Question

The manager of the local grocery store has determined that, on average, 4 customers use the service desk every half-hour. Assume that the number of customers using the service desk has a Poisson distribution. What is the probability that during a randomly selected half-hour period, exactly 2 customers use the service desk

Answer 1

Answer:

There is a 14.65% probability that during a randomly selected half-hour period, exactly 2 customers use the service desk.

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 time interval.

The manager of the local grocery store has determined that, on average, 4 customers use the service desk every half-hour.

This means that $\mu = 4$

What is the probability that during a randomly selected half-hour period, exactly 2 customers use the service desk?

This is P(X = 2). So

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

$P(X = 2) = \frac{e^{-4}*(4)^{2}}{(2)!} = 0.1465$

There is a 14.65% probability that during a randomly selected half-hour period, exactly 2 customers use the service desk.