Answer:
26.76% probability that a randomly chosen golfer's score is above 70 strokes.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 68.2, \sigma = 2.91[/tex]
What is the probability that a randomly chosen golfer's score is above 70 strokes?
This is 1 subtracted by the pvalue of Z when X = 70. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{70 - 68.2}{2.91}[/tex]
[tex]Z = 0.62[/tex]
[tex]Z = 0.62[/tex] has a pvalue of 0.7324.
So there is a 1-0.7324 = 0.2676 = 26.76% probability that a randomly chosen golfer's score is above 70 strokes.
