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# Assign5 - STA 3032 Assignment 5 Problem 5.10 a Probability...

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STA 3032 – Assignment 5 Problem 5.10: a.) Probability that a tire lasts at most 10000 miles. [ ] 39347 . 0 1 20 1 ) 10 Pr( 5 . 0 10 0 20 / 10 0 20 / = - = - = = - - - e e dx e X x x b.) Probability that a tire lasts anywhere from 16000 to 24000 miles. [ ] 148135 . 0 20 1 ) 24 16 Pr( 2 . 1 8 . 0 24 16 20 / 24 16 20 / = - = - = = - - - - e e e dx e X x x c.) Probability that a tire lasts at least 30000 miles. [ ] 22313 . 0 20 1 ) 30 Pr( 5 . 1 30 20 / 30 20 / = = - = = - - - e e dx e X x x Problem 5.20: a.) Pr(0 Z 2.7) = 0.9965 – 0.5000 = 0.4965 b.) Pr (1.22 Z 2.43) = 0.9925 – 0.8888 = 0.1037 c.) Pr(-1.35 Z -0.35) = 0.3632 – 0.0885 = 0.2747 d.) Pr(-1.70 Z 1.35) = 0.9115 – 0.0446 = 0.8669 Problem 5.24: Normal r.v. with mean 16.2 and variance 1.5625. a.) Probability r.v. is greater than 16.8. 3156 . 0 6844 . 0 1 ) 48 . 0 Pr( 1 ) 48 . 0 Pr( 25 . 1 2 . 16 8 . 16 Pr ) 8 . 16 Pr( = - = < - = = - = Z Z Z X b.) Probability r.v. is less than 14.9. 1492 . 0 ) 04 . 1 Pr( 25 . 1 2 . 16 9 . 14 Pr ) 9 . 14 Pr( = - < = - < = < Z Z X c.) Probability r.v. is between 13.6 and 18.8.

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9624 . 0 0188 . 0 9812 . 0 ) 08 . 2 Pr( ) 08 . 2 Pr( 25 . 1 2 . 16 8 . 18 25 . 1 2 . 16 6 . 13 Pr ) 8 . 18 6 . 13 Pr( = - = - - = = - - = Z Z Z X d.) Probability r.v. is between 16.5 and 16.7. 0606 . 0 5948 . 0 6554 . 0 ) 24 . 0 Pr( ) 4 . 0 Pr( 25 . 1 2 . 16 7 . 16 25 . 1 2 . 16 5 . 16 Pr ) 7 . 16 5 . 16 Pr( = - = < - < = - < < - = < < Z Z Z X Problem 5.26: a.) Probability it will take longer than 258.3 sec to pop. 9332 . 0 0668 . 0 1 ) 5 . 1 Pr( 1 ) 5 . 1 Pr( 10 3 . 273 3 . 258 Pr ) 3 . 258 Pr( 3 . 273 2 . 9 5 . 282 92 . 0 10 5 . 282 8212 . 0 10 5 . 282 Pr = - = - < - = - = - =
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