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2 years ago

Anyone know how to do this? It missing one bit of info the total amount of pigs is 450

Mathematics

1 answer:

Tju [1.3M]2 years ago

6 0

Answer:

Step-by-step explanation:

Number of pigs with John = x

Number of pigs with Taylor = x - 7

Number of pigs with Mark = 2x + 6

Total pigs = 459

x + x - 7 + 2x + 6 = 459

x+ x + 2x - 7 + 6 = 459

4x - 1 = 459

4x = 459 + 1

4x = 460

x = 460/4

x = 115

Number of pigs with John = 115

Number of pigs with Taylor = 115 - 7 = 108

Number of pigs with Mark = 2*115 + 6 = 230 + 6 = 236

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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 $P(X \geq 1)$

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

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

8 0

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