Question

please answer all parts.

The drive distance (in yards) for professional golfers is normally distributed with a mean of 287.47 yards and a standard deviation of 12.887 yards. a. Between what two values would we expect 95% of Bubba Watson's drive distances to fall? o betweern and yards b. What percentage of golfers has a drive distance between 280 and 305 yards? percentage- % (make sure to change your decimal to a percentage) o C. The top 14% of golfers have drive distances of what length? o distance yards

Answer 1

A) P(X < x) = 0.025

Or, P((X - $\mu$)/$\sigma$ < (x - $\mu$)/$\sigma$) = 0.025

Or, P(Z < z) = 0.025

Or, z = -1.96

Or, (x - 287.47)/12.887 = -1. 96

Or, x = -1.96 * 12.887 + 287.47

Or, x = 262.211

P(X > x) = 0.025

Or, P(Z > z) = 0.025

Or, P(Z < z) = 0.975

Or, z = 1.96

Or, (x - 287.47)/12.887 = 1.96

Or, x = 1.96 * 12.887 + 287.47

Or, x = 312.729

Between 262.211 and 312.729 yards.

B) P(280 < X < 305)

= P((280 - $\mu$)/$\sigma$ < (X - $\mu$)/$\sigma$ < (305 - $\mu$)/$\sigma$)

= P((280 - 287.47)/12.887 < Z < (305 - 287.47)/12.887)

= P(-0.58 < Z < 1.36)

= P(Z < 1.36) - P(Z < -0.58)

= 0.9131 - 0.2810

= 0.6321 = 63.21%

C) P(X > x) = 0.14

Or, P(Z > z) = 0.14

Or, P(Z < z) = 0.86

Or, z = 1.08

Or, (x - 287.47)/12.887 = 1.08

Or, x = 1.08 * 12.887 + 287.47

Or, x = 301.388