Suppose the velocities of golf swings for amateur golfers are normally distributed with a mean of 95 mph and a standard deviation of 3.9 mph. what is the difference in velocities between a golfer whose z-score is 0 and another golfer whose z-score is −1? 1.95 mph 2.9 mph 3.9 mph 7.8 mph

To get the difference between the golfers we proceed as follows;
the z-score is given by:
z=(x-μ)/σ
the score for golfer with z-score of 0 will be:

μ- mean
σ- standard deviation
0=(x-95)/3.9
the value of x will be
0=x-95
x=95

the score for the golfer with z-score of -1 is:
-1=(x-95)/3.9
-3.9=x-95
x=-3.9+95
x=91.1
the difference in the score of the golfers will be:
95-91.1
=3.9
the answer is C] 3.9 mph

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