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 24.7 years. The standard deviation of the distribution is 2.8 years. For a random sample of 60 men, what is the likelihood that the age when they were first married is less than 25.2 years

Answer 1

Answer:

The likelihood is $P(X < 25.2) = 0.91668$

Step-by-step explanation:

From the question we are told that

     The population mean is  $\mu = 24.7 \ years$

      The standard deviation is  $\sigma = 2.8 \ years$

      The sample size is $n = 60 \ men$

       The  consider random value is  x =  25.2  years

Given that mean age is normally distributed, the likelihood that the age when they were first married is less than x is mathematically represented as

          $P(X < x) = P( \frac{X - \mu }{\sigma_{\= x }} < \frac{x - \mu }{\sigma_{\= x }} )$

Generally  $\frac{X - \mu }{ \sigma_{\= x}} = Z (The \ standardized \ value \ of \ X )$

So

      $P(X < x) = P(Z< \frac{x - \mu }{\sigma_{\= x }} )$

Where $\sigma_{\= x }$ is the standard error of the sample mean which mathematically evaluated as

       $\sigma_{\= x } = \frac{ \sigma }{\sqrt{n} }$

substituting values

      $\sigma_{\= x } = \frac{ 2.8 }{\sqrt{ 60 } }$

      $\sigma_{\= x } = 0.3615$

So  

     $P(X < 25.2) = P(Z< \frac{ 25.2 - 24.7 }{0.3615} )$

     $P(X < 25.2) = P(Z< 1.3831 )$

From z-table the value for  P(Z<    1.3831  ) is $P(Z < 1.3831 ) = 0.91668$

  So

        $P(X < 25.2) = 0.91668$