Question

Historically, the average score of PGA golfers for one round is 66.4 with a standard deviation...

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:

1)

1.0063

2)

The sample mean will never fall in this range.

3)

0.0577

4)

0.6089

5)

0.3911

Homework Answers

Answer #1

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.

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