Question 4: My dog Peter likes to lick children. At the Statistics Department ho

Question

Question 4: My dog Peter likes to lick children. At the Statistics Department holiday party, he stands near the cookie table, where children arrive according to a Poisson process with rate per hour. Assume Peter licks every child he encounters. (a) (10 pts) What is the probability that Peter licks 3 or more children in the first 2 hours of the party? (b) (10 pts) Sometimes children scream when they get licked. If each child screams with probability 1/2 independently (and independent of the number of children that get licked), use the law of iterated expectations to find the expected number of children that scream in the first two hours.

Explanation / Answer

Question 4

= 3 per hour

(a) Expected number of children arrive in next 2 hours = 2 * 3 = 6

So if X is the number of children arrive in next 2 hours.

Pr(X >= 3) = POISSON (X > = 3; 6) = 1 - POISSON (X < 3 ; 6) = 1 - 0.0620 = 0.938

(b) Here as Pr(A child will cry when licked) = 0.5

so here the expected number of child who will arrive in next 2 hours = 3 * 2 = 6

So expected number of child who will cry when licked = 6 * 0.5 = 3

(c) Number of children arrive at cookie table in half an hour = 3 * 0.5 = 1.5

Number of children arrive at soda table in half an hour = 5 * 0.5 = 2.5

so expected number of children will be licked by peter = 1.5 + 2.5 = 4

Pr(X > 2) = POISSON (X > 2 ; 4) = 1 - POISSON (X < = 2; 4) = 1 - 0.2381 = 0.7619

(d) Here the expected number of time in which peter get  children will be licked by peter in one hour = 4 * 2 = 8

so expected time that peter will lick 10 children = 10/8 = 1.25 hour

So , I and peter could expect to go home at 10: 15 PM

Variance of time for one person to be licked. = 1/8  Hours

Varaince of time for 10 person to be licked. = 10/8 = 1.25 Hours2

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