Question

The average score of all golfers for a particular course has a mean of 75 and a standard deviation of 4.5. Suppose 81 golfers played the course today. Find the probability that the average score of the 81 golfers exceeded 76. Group of answer choices

Answer 1

Answer:

The probability that the average score of the 81 golfers exceeded 76

P(x⁻≤ 76) = 0.9772    

Step-by-step explanation:

Step(i):-

Given mean of the Population (μ) = 75

Given standard deviation of the Population (σ) = 4.5

Given size 'n' =81

Let 'X' be the random variable in Normal distribution

$Z = \frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{76-75}{\frac{4.5}{\sqrt{81} } }$

Z = 2

Step(ii):-

The probability that the average score of the 81 golfers exceeded 76

P(x⁻≤ 76) = P( Z≤ 2)

               = 1 - P(Z>2)

              = 1 - ( 0.5 - A(2))

              = 0.5 + A(2)

            = 0.5 +0.4772

            = 0.9772

The probability that the average score of the 81 golfers exceeded 76

P(x⁻≤ 76) = 0.9772