The average score of all golfers for a particular course has

The average score of all golfers for a particular course has a mean of 60 and a standard deviation of 3.5. Suppose 49 golfers played the course today. Find the probability that the average score of the 49 golfers exceeded 61. Roun to four decimal places.

A. .1293

B. .3707

C. .4772

D. .0228

Solution

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    61      
u = mean =    60      
n = sample size =    49      
s = standard deviation =    3.5      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    2      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2   ) =    0.022750132 = 0.0228 [OPTION D]