poisson distribution

Question

The time between calls to a corporate office is exponentially distributed with a mean
of 10 minutes.
a) What is the probability that there have call in one-half hour?

b) What is the probability that there are no calls within one-half hour?

c) Determine x such that probability that there are no calls within x hours is 0.02.

d) If four non-overlapping one-half-hour intervals are selected, what is the probability
that none of these intervals contains any call?

e) State the relationship between the results in part (a) and (b).

f) What is the name applied to the distribution and parameter values of the time until
12th call?

g) What is the mean time until the 12th call?

Answer 1

PDF: f(X)=(1/mu)e^-(X/mu); lambda=1/mu where mu is the mean time between events, so lambda is the number of events in unit time. X>0. f(X)=0 when X=0.

CDF: F(X)=1-e^-(X/mu), where mu=E(X) (expectation)

a) mu=10 mins=1/3 of a half-hour. Lambda=3.

f(1)=3e^-3=0.1494 or 14.94%

b) f(X)=0 when X=0 by definition, so we cannot use f(X) directly, so we use lambda=6, the number of calls within an hour, and then we can calculate the probability using the CDF function F(X): P(X>1/2)=1-P(X<=1/2)=1-F(1/2)=1-(1-e^-(6/2))=e^-3=0.0498 approx, probability of no calls within half an hour.

c) If probability is 0.02 of no calls in x hours we use the same logic as in b):

Again we start with P(X>x)=1-P(X<=x)=1-F(x)=1-(1-e^-(6x))=e^-(6x)=0.02, so, taking natural logs: -6x=ln(0.02)=-3.912, so x=0.652hr=39.12 mins approx.

d) We know from (b) the probability of no calls within a half-hour is e^-3. The same probability applies to each of the non-overlapping half-hour periods, so the combined probability is (e^-3)^4=e^-12=0.000006144 approx.

e) There is a factor of 3 between (a) and (b): (a)=3(b).

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