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 :

Answer:

The Probability that the sample average was 289 yards or less

P(x⁻≤ 289) = P( Z≤ -1.909) =  0.0287

Step-by-step 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

[tex]Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }[/tex]

Let  X⁻ = 289

[tex]Z = \frac{289 -292.5}{\frac{14,2}{\sqrt{60} } }[/tex]

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

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