Question

Scores turned in by an amateur golfer at the Bonita Fairways Golf Course in Bonita Springs, Florida, during and are as follows: 2018 Season: 74 78 79 77 75 73 75 77 2019 Season: 71 70 75 77 85 80 71 79 a. Use the mean (to the nearest whole number) and standard deviation (to decimals) to evaluate the golfer's performance over the two-year period. Mean Standard deviation Mean Standard deviation b. What is the primary difference in performance between and

Answer 1

Answer:

Step-by-step explanation:

The mean for 2018 scores is:

$M ean ( \overline X) = \dfrac{\sum x_i}{n}$

$M ean ( \overline X) = \dfrac{74 + 78+79+77+75+73+75+77}{8}$

$M ean ( \overline X) = \dfrac{608} {8}$

$\mathbf{M ean ( \overline X) = 76}$

The standard deviation for 2018 scores is:

$s = \sqrt{\dfrac{\sum ( x_i - \bar x)^2}{n- 1} }$

$s = \sqrt{\dfrac{ ( 74 -76)^2+( 78 -76)^2+( 79 -76)^2+...+( 73 -76)^2+( 75 -76)^2+( 77 -76)^2}{8- 1} }$

$s = \sqrt{\dfrac{30}{7} }$

s = 2.07

The mean and standard deviation for the year 2018 scores is:

                                     2018

Mean                                 76

Standard deviation       2.07

The mean for 2019 scores is:

$M ean ( \overline X) = \dfrac{\sum x_i}{n}$

$M ean ( \overline X) = \dfrac{71 + 70+75+77+85+80+71+79}{8}$

$M ean ( \overline X) = \dfrac{608} {8}$

$\mathbf{M ean ( \overline X) = 76}$

The standard deviation for 2019 scores is:

$s = \sqrt{\dfrac{\sum ( x_i - \bar x)^2}{n- 1} }$

$s = \sqrt{\dfrac{ ( 71 -76)^2+( 70 -76)^2+( 75-76)^2+...+( 80 -76)^2+( 71 -76)^2+( 79 -76)^2}{8- 1} }$

$s = \sqrt{\dfrac{194}{7} }$

s = 5.26

The mean and standard deviation for the year 2019 scores is:

                                     2019

Mean                                 76

Standard deviation       5.26

Difference: Both the two years have the same mean but the 2019 scores have higher variation as compared to 2018 scores.

Thus, the variation in scores was higher in 2019.