Question

If the mean GPA among students is 3.25 with a standard deviation of 0.75, what is the probability that a random sample of 300 students will have a mean GPA greater than 3.30

Answer 1

Answer:

The value is  $P(X > 3.30) = 0.12405$

Step-by-step explanation:

From the question we are told that

   The mean GPA is  $\mu = 3.25$

   The standard deviation is $\sigma = 0.75$

    The sample size is  n  =  300

Generally the standard error of mean is mathematically represented as

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

=>   $\sigma_{\= x} = \frac{0.75}{\sqrt{300} }$

=>   $\sigma_{\= x} = 0.0433$

Generally the  probability that a random sample of 300 students will have a mean GPA greater than 3.30 is mathematically represented as

     $P(X > 3.30) = P(\frac{X - \mu}{\sigma_{\= x}} > \frac{3.30 -3.25}{ 0.0433} )$

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

    $P(X > 3.30) = P(Z> 1.155 )$

From the z table the probability of  (Z >  1.155 ) is

     $P(Z> 1.155 ) = 0.12405$

   $P(X > 3.30) = 0.12405$