We want to compare the grades on an exam in an intro statist

We want to compare the grades on an exam in an intro statistics course for biology majors versus public health majors. Suppose 72% percent of the class are biology majors. while the remaining 28% are public health majors. The biology majors earned grades that are normally distributed, with a mean of 84 and a standard deviation of 10. The public health majors earned grades that are also normally distributed, with a mean of 87 and a standard deviation of 14. What proportion of the class got an A on the test (scored at least 90)?

Solution

For biology majors:

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 =    84      
          
s = standard deviation =    10      
          
Thus,          
          
z = (x - u) / s =    0.6      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.6   ) =    0.274253118

************

For public health majors:

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 =    87      
          
s = standard deviation =    14      
          
Thus,          
          
z = (x - u) / s =    0.214285714      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.214285714   ) =    0.415162129

***************

Thus,

P(at least 90) = P(bio) P(at least 90|bio) + P(pub) P(at least 90|pub)

= 0.72(0.274253118) + 0.28(0.415162129)

= 0.313707641 [ANSWER]