Question

1. When professional golfer Jordan Spieth hits his driver, the distance the ball travels can be modeled by

Answer 1

Answer:

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 = $\frac{290-304}{8}$ = -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 = $\frac{x-304}{8}$

- 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