Mathematics, 13.07.2021 15:20 kavron2322

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?

Answers: 2

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Answers

Answer from: lexieprochaskaaa

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

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

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

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

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.

Answer from: Quest

Dbc if u do the math and stuff that what i got let me know if i was wrong

Answer from: Quest

starting with:

5 + x - 12 = 2x - 7

5 + x - 12 - x = 2x - 7 - x (subtract x from both sides)

5 - 12 = x - 7 (combine and cancel x's from both sides)

5 - 12 + 7 = x - 7 + 7 (add 7 to both sides)

0 = x (add numbers together on both sides)

so in order for the equation to be true, x must be 0.

trying x = -0.5

5 + x - 12 = 2x - 7 (original equation)

5 + -0.5 - 12 = 2(-0.5) - 7 (plug in value of x)

5 + -0.5 - 12 = -1 - 7 (did multiplication)

-7.5 = -8 (did addition on both sides)

since -7.5 is obviously not equal to -8, the value of -0.5 for x is wrong.

trying x = 0

5 + x - 12 = 2x - 7 (original equation)

5 + 0 - 12 = 2(0) - 7 (plug in value of x)

5 + 0 - 12 = 0 - 7 (did multiplication)

-7 = -7 (did addition on both sides)

since -7 equals -7, that demonstrates that the value of 0 for x is correct.

trying x = 1

5 + x - 12 = 2x - 7 (original equation)

5 + 1 - 12 = 2(1) - 7 (plug in value of x)

5 + 1 - 12 = 2 - 7 (did multiplication)

-6 = -5 (did addition on both sides)

since -6 is obviously not equal to -5, the value of 1 for x is wrong.

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