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) = 0.09. Determine the mean, E(X), and variance, V(X). Round your answers to two decimal places (e.g. 98.76).

Answer 1

Answer:

$E(X) = V(X) = 2.41$

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

In the Poisson distribution, the mean is the same as the variance.

P(X = 0) = 0.09

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-\mu}*\mu^{0}}{(0)!} = e^{-\mu}$

So

$e^{-\mu} = 0.09$

We apply ln to both sides, which is the inverse operation of the exponential.

So

$\ln{e^{-\mu}} = \ln{0.09}$

$-\mu = -2.41$

Multiplying by -1

$\mu = 2.41$

So

$E(X) = V(X) = 2.41$