Question

In: Statistics and Probability

Scores turned in by an amateur golfer at the Bonita Fairways Golf Course in Bonita Springs,...

Scores turned in by an amateur golfer at the Bonita Fairways Golf Course in Bonita Springs, Florida, during 2005 and 2006 are as follows: 2005 Season 73 77 78 76 74 72 74 76 2006 Season 70 69 74 76 84 79 70 78 Calculate the mean (0 decimals) and the standard deviation (to 2 decimals) of the golfer's scores, for both years. 2005 Mean Standard deviation 2006 Mean Standard deviation What is the primary difference in performance between 2005 and 2006? What improvement, if any, can be seen in the 2006 scores?

Solutions

Expert Solution

By observing the scores turned in by the amateur golfer during 2005 and 2006 it can be be said that the average scores he obtained is exactly same which is 75, while it can also be seen that the golfer has a higher sample standard deviation in 2006 in comparison of what he scored in 2005, a higher standard deviation corresponds lower consistency which can also be interpreted as the the golfer scoring good in some games (above average) and also at the same time scoring bad ( below average in few other games).


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