Question

Bob's golf score at his local course follows the normal distribution with a mean of 92.1 and a standard deviation of 3.8. What is the probability that the score on his next round of golf will be between 82 and​ 89?

Answer 1

Answer:

The probability is 0.20

Step-by-step explanation:

a) Lets revise how to find the z-score

- The rule the z-score is z = (x - μ)/σ , where

# x is the score

# μ is the mean

# σ is the standard deviation

* Lets solve the problem

- Bob's golf score at his local course follows the normal distribution

- The mean is 92.1

- The standard deviation is 3.8

- The score on his next round of golf will be between 82 and​ 89

- Lets find the z-score for each case

# First case

∵ z = (x - μ)/σ

∵ x = 82

∵ μ = 92.1

∵ σ = 3.8

∴ $z=\frac{82-92.1}{3.8}=\frac{-10.1}{3.8}=-2.66$

# Second case

∵ z = (x - μ)/σ

∵ x = 89

∵ μ = 92.1

∵ σ = 3.8

∴ $z=\frac{89-92.1}{3.8}=\frac{-3.1}{3.8}=-0.82$

- To find the probability that the score on his next round of golf will

  be between 82 and​ 89 use the table of the normal distribution

∵ P(82 < X < 89) = P(-2.66 < z < -0.82)

∵ A z-score of -2.66 the value is 0.00391

∵ A z-score of -0.82 the value is 0.20611

∴ P(-2.66 < z < -0.82) = 0.20611 - 0.00391 = 0.2022

* The probability is 0.20