Suggested Solution of Class Test

# Suggested Solution of Class Test - Given that f Y Y =...

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THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT1306 INTRODUCTORY STATISTICS (SEMESTER 1 2008/2009) Suggested Solution of Class Test Question 1 correct is 2 6 = 1 3 . (b) From (a), the probability that he/she will check the correct answer is P ( check correct answer ) = P ( check correct answer j ) P ( ) + P ( check correct answer j second answer is correct ) P ( second answer is correct ) + P ( check correct answer j third answer is correct ) P ( third answer is correct ) = 1 3 1 3 + 1 3 1 3 + 1 3 1 3 = 1 3 where the law of total probability is used here. Let Y be the number of correct answers. Then, Y ± Bin ( n = 8 ; p = 1 = 3) : The corresponding probability is P ( Y = 4) = 8 4 1 3 ± 4 2 3 ± 4 = 1120 6561 = 0 : 1707 : Question 2 Let Y be the number of shots on which the third missing occurs. Then, Y ± NegBin ( r = 3 ; p = 0 : 06) : The corresponding probability is P ( Y = 15) = 15 ² 1 3 ² 1 ± (0 : 06) 3 (1 ² 0 : 06) 15 3 = 9 : 3547 10 3 : 1

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Unformatted text preview: Given that f Y ( Y ) = 2 &ye & &y 2 for y > , the mean of Y is E ( Y ) = 2 & Z 1 y 2 e & &y 2 dy (by letting u = &y 2 and du = 2 &ydy ) = 1 & 1 = 2 Z 1 u 1 = 2 e & u du = 1 & 1 = 2 & & 3 2 ± = 1 & 1 = 2 & 1 2 ± & & 1 2 ± = 1 2 r ± & : To obtain the variance of Y , we need E ( Y 2 ) E ( Y 2 ) = 2 & Z 1 y 3 e & &y 2 dy = Z 1 ² u & ³ e & u du = 1 & &(2) = 1 & : Hence, the variance of Y is V ar ( Y ) = E ( Y 2 ) & [ E ( Y )] 2 = 1 & & ± 4 & = 1 & ² 1 & ± 4 ³ : Question 4 Let Y be the mileage (in thousands of miles) that car owners get with a certain kind of radial tire. Then, Y ± Exp ( ² = 1 = 40) : (a) P ( Y ² 20) = Z 1 20 1 40 e & x 40 dx = ´ & e & x 40 µ 1 20 = e & 1 = 2 2 = 0 : 6065 : (b) P ( Y & 30) = Z 30 1 40 e & x 40 dx = & ± e & x 40 ± 30 = 1 ± e & 3 4 = 0 : 5276 : 3...
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Suggested Solution of Class Test - Given that f Y Y =...

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