Historically, the average score of PGA golfers for one round is 66.4 with a standard deviation of 1.79. A random sample of 72 golfers is taken. What is the probability that the sample mean is between 66.14 and 66.4?
Question 9 options:









Solution :
Given that,
mean = = 66.4
standard deviation = = 1.79
_{} = / n = 1.79 / 72 = 0.2110
= P[(66.14  66.4) /0.2110 < (  _{}) / _{} < (66.4  66.4 ) / 0.2110)]
= P(1.23 < Z < 0)
= P(Z < 0)  P(Z < 1.23)
= 0.5  0.1093
= 0.3907
The sample mean will never fall in this range.
Get Answers For Free
Most questions answered within 1 hours.