The Mean and the Standard DeviationLast year, golfers at Smith Ranch golf course had a mean score of 41 and a standard deviation of 2 on the first nine holes of the course, and a mean of 38 and a...

The Mean and the Standard Deviation

Last year, golfers at Smith Ranch golf course had a mean score of 41 and a standard deviation of 2 on the first nine holes of the course, and a mean of 38 and a standard deviation of 1.8 on the last nine holes.  One afternoon, Nathan scored 39 on the first nine holes and 37 on the last nine.  On which half of the course did Nathan golf better, relative to the other golfers?  Remember that in golf it is better to have a low score.  Explain.

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embizze | High School Teacher | (Level 1) Educator Emeritus

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We can compare Nathan's scores by converting them to standard normal scores using `z=(x-bar(x))/s` where `bar(x)` is the mean and s is the standard deviation.

On the first nine holes Nathan's score of 39 converts to `z=(39-41)/2=-1`

On the second nine holes Nathan's score of 37 converts to `z=(37-38)/1.8=-5/9=-.bar(5)`

So on teh first 9 holes Nathan's score is a full standard deviation below the mean, while on the second nine holes his score was only `5/9` of a standard deviation below the mean.

So Nathan did better on the 1st nine holes.

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