Answer to Question #136726 in Statistics and Probability for NOMHLE

Question

2.. A BTC operator on the average handles 20 calls every 10 minutes. It is known that
the number of calls received in a time interval follows Poisson distribution. Find the
probability that the operator will
a) be busy in the next one minute
b) handle at least two calls in the next two minutes
c) handle at least one call in the next 90 seconds.

Answer 1

Let $X=$ the number of calls received in a time interval $t$ : $X\sim Po(\lambda t)$

$$P(X=x)=\dfrac{e^{-\lambda t}(\lambda t)^x}{x!}, x=0,1,2,...$$

$\lambda=\dfrac{20}{600 s}=\dfrac{1}{30}s^{-1}$

a) $t=60s$

$\lambda t=\dfrac{1}{30}s^{-1}(60s)=2$

$P(X=0)=\dfrac{e^{-2}(2)^0}{0!}=e^{-2}\approx0.135335$

b) $t=120s$

$\lambda t=\dfrac{1}{30}s^{-1}(120s)=4$

$P(X\geq2)=1-P(X=0)-P(X=1)=$

$=1-\dfrac{e^{-4}(4)^0}{0!}-\dfrac{e^{-4}(4)^1}{1!}=1-5e^{-4}\approx0.908422$

c) $t=90s$

$\lambda t=\dfrac{1}{30}s^{-1}(90s)=3$

$P(X\geq1)=1-P(X=0)=$

$=1-\dfrac{e^{-3}(3)^0}{0!}=1-e^{-3}\approx0.950213$