A report by the US government states that 85,5% of citizens have a cell phone. You want to calculate the probability that exactly 36 out of 48 people in your study group have cell phones.

Answer:

0.0214 = 2.14% probability that exactly 36 out of 48 people in your study group have cell phones.

Step-by-step explanation:

For each citizen, there are only two possible outcomes. Either they have a cellphone, or they do not. The probability of a citizen having a cellphone is independent of any other citizen. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

A report by the US government states that 85,5% of citizens have a cell phone.

This means that [tex]p = 0.855[/tex]

You want to calculate the probability that exactly 36 out of 48 people in your study group have cell phones.

This is [tex]P(X = 36)[/tex] when [tex]n = 48[/tex]. So

Answer:0.0214 = 2.14% probability that exactly 36 out of 48 people in your study group have cell phones.

Step-by-step explanation:For each citizen, there are only two possible outcomes. Either they have a cellphone, or they do not. The probability of a citizen having a cellphone is independent of any other citizen. This means that the binomial probability distribution is used to solve this question.

Binomial probability distributionThe binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

A report by the US government states that 85,5% of citizens have a cell phone.This means that [tex]p = 0.855[/tex]

You want to calculate the probability that exactly 36 out of 48 people in your study group have cell phones.This is [tex]P(X = 36)[/tex] when [tex]n = 48[/tex]. So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 36) = C_{48,36}.(0.855)^{36}.(1-0.855)^{12} = 0.0214[/tex]

0.0214 = 2.14% probability that exactly 36 out of 48 people in your study group have cell phones.