Question

A bank gets an average of 12 customers per hour. Assume the variable follows a Poisson distribution. Find the probability that there will be 4 or more customers at this bank in one hour.

Answer 1

Using the Poisson distribution, there is a 0.9978 = 99.78% probability that there will be 4 or more customers at this bank in one hour.

What is the Poisson distribution?

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

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

The parameters are:

• x is the number of successes

• e = 2.71828 is the Euler number

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

A bank gets an average of 12 customers per hour, hence the mean is $\mu = 12$.

The probability that there will be 4 or more customers at this bank in one hour is:

$P(X \geq 4) = 1 - P(X < 4)$

In which:

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

Then:

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

$P(X = 0) = \frac{e^{-12}12^{0}}{(0)!} \approx 0$

$P(X = 1) = \frac{e^{-12}12^{1}}{(1)!} \approx 0$

$P(X = 2) = \frac{e^{-12}12^{2}}{(2)!} = 0.0004$

$P(X = 3) = \frac{e^{-12}12^{3}}{(3)!} = 0.0018$

Then:

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0 + 0 + 0.0004 + 0.0018 = 0.0022.

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.0022 = 0.9978$

0.9978 = 99.78% probability that there will be 4 or more customers at this bank in one hour.

More can be learned about the Poisson distribution at https://brainly.com/question/13971530

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