Question

Historically, the average score of PGA golfers for one round is 65.9 with a standard deviation of 3.56. A random sample of 104 golfers is taken. What is the probability that the sample mean is greater than 65.6?

Answer 1

X : score of PGA golfers for one round

Average score of PGA golfers for one round: = 65.9 and standard deviation: =3.56

Sample size : n = 104

Sample mean follows normal distribution with mean = 65.9 and standard deviation :

  probability that the sample mean is greater than 65.6 = P( > 65.6)

P( > 65.6) = 1-P(65.6)

Z-score for 65.6 = (65.6 - 65.9)/0.3491 = -0.86

From standard normal tables,

P(Z-0.86) = 0.1949

P(65.6)=P(Z-0.86) = 0.1949

P( > 65.6) = 1-P(65.6) = 1-0.1949=0.8051

Probability that the sample mean is greater than 65.6 = P( > 65.6) = 0.8051

Probability that the sample mean is greater than 65.6 = 0.8051