in a class of 500 students the mean exam grade was 86 with a

in a class of 500 students the mean exam grade was 86 with a standard deviation of 10.1. scores were skewed to the right. If a random sample of 30 students were drawn from this population what is the probablity that their sample mean would be greater than 90 points?

a. .142

b. .4

c. .9858

d. .0339

Solution

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    90      
u = mean =    86      
n = sample size =    30      
s = standard deviation =    10.1      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    2.169198248      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.169198248   ) =    0.015033818 [ANSWER]

This answer is not here. However, its complement is here, 0.9858. Please report this problem, as the correct answer is really the one above.