Answer to Question #212517 in Statistics and Probability for Safa Ahmed Alrawah

Question

Suppose that the number of customers that enter a bank in an hour is a Poisson random variable, and suppose that P(X =0) = 0.02.

i. What is the probability that at most one person enter the bank in one hour?

ii. Determine the mean and standard deviation of X

Answer 1

PMF of a Poisson distribution is $p(x;\lambda) = \frac{e^{-\lambda}\lambda^{x}} {x!} for x = 0, 1, 2, \cdots$

$P(x=0)= \frac{e^{-\lambda}\lambda^{0}} {0!}$

$e^{-\lambda} =0.02$

$\lambda=-\ln0.02$

i. $P(\le1)$

$P(x\le1)=P(x=0)+P(x=1)$

$=0.02+\frac{e^{\ln{0.02}}(-\ln0.02)} {1!}$

$=0.02+0.02\times3.9120$

$=0.09824$

The excel function =POISSON.DIST(1,-LN(0.02),TRUE) can be used to obtain the answer.

ii. Mean and standard deviation of x

The mean and variance of a Poisson distribution is $\lambda$

Thus,

The mean is $-\ln(0.02)=3.912$ and

Standard deviation is $\sqrt{-\ln(0.02)}=\sqrt{3.912}=1.978$