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nydimaria [60]
3 years ago
9

A website manager has noticed that during the evening​ hours, about 5 people per minute check out from their shopping cart and m

ake an online purchase. She believes that each purchase is independent of the others and wants to model the number of purchases per minute. ​a) What model might you suggest to model the number of purchases per​ minute? ​b) What is the probability that in any one minute at least one purchase is​ made? ​c) What is the probability that seven people make a purchase in the next four ​minutes?
Mathematics
1 answer:
Over [174]3 years ago
8 0

Answer:

a) Poisson distribution

b) 99.33% probability that in any one minute at least one purchase is​ made

c) 0.05% probability that seven people make a purchase in the next four ​minutes

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.

5 people per minute check out from their shopping cart and make an online purchase.

This means that \mu = 5

a) What model might you suggest to model the number of purchases per​ minute? ​

The only information that we have is the mean number of an event(purchases) in a time interval. Each event is also independent fro each other. So you should suggest the Poisson distribution to model the number of purchases per​ minute.

b) What is the probability that in any one minute at least one purchase is​ made? ​

Either no purchases are made, or at least one is. The sum of the probabilities of these events is 1. So

P(X = 0) + P(X \geq 1) = 1

We want to find P(X \geq 1)

So

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

In which

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

P(X = 0) = \frac{e^{-5}*(5)^{0}}{(0)!} = 0.0067

1 - 0.0067 = 0.9933.

99.33% probability that in any one minute at least one purchase is​ made

c) What is the probability that seven people make a purchase in the next four ​minutes?

The mean is 5 purchases in a minute. So, for 4 minutes

\mu = 4*5 = 20

We have to find P(X = 7).

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

P(X = 0) = \frac{e^{-20}*(20)^{7}}{(7)!} = 0.0005

0.05% probability that seven people make a purchase in the next four ​minutes

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