Answer :
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 = [tex]\frac{290-304}{8}[/tex] = -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 = [tex]\frac{x-304}{8}[/tex]
- 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