Question

The drive distance (in yards) for professional golfers is normally distributed with a mean of 318.36 yards and a standard deviation of 10.435 yards.

1. Between what two values would we expect 95% of Bubba Watson's drive distances to fall?
   • between_____ and________ yards
2. What percentage of golfers has a drive distance between 280 and 305 yards?
   • z1 = _____
   • z2 = _____
   • percentage = _________% (make sure to change your decimal to a percentage!)
3. The top 10% of golfers have drive distances of what length?
   • z = _____
   • distance =___________ yards

Answer 1

Part a)

X ~ N ( µ = 318.36 , σ = 10.435 )
P ( a < X < b ) = 0.95
Dividing the area 0.95 in two parts we get 0.95/2 = 0.475
since 0.5 area in normal curve is above and below the mean
Area below the mean is a = 0.5 - 0.475
Area above the mean is b = 0.5 + 0.475
Looking for the probability 0.025 in standard normal table to calculate critical value Z = -1.96
Looking for the probability 0.975 in standard normal table to calculate critical value Z = 1.96
Z = ( X - µ ) / σ
-1.96 = ( X - 318.36 ) / 10.435
a = 297.9074
1.96 = ( X - 318.36 ) / 10.435
b = 338.8126
P ( 297.9074 < X < 338.8126 ) = 0.95

Part b)

X ~ N ( µ = 318.36 , σ = 10.435 )
P ( 280 < X < 305 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 280 - 318.36 ) / 10.435
Z1 = -3.68
Z = ( 305 - 318.36 ) / 10.435
Z2 = -1.28
P ( -3.68 < Z < -1.28 )
P ( 280 < X < 305 ) = P ( Z < -1.28 ) - P ( Z < -3.68 )
P ( 280 < X < 305 ) = 0.1002 - 0.0001
P ( 280 < X < 305 ) = 0.1001

percentage = 10.01_%

Part c)

X ~ N ( µ = 318.36 , σ = 10.435 )
P ( X > x ) = 1 - P ( X < x ) = 1 - 0.1 = 0.9
To find the value of x
Looking for the probability 0.9 in standard normal table to calculate critical value Z = 1.28

Z = 1.28

Z = ( X - µ ) / σ
1.2816 = ( X - 318.36 ) / 10.435
X = 331.7335
P ( X > 331.7335 ) = 0.1