Question

The number of entries made in accounts receivable is a poisson distribution with average 2 per day .What is the probability that on a specified day at least 3 entries will be received?     

Answer 1

Given,

$Mean\ \mu =2$

and we know that,

$For\ Poisson\ Distribution\\P(x)=\dfrac{\mu^xe^{-\mu}}{x!}\\Putting\ \mu=2\\P(x)=\dfrac{2^xe^{-2}}{x!}$

Now, Probability that on a specified day at least 3 entries will receive = $P(x\geq 3)$

$P(x\geq 3)=1-[P(0)+P(1)+P(2)]$

$=1-\dfrac{1}{e^2}[\dfrac{2^0}{0!}+\dfrac{2^1}{1!}+\dfrac{2^2}{2!}]\\\ \\=1-\dfrac{1}{e^2}[1+2+2]\\\ \\=1-\dfrac{5}{e^2}=0.323$

Hence, $\boxed{P(x\geq 3)=0.323}$