Question

The average score of all golfers for a particular course has a mean of 61 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 62.
A) .3707

B) .4772

C) .1293

D) .0228

Answer 1

Answer:

The probability that the average score of the 49 golfers exceeded 62 is 0.3897

Step-by-step explanation:

The average score of all golfers for a particular course has a mean of 61 and a standard deviation of 3.5

$\mu = 61$

$\sigma = 3.5$

We are supposed to find he probability that the average score of the 49 golfers exceeded 62.

Formula : $Z=\frac{x-\mu}{\sigma}$

$Z=\frac{62-61}{3.5}$

$Z=0.285$

Refer the z table for p value

p value = 0.6103

P(x>62)=1-P(x<62)=1-0.6103=0.3897

Hence the probability that the average score of the 49 golfers exceeded 62 is 0.3897