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

Question

A website manager has noticed that during the evening​ hours, about 5 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. ​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?

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Melody 3 months 2021-10-18T13:34:09+00:00 1 Answer 0 views 0

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    2021-10-18T13:35:56+00:00

    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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