1. when professional golfer jordan spieth hits his driver, the distance the ball travels can be modeled by

distribution with mean 304 yards and standard deviation 8 yards.

a. un a specific hole, jordan would need to hit the ball at least 290 yards to have a clear second shot that

avoids a large group of trees. what percent of spieth's drives travel at least 290 yards?

b. on another golf hole. spieth has the opportunity to drive the ball onto the green if he hits the ball a distance

in the top 10% of all his drives. how far does the ball have to go?

a. About 96% of Spieth's drives travel at least 290 yards

b. The ball has to travel about 314.24 yards

Step-by-step explanation:

* Let us explain how to solve the problem

- The distance the ball travels can be modeled by  distribution with

  mean 304 yards and standard deviation 8 yards

a.

- Jordan would need to hit the ball at least 290 yards to have a clear

 second shot that avoids a large group of trees

- We need to find the percent of Spieth's drives travel at least

   290 yards

∵ The mean μ = 304 yards

∵ The standard deviation σ = 8 yards

∵ The distance x = 290

∵ P(x ≥ 290) = P(z ≥ z)

- We need to find z score

∵ z-score = (x - μ)/σ

∴ z = = -1.75

* Let us use the normal distribution table to find the corresponding

 area of z = -1.75

∵ P(-z) = 1 - P(z)

∵ P(z ≥ -1.75) = 1 - 0.04006 = 0.95994

∴ P(x ≥ 290) = 0.96 = 96%

* About 96% of Spieth's drives travel at least 290 yards

b.

- On another golf hole, Spieth has the opportunity to drive the ball

  onto the green if he hits the ball a distance  in the top 10% of all

  his drives

- We need to find how far the ball has to travel

- Let us find the z-score from the normal distribution table for the 10%

  to the right or 90% to the left

∵ The area which equivalent to 0.9 ≅ 0.89973

∴ z = 1.28

∵ z-score = (x - μ)/σ

∴ 1.28 =

- Multiply both sides by 8

∴ 10.24 = x - 304

- Add 304 for both sides

∴ x = 314.24 yards

* The ball has to travel about 314.24 yards