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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?

Respuesta :

Answer:

[tex] z= \frac{25-24.6}{\frac{2.8}{\sqrt{66}}}=1.16[/tex]

So we want to find this probability:

[tex] P(z<1.16)[/tex]

And using the normal standard distirbution or excel we got:

[tex] P(z<1.16)=0.877[/tex]

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:

[tex]X \sim N(24.6,2.8)[/tex]  

Where [tex]\mu=24.6[/tex] and [tex]\sigma=2.8[/tex]

We are interested on this probability

[tex]P(\bar X<25)[/tex]

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

[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

If we find the z score for 25 we got:

[tex] z= \frac{25-24.6}{\frac{2.8}{\sqrt{66}}}=1.16[/tex]

So we want to find this probability:

[tex] P(z<1.16)[/tex]

And using the normal standard distirbution or excel we got:

[tex] P(z<1.16)=0.877[/tex]

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