The probability that the average score of the golfers exceeded 62 = 0.0228
Given the mean score of all golfers, μ = 72
and the Standard Deviation, σ = 4
Number of golf players, n = 64
We have to find the probability that the average score of the golfers exceed 73. i.e., x = 73
This follows the Normal Distribution.
Hence the z-score, Z = (x - μ)/(σ/√n)
= (73 - 72)√64/4
= 2
Probability that the average score of the golfers exceeded 73 = P(Z > 2)
= 0.0228 [from the z-tables]
The question is incomplete. Find the complete question below:
t\The average score of all golfers for a particular course has a mean of 72 and a standard deviation of 4. Suppose 64 golfers played the course today. Find the probability that the average score of the golfers exceeded 73 .
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