points ) a. 30; b. 14.32; c. 40.24; d. 16.84. 5.36. ( 9 points ) Let X = score on the Dental Anxiety Scale, then X N (11 , 3 . 5 2 ) . a. z = 16 - 11 3 . 5 1 . 43 . b. P (10 < X < 15) = P 10 - 11 3 . 5 < X - 11 3 . 5 < 15 - 11 3 . 5 = P ( - 0 . 29 < Z < 1 . 14) = 0 . 1141 + 0 . 3729 = 0 . 4870 . c. P ( X > 17) = P X - 11 3 . 5 > 17 - 11 3 . 5 = P ( Z > 1 . 71) = 0 . 5 - 0 . 4564 = 0 . 0436 . Note: It is also reasonable to compute P (17 < X 20) in part c considering that the score is between 0 and 20. 5.38. ( 6 points ) a. P ( X m 50) = P Z 50 - 19 65 P ( Z > 0 . 48) = 0 . 5 - 0 . 1844 = 0 . 3156 . b. P ( X v 50) = P Z 50 - 7 49 P ( Z > 0 . 88) = 0 . 5 - 0 . 3106 = 0 . 1894 . 5.44. ( 12 points ) Let X = the startle response, then X N (37 . 9 , 12 . 4 2 ) . 1

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a. P (40 < X < 50) = P 40 - 37 . 9 12 . 4 < X - 37 . 9 12 . 4 < 50 - 37 . 9 12 . 4 = P (0 . 17 < Z < 0 . 98) = 0 . 3365 - 0 . 0675 = 0 . 2690 . b. P ( X < 30) = P X - 37 . 9 12 . 4 < 30 - 37 . 9 12 . 4 = P ( Z < - 0 . 64) = 0 . 5 - 0 . 2389 = 0 . 2611 . c. We want to find an interval for X , denoted by ( x 1 , x 2 ), which is centered around the mean of X (37.9), such that, P ( x 1 < X < x 2 ) = 0 . 95 . That is, P ( x 1 < X < x 2 ) = P x 1 - 37 . 9 12 . 4 < X - 37 . 9 12 . 4 < x 2 - 37 . 9 12 . 4 = P ( - z 0 < Z < z 0 ) , = 0 . 95 . In 5.28 part b , we found that P ( - 1 . 96 < Z < 1 . 96) = 0 . 95 . Hence z 0 above is 1.96. It follows that x 1 - 37 . 9 12 . 4 = - 1 . 96 x
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hw6solu_STAT515Fall2011 - Solution to Homework 6(STAT515...

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