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?

A 1.95 mph

B 2.9mph

C 3.9mph

D 7.8mph

Nuggetguy Mar 27, 2019

#1**+2 **

z score = [ raw score - mean ] / std deviation

So...for the fiirst

[ raw score - 95 ] / 3.9 = 0 multiply throughh by 3.9

raw score- 95 = 0

Raw score = 95 = 95 mph

For the second, we have

[ raw score - 95 ] / 3.9 = -1 multiply through by 3.9

raw score - 95 = -3.9 add 95 to both sides

raw score = 91.1 = 91.1 mph

So....the difference in speeds is

95 - 91.1 =

3.9 mph

CPhill Mar 27, 2019