Mathematics, 05.03.2020 17:24 eriks7779

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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Answer from: kay4358

The Poisson distribution model is used when the data consist of counts of occurrences.

a) The Poisson distribution model is used when the data consist of counts of occurrences.

b) Given that: λ (mean number of occurrence) = 5 people per minute, hence:

$P(X\ge 1)=1-P(X=0)=1-\frac{e^{-\lambda }\lambda^x}{x!}= 1-\frac{e^{-5 }5^0}{5!}=0.9999$

The probability that in any one minute at least one purchase is made is 0.9999.

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Answer from: amoore51

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 $P(X \geq 1)$

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

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

Answer from: Quest

34.5 cm

step-by-step explanation:

generally for a circular segment of angle, θ° and radius r

arc length

= (θ/360) x circumference

= (θ/360) x 2πr

in our case, θ = 60°, π=3.142 and r = 33 cm

substitute into above formula

arc length

= (60/360) x 2(3.142)( 33) = 34.5 cm

Answer from: Quest

could you provide figure.

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