Question

: Customers arrive at a shopping mart randomly at a rate of 7 per hour.

a) Find the probability that during any 90 minutes’ period, the number of customers arriving at the shopping mart is exactly 8

b) If a customer arrives at 11. 30 am, find the probability that the next patient arrives before

11. 45 am

Answer 1

Using the Poisson distribution, it is found that the probabilities are given by:

a) 0.1009 = 10.09%.

b) 0.8262 = 82.62%.

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.

Item a:

The mean is of 7 customers for 60 minutes, hence, for 90 minutes = 1.5 x 60 minutes, the mean is given by:

$\mu = 1.5 \times 7 = 10.5$

The probability of exactly 8 customers is of P(X = 8), hence:

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

$P(X = 8) = \frac{e^{-10.5}10.5^{8}}{(8)!} = 0.1009$

Item b:

Probability of at least one patient in 15 minutes, hence:

$P(X \geq 1) = 1 - P(X = 0)$

For 15 minutes, the mean is given by:

$\mu = 0.25 \times 7 = 1.75$

Then:

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

$P(X = 8) = \frac{e^{-1.75}1.75^{0}}{(0)!} = 0.1738$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.1738 = 0.8262$

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