# mid1_06_sol - MATH 306 MIDTERM EXAM I SPRING 2004-2005...

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MATH 306 MIDTERM EXAM I - SPRING 2004-2005, April 9,2005 5 questions, total 105 pts . Time allowed : 1 hr. 45 min. Question 1. a) The amount of liquid in a certain type of soft drink bottles is a random variable with mean 200 milliliters and standard deviation of 15 milliliters. What is the probability that in a random sample of 36 bottles the total amount of liquid will be at least 7350 milliliters? 6pts . b) In a random sampling from a normal distribution the sample size is n = 11 and σ = 0 . 8 . Find P ( 11 i =1 ( Y i - ¯ Y ) 2 3 . 1137) . 6pts . c) Let T be a t - random variable with ν = 13 degrees of freedom. Find a number k such that P ( k < T < 1 . 35) = 0 . 875 . 6pts . Solution: a) P ( 36 i =1 Y i 7350) = P ( ¯ Y 7350 36 = P ( ¯ Y 204 . 1667) = P ( 204 . 1667 - 200 15 6 ) = P ( Z 1 . 67) = 0 . 0475 . b) P ( i = 1 11 ( Y | i - ¯ Y ) 2 (0 . 8) 2 3 . 1137 0 . 64 ) = P ( χ 2 10 4 . 8651) = 1 - 0 . 90 = 0 . 10 (Chi-square table : χ 2 0 . 90;10 = 4 . 86518) . c) From t-table t 0 . 10;13 = 1 . 35 , k = - t 0 . 025;13 , because P ( - t 0 . 025;13 < T 13 < t 0 . 10;13 ) = 0 . 875 , from the table k = - 2 . 160 . Question 2. Let Y 1 , Y 2 , ··· , Y 10 be a random sample of size 10 from a normal population with mean 0 and variance 1 . Let ¯ Y denote the sample mean, let W = 10 i =1 Y 2 i and U = 10 i =1 ( Y i - ¯ Y ) 2 . Furthermore let Z be a standard normal random variable independent of the sample random variables Y i . a)

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## This note was uploaded on 03/17/2012 for the course MATHEMATIC 306 taught by Professor U during the Spring '12 term at Sabancı University.

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mid1_06_sol - MATH 306 MIDTERM EXAM I SPRING 2004-2005...

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