Question

You find that you receive on average about 3 pieces of junk mail per week. Find the probability that the number of pieces of junk mail you receive next week will be (a) none at all. (b) exactly four. (c) no more than two. (d) more than two g

Answer 1

a) 0.0498 = 4.98% probability that the number of pieces of junk mail you receive next week will be none at all.

b) 0.1680 = 16.80% probability that the number of pieces of junk mail you receive next week will be exactly four.

c) 0.4232 = 42.32% probability that the number of pieces of junk mail you receive next week will be no more than two.

d) 0.5768 = 57.68% probability that the number of pieces of junk mail you receive next week will be more than two.

Explanation:

We have only the mean, which means that we use the Poisson distribution to solve this question.

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) = (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.

You find that you receive on average about 3 pieces of junk mail per week.

This means that
`\mu = 3`

(a) none at all.

This is P(X = 0).

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

`P(X = 0) = (e^(-3)*3^(0))/((0)!) = 0.0498`

0.0498 = 4.98% probability that the number of pieces of junk mail you receive next week will be none at all.

(b) exactly four.

This is P(X = 4).

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

`P(X = 4) = (e^(-3)*3^(4))/((4)!) = 0.1680`

0.1680 = 16.80% probability that the number of pieces of junk mail you receive next week will be exactly four.

(c) no more than two.

This is:

`P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)`

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

`P(X = 0) = (e^(-3)*3^(0))/((0)!) = 0.0498`

`P(X = 1) = (e^(-3)*3^(1))/((1)!) = 0.1494`

`P(X = 2) = (e^(-3)*3^(2))/((2)!) = 0.224`

`P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0498 + 0.1494 + 0.224 = 0.4232`

0.4232 = 42.32% probability that the number of pieces of junk mail you receive next week will be no more than two.

(d) more than two

`P(X > 2) = 1 - P(X \leq 2) = 1 - 0.4232 = 0.5768`

0.5768 = 57.68% probability that the number of pieces of junk mail you receive next week will be more than two.