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?

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

Let  X⁻ = 289

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

3-4

step-by-step explanation:

the answer is 3-4 because both 45 and 60 has a multiple of 15.

45/3=15

60/4=15

therefore the odds against selecting a red is 4-3

2160ft3

step-by-step explanation: