Question

In the United States, the mean age of men when they marry for the first time follows the normal distribution with a mean of 24.6 years. The standard deviation of the distribution is 2.8 years. For a random sample of 66 men, what is the likelihood that the age when they were first married is less than 25 years?

Answer 1

`z= (25-24.6)/((2.8)/(√(66)))=1.16`

So we want to find this probability:

`P(z<1.16)`

And using the normal standard distirbution or excel we got:

`P(z<1.16)=0.877`

Explanation:

Let X the random variable that represent the age of men married of a population, and for this case we know the distribution for X is given by:

`X \sim N(24.6,2.8)`

Where
`\mu=24.6` and
`\sigma=2.8`

We are interested on this probability

`P(\bar X<25)`

The sample size is n =66. We can use the z score to solve this problem:

`z=(x-\mu)/((\sigma)/(√(n)))`

If we find the z score for 25 we got:

`z= (25-24.6)/((2.8)/(√(66)))=1.16`

So we want to find this probability:

`P(z<1.16)`

And using the normal standard distirbution or excel we got:

`P(z<1.16)=0.877`