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