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

Answer:

$z= \frac{25-24.6}{\frac{2.8}{\sqrt{66}}}=1.16$

So we want to find this probability:

$P(z$

And using the normal standard distirbution or excel we got:

$P(z$

Step-by-step 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

$P(\bar X$

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

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

If we find the z score for 25 we got:

$z= \frac{25-24.6}{\frac{2.8}{\sqrt{66}}}=1.16$

So we want to find this probability:

$P(z$

And using the normal standard distirbution or excel we got:

$P(z$