Question

A website manager has noticed that during the evening​ hours, about 3.23.2 people per minute check out from their shopping cart and make an online purchase. She believes that each purchase is independent of the others and wants to model the number of purchases per minute.
1. What model might you suggest to model the number of purchases per​ minute?

a. Binomial
b. Uniform
c. Poisson
d. Geometric

2. What is the probability that in any one minute at least one purchase is​ made?
3. What is the probability that no one makes a purchase in the next 2​ minutes?

Answer 1

Answer:

1.  c. Poisson

2. 0.9592 = 95.92% probability that in any one minute at least one purchase is​ made.

3. 0.0017 = 0.17% probability that no one makes a purchase in the next 2​ minutes.

Step-by-step explanation:

We have only the mean, which means that the Poisson distribution is used to solve this question, and thus the answer to question 1 is given by option c.

Poisson distribution:

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.

Mean of 3.2 minutes:

This means that $\mu = 3.2n$, in which n is the number of minutes.

2. What is the probability that in any one minute at least one purchase is​ made?

$n = 1$, so $\mu = 3.2$.

This probability is:

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

In which

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

$P(X = 0) = \frac{e^{-3.2}*3.2^{0}}{(0)!} = 0.0408$

So

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0408 = 0.9592$

0.9592 = 95.92% probability that in any one minute at least one purchase is​ made.

3. What is the probability that no one makes a purchase in the next 2​ minutes?

2 minutes, so $n = 2, \mu = 3.2(2) = 6.4$

This probability is P(X = 0). So

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

$P(X = 0) = \frac{e^{-6.4}*6.4^{0}}{(0)!} = 0.0017$

0.0017 = 0.17% probability that no one makes a purchase in the next 2​ minutes.