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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

 Mar 27, 2019
 #1
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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

 

 

cool cool cool

 Mar 27, 2019
 #2
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+2

A z-score of 0 is 0 s.d. from the mean...so it eqauls the mean of 95

A z-score of -1 is one standard deviation below the mean   so   -3.9 below the mean ....and below the other golfer.     3.9 mph

 Mar 27, 2019

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