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It is estimated that 0.1% of the callers to the billing department of a local telephone...

It is estimated that 0.1% of the callers to the billing department of a local telephone company will receive a busy signal. What is the probability that of today's 1230 callers at least 5 received a busy signal? Use the poisson approximation to the binomial.

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Expert Solution

p: Probability that the callers to the to the billing department of a local telephone company will receive a busy signal = 0.1/100=0.001

Today's number of callers : n=1230

X : Number of callers will receive a busy signal

Possion distribution as an approximation to Binomial

mean number of callers to receive a busy signal for today's 1230 callers : = np=1230x0.001=1.23

X follows Possion distribution (as an approximation to Binomial) with =1.23

Probability that 'x' callers will receive a busy signal today is given

Probability that at least 5 recieved a busy signal = P(X5) = 1 - [P(X=0]+P(X=1]+P(X=2]+P(X=3]+P(X=4)]

[P(X=0]+P(X=1]+P(X=2]+P(X=3]+P(X=4)] = 0.29229+0.35952+0.22110+0.09065+0.02788=0.99144

P(X5) = 1 - [P(X=0]+P(X=1]+P(X=2]+P(X=3]+P(X=4)] =1 - 0.99144 = 0.00856

Probability that at least 5 recieved a busy signal = 0.00856

orchestra answered 2 years ago
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