Question

during 2017 and 2018 are as follows 2017 Season 2018 Season 70 60 (a) Use the mean and standard deviation to evaluate the golfer's performance over the two-year period. Compute the mean and standard deviation for the 2017 season. (Round the standard deviation to two decimal places.) mean sample standard deviation Compute the mean and standard deviation for the 2016 season (Round the standard deviation to two decimal places) mean sample standard deviation (6) What is the primary difference in performance between 2017 and 2018? The mean score in 2018 - Se e the mean score in 2017. The standard deviation for the scores in 2018 is What improvement, any can be seen in the 2018 Co the standard deviation for scores in 2017. ? In 2018, Sate of the ght scores were better than any score reported for 2017

Answer 1

2017 season data (x-x)2 -2 | 73 77 78 76 14 72 74 76 0 600 n=8 Sample mean = 2* = 75 wn |(x-x)2 Standard deviation =) www.v n -1 = 2.0702

2018 season data (x-x) (X-X) 25 70 -5 -6 69 74 76 84 81 79 70 16 25 3 9 6000 194 Sample mean = <x = 75 wen (x - 2)2 Standard deviation = www. V n-1 = 5.2644

a) Mean and standard deviation for the data 2017 Sample mean = 2x = 75 wwwww n [(x- x)² Standard deviation = . -= 2.07 www v n-1 Mean and standard deviation for the data 2018 Sample mean = 2y = 75 n (x - 2)2 Standard deviation = wwwww V n-1 = 5.26 b) The mean score in 2018 is same as the mean score in 2017the standard deviation for the score in 2018 is more than the standard deviation in 2017 in 2018, Standard deviation of the eight scores were better than any score reported for 2017