Q:

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?

Accepted Solution

A:
Answer:The probability is 0.20Step-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∴ [tex]z=\frac{82-92.1}{3.8}=\frac{-10.1}{3.8}=-2.66[/tex]# Second case∵ z = (x - μ)/σ∵ x = 89∵ μ = 92.1∵ σ = 3.8∴ [tex]z=\frac{89-92.1}{3.8}=\frac{-3.1}{3.8}=-0.82[/tex]- 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