Question

A population is normally distributed with a mean of 100 and a standard deviation of 10. For samples of size 25, what is the probability of randomly sampling and finding a sample mean of 99 or more

Answer 1

Answer:

The value is  $P( \= X \ge 99) = 0.69146$

Step-by-step explanation:

From the question we are told that

    The  mean is  $\mu = 100$

     The standard deviation is  $\sigma = 10$

       The  sample size is  n  =  25

The standard deviation of the sampling distribution is  

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

=>      $\sigma_{\= x} = \frac{10}{\sqrt{25} }$

=>      $\sigma_{\= x} = \frac{10}{5 }$

=>      $\sigma_{\= x} = 2$

Generally the probability of randomly sampling and finding a sample mean of 99 or more is mathematically represented as

        $P( \= X \ge 99) = 1 - P(\= X < 99)$

Here  

       $P(\= X < 99) = P(\frac{\= X - \mu }{\sigma } < \frac{99 - 100 }{2} )$

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

=>     $P(\= X < 99) = P(Z < -0.5 )$

From the z table probability of  (Z <  -0.5  ) is  

     $P(Z < -0.5 ) = 0.30854$

Generally  

      $P( \= X \ge 99) = 1 - 0.30854$

=>    $P( \= X \ge 99) = 0.69146$