STAT 333 tutorial1sol

# STAT 333 tutorial1sol - Stat 333 Tutorial 1 Solutions Only...

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Unformatted text preview: Stat 333 - Tutorial 1 Solutions Only ﬁnal answers will be given for questions which have been addressed in the tutorial. The solution to the ﬁnal part of question 6 is provided. 1. i) P (A ∪ B ) = 0.45 ¯¯ ii) P (A ∪ B ) = 1 2. i) P (A ∩ B ) = 0.15 ii) P (B |A) = 3. ¯ iii) P (B |A) = 1 3 11 4 9 p1 +p2 2 4. a) 1 − b) 5. a) 0.002930 c) 0.156250 (1−p1 )n +(1−p2 )n 2 c) (1−p1 )n (1−p1 )n +(1−p2 )n b) 0.05840 d) 0.156250 6. a) 0.45 b) 0.2 c) 2/3 d) -0.375 Find V ar(X + Y ): (a) We can ﬁnd the marginal probabilities by calculating the row and column totals. x=0 0 0.05 0.1 0.05 0.2 y=0 1 2 3 P (X = x) 1 0.05 0.1 0.1 0.05 0.3 2 0.1 0.1 0.1 0 0.3 3 0.05 0.1 0.05 0 0.2 P (Y = y ) 0.2 0.35 0.35 0.1 1 There are two ways to approach this question: i. Method 1: Use the formula V ar(X + Y ) = V ar(X ) + V ar(Y ) + 2Cov (X, Y ). We have 3 2 2 3 2 2 V ar(X ) = E (X ) − E (X ) = x · P (X = x) − x=0 3 V ar(Y ) = E (Y 2 ) − E 2 (Y ) = x · P (X = x) = 1.05 x=0 y 2 · P (Y = y ) − y =0 2 3 y · P (Y = y ) = 0.8275 y =0 and so V ar(X + Y ) = 1.05 + 0.8275 + 2(−0.375) = 1.1275. ii. Method 2: Consider the random variable Z = X + Y and ﬁnd its probability mass function. We have 01 2 3 4 5 6 z P (Z = z ) 0 0.1 0.3 0.3 0.25 0.05 0 Therefore, V ar(X + Y ) = V ar(Z ) = E (Z 2 ) − E 2 (Z ) 6 2 6 2 z · P (Z = z ) − = z =0 1 z · P (Z = z ) z =0 = 1.1275 ...
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## This note was uploaded on 01/21/2012 for the course STAT 333 taught by Professor Chisholm during the Fall '08 term at Waterloo.

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