Question

The test results for a class with 24 students are normally distributed with a mean of 82 and a standard deviation of 8.9. What is the probability that a randomly chosen student earned a B or higher on the test (score of 80 or higher)?

Answer 1

Answer:

The value is  $P( X > 80 ) = 0.58901$

Step-by-step explanation:

From the question we are told that

    The sample size is  n  =  24

    The mean is  $\mu = 82$  

    The standard deviation is  $\sigma = 8.9$

   

Generally the  probability that a randomly chosen student earned a B or higher on the test (score of 80 or higher )  is mathematically represented as

         $P( X > 80 ) = P( \frac{ X - \mu }{\sigma } > \frac{ 80 - 82 }{ 8.9 } )$

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

 =>  $P( X > 80 ) = P(Z > -0.225 )$

From the z table  the area under the normal curve to the right  corresponding to    -0.225  is  

     $P(Z > -0.225 ) = 0.58901$

=>   $P( X > 80 ) = 0.58901$