Question

The mean number of words per minute (WPM) typed by a speed typist is 149149 with a standard deviation of 1414 WPM. What is the probability that the sample mean would be greater than 147.8147.8 WPM if 8888 speed typists are randomly selected

Answer 1

Answer:

The probability is  $P(\= X > x ) = 0.78814$

Step-by-step explanation:

From the question we are given that

     The population  mean is  $\mu = 149$

     The standard deviation is  $\sigma = 14$

     The random number    $x = 147.81$

      The sample  size is  $n = 88$

The probability that  the sample mean would be greater than $P(\= X > x ) = P( \frac{ \= x - \mu }{\sigma_{\= x} } > \frac{ x - \mu }{\sigma_{\= x} } )$

Generally the z- score of this normal distribution is mathematically represented as

       $Z = \frac{ \= x - \mu }{\sigma_{\= x} }$

Now

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

substituting values

         $\sigma_{\= x } = \frac{14 }{\sqrt{88} }$

         $\sigma_{\= x } = 1.492$

Which implies that

         $P(\= X > x ) = P( Z > \frac{ 147.81 - 149 }{ 1.492} )$

        $P(\= X > x ) = P( Z > -0.80 )$

Now from the z-table  the  probability is  found to be

        $P(\= X > x ) = 0.78814$