Homework Solutions 05

Homework Solutions 05 - (W4150 Intro to Probability and...

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Solutions to Homework #5 (W4150 ‘Intro to Probability and Statistics’, S04) Sec. 5.3. (4) (a) P(X=2) = C 5 2 (3/4) 2 (1/4) 3 = .0879 ; or you can use tables/Excel to find the same value by taking difference P(X=2) = P(X< 2) – P(X< 1) = 2 0 b(x ;5,.75) - 1 0 b(x ;5,.75) (b) P(X< 3) = 3 0 b(x ;5,.75) = .3672 Sec.5.3. (7) (a) p = .7; n = 10 ± P(X<n/2) = P(X<5) = P(X< 4) = 4 0 b(x ;10,.7) = .0473 (b) p = .7; n = 20 ± P(X<n/2) = P(X<10) = P(X< 9) = 9 0 b(x ;20,.7) = .0171 Sec. 5.3. (15) P(inoculated mouse is protected) = p = .6; n = 5. Let X = {# of mice that didn’t get ill} (a) P(none of the mice is infected} = P(X = n = 5) = 5 0 b(x ;5,.6) - 4 0 b(x ;5,.6) = .0778 (b) P(fewer than 2 contract the disease) = P(X> 4) = 1 – P(X<4) = 1 - 3 0 b(x ;5,.6) = .337 (c) P(more than 3 contract the disease) = P(X< 1) = 1 0 b(x ;5,.6) = .087 Comment : if you choose X to be number of ill mices than you need to preserve «connection» between the notion of ‘success’ and it’s probability. Namely, if you adopt the above
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Homework Solutions 05 - (W4150 Intro to Probability and...

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