A PGA tournament organizer is attempting to determine whether hole (pin) placeme

Question

A PGA tournament organizer is attempting to determine whether hole (pin) placement has a significant impact on the average number of strokes for the 13th hole on a given golf course. Historically, the pin has been placed in the front right corner of the green, and the historical mean number of strokes for the hole has been 4.25, with a standard deviation of 1.6 strokes. On a particular day during the most recent golf tournament, the organizer placed the hole (pin) in the back left corner of the green. 64 golfers played the hole with the new placement on that day. Determine the probability of the sample average number of strokes less than 4.75.

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean ( u ) = 4.25
standard Deviation ( sd )= 1.6/ Sqrt ( 64 ) =0.2
sample size (n) = 64
P(X < 4.75) = (4.75-4.25)/1.6/ Sqrt ( 64 )
= 0.5/0.2= 2.5
= P ( Z <2.5) From Standard NOrmal Table
= 0.9938

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