Question

The average score of all golfers for a particular course has a mean of 75 and a standard deviation of 3.5. Suppose 49 golfers played the course today. Find the probability that the average score of the 49 golfers exceeded 76.

Answer 1

Answer:

The probability that the average score of the 49 golfers exceeded 76 is 0.0228

Step-by-step explanation:

To find the probability that the average score of the 49 golfers exceeded 76, we need to calculate z-score of the score 76.

Z-score can be calculated as follows:

z=$\frac{X-M}{\frac{s}{\sqrt{N} } }$ where

• X=76
• M is the mean score of all golfers (75)
• s is the standard deviation of the scores (3.5)
• N is the sample size (49)

then z(76)=$\frac{76-75}{\frac{3.5}{\sqrt{49} } }$ = 2

By looking at the z-table we find that the probability of z-values of scores smaller or equal to 2 is Pr(Z<2)=0.9772. Therefore

The probability that the average score of the 49 golfers exceeded 76 is 1-0.9772=0.0228