Question

The drive distance (in yards) for professional golfers is normally distributed with a mean of 277.21 yards and a standard deviation of 5.879 yards.
a. Between what two values would we expect 95% of Bubba Watson's drive distances to fall?
between_____________and__________ yards
b. 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!)
c. The top 14% of golfers have drive distances of what length?
z = __________
distance = ________ yards

Answer 1

Solution :

Given that ,

mean = = 277.21

standard deviation = = 5.879

a) Using standard normal table,

P( -z < Z < z) = 95%

= P(Z < z) - P(Z <-z ) = 0.95

= 2P(Z < z) - 1 = 0.95

= 2P(Z < z) = 1 + 0.95

= P(Z < z) = 1.95 / 2

= P(Z < z) = 0.975

= P(Z < 1.96) = 0.975

= z  ± 1.96

Using z-score formula,

x = z * +

x = -1.96 * 5.879 + 277.21

x = 265.69

Using z-score formula,

x = z * +

x = 1.96 * 5.879 + 277.21

x = 288.73

The 95 % values between 265.69 and 288.73 yards

b) P(280 < x < 305 ) = P[(280 - 277.21) / 5.879) < (x - ) /  < (305 - 277.21) / 5.879) ]

= P( 0.47 < z < 4.73)

= P(z < 4.73) - P(z < 0.47)

Using z table,

= 1 - 0.6808

= 0.3192

Z1 = 4.73

Z2 = 0.47

The percentage is = 31.92%

c) Using standard normal table,

P(Z > z) = 14%

= 1 - P(Z < z) = 0.14  

= P(Z < z) = 1 - 0.14

= P(Z < z ) = 0.86

= P(Z < 1.08) = 0.86  

z = 1.08

Using z-score formula,

x = z * +

x = 1.08 * 5.879 + 277.21

x = 283.56

z = 1.08

distance = 283.56 yards