16. The number of customers arriving per hour at a certain automobile service fa

Question

16. The number of customers arriving per hour at a certain automobile service facility is assumed to follow a Poisson distribution with mean =7 (a) Compute the probability that more than 8 customers will arrive in a 2-hour period. (b) What is the mean number of arrivals during a 2-hour period? Click here to view page 1 of the table of Poisson probability sums. Click here to view page 2 of the table of Poisson probability sums (a) The probability that more than 8 customers will arrive is (Round to four decimal places as needed.) (b) The mean number of arrivals is (Type an integer or a decimal. Do not round.) 7: Table of Poisson probability sums (page 1)

Explanation / Answer

16)

Here, m= 7

a) now, m = 2*7 = 14

P(X > 8) = 1 - P(X<=8)
P(X<=8) = m^x * e^-m / x!
= 14^8 * e^-14/8! + 14^7 * e^-14/7! + 14^6 * e^-14/6! + 14^5 * e^-14/5! + 14^4 * e^-14/4! + 14^3 * e^-14/3! + 14^2 * e^-14/2! + 14^1* e^-14/1!
= 0.06205

P(X > 8) = 1 - P(X<=8) = 1 - 0.06205 = 0.9379

b)

mean = 7 * 2 = 14

#17.

a)
As per binomial distribution,
P(X=r) = nCr * p^r * (1-p)^(n-r)

Here p = 0.02
P(X >= 3) = 1 - P(X <= 2)

P(X <= 2) = 100C0 * 0.02^0 * 0.98^100 + 100C1 * 0.02^1 * 0.98^99 + 100C2 * 0.02^2 * 0.98^98 = 0.6767

P(X >= 3) = 1 - 0.6767 = 0.3233

Hence the probability of rejection is 0.3233

b)
As per binomial distribution,
P(X=r) = nCr * p^r * (1-p)^(n-r)

Here p = 0.05

P(X <= 2) = 100C0 * 0.05^0 * 0.95^100 + 100C1 * 0.05^1 * 0.95^99 + 100C2 * 0.05^2 * 0.95^98 = 0.1183

Hence the probability of acceptance is 0.1183

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