Service calls come to a maintenance counter to a poisson process, and on average, 2.7 calls are recieved per minute. Find the probability that
(a) No more than 4 calls come in any minute
(b) Fewer than 2 calls in any minute
(c) More than 10 calls come in a 5 min period
My attempt:
(a) $P(x \leq 4) =\sum_{x=0}^{4} \frac{e^{-2.7}(2.7^x)}{x!} = (\text{ how to calculate?})$
(b) $P(x \leq 1) = \sum_{x=0}^{1} \frac{e^{-2.7}(2.7^x)}{x!} = \text{ how to calculate? }$
(c) $P(x > 10) = 1 - P(x \leq 10) = 1 - \sum_{x=0}^{10} \frac{e^{-(2.7\cdot 5)}(2.7\cdot5)^x}{x!}$
Not sure if I'm doing it right
Yes, your are doing correct.
Let random variable $X$ denotes number of service calls in $t$ minutes.
Arrival rate of service calls is $\lambda=2.7$ per minute. $X$ has a poisson distribution with parameter $\mu=\lambda t =2.7t$
$$P(X=x)=e^{-\lambda t}\frac{(\lambda t)^x}{x!},\ x=0,1,2,3,4....$$
a) Probability that there will be no more than $4$ service calls in any minute of time. Here $t=1$ and $\mu=2.7$ $$P(X\le4)=\sum _{ x=0 }^{ 4 }{ e^{-2.7}\frac{(2.7)^x}{x!}}=0.863 $$
b)Here $t=1$, $\mu=2.7$ $$P(X<2)=\sum _{ x=0 }^{ 1 }{ e^{-2.7}\frac{(2.7)^x}{x!} }=0.2487 $$
c) Here $t=5$, $\mu=(2.7)(5)=13.5$ $$P(X>10)=1-P(X\le10)$$ $$=1-\sum _{ x=0 }^{ 10 }{e^{-13.5}\frac{(13.5)^x}{x!}}$$ $$P(X>10)=(1-0.2112)=0.789$$