Question

Suppose the average driving distance for last year's Player's Champion Golf Tournament in Ponte Vedra, FL, was 292.5 yards with a standard deviation of 14.2 yards. A random sample of 60 drives was selected from a total of 4,244 drives that were hit during this tournament. What is the probability that the sample average was 289 yards or less?

Answer 1

The Probability that the sample average was 289 yards or less

P(x⁻≤ 289) = P( Z≤ -1.909) = 0.0287

Explanation:

step(i):-

Mean of the Population = 292.5 yards

Standard deviation of the Population = 14.2 yards

sample size 'n' =60 drives

N = 4244 drives

Step(ii):-

Let X⁻ be random sample average

`Z = (x^(-) -mean)/((S.D)/(√(n) ) )`

Let X⁻ = 289

`Z = (289 -292.5)/((14,2)/(√(60) ) )`

Z = - 1.909

The Probability that the sample average was 289 yards or less

P(x⁻≤ 289) = P( Z≤ -1.909)

= 0.5 -A(1.909)

= 0.5 -0.4713

= 0.0287

Conclusion:-

The Probability that the sample average was 289 yards or less = 0.0287