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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