Complete question is;
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
A) Calculate the mean (to the nearest whole number) and the standard deviation (to decimals) of the golfer's scores, for both years.
B) What is the primary difference in performance between 2005 and 2006? What improvement,
if any, can be seen in the 2006 scores?
Answer:
A) 2006 mean = 75
2005 mean = 75
2006 standard deviation = 5.2644
2005 standard deviation = 2.0702
B)The primary difference is that variation is higher in the 2006 season than the 2005 season.
Explanation:
A) Mean is the sum of all scores divided by the number of scores.
Thus;
μ_2005 = (73 + 77 + 78 + 76 + 74 + 72 +74 + 76)/8 = 75
Similarly;
μ_2006 = (70 + 69 + 74 + 76 + 84 + 79 + 70 + 78)/8 = 75
Now, variance is calculated by the sum of the square of mean deviations divided by (n - 1)
Thus;
2005 Variance = ((73-75)² + (77-75)² + (78-75)² + (76-75)² + (74-75)² + (72-75)² + (74-75)² + (76-75)²)/(8-1) = 4.2857
2006 Variance = ((70-75)² + (69-75)² + (74-75)² + (76-75)² + (84-75)² + (79-75)² + (70-75)² + (78-75)²)/(8 - 1) = 27.7143
Now, standard deviation is the square root of variance.
Thus;
2005 standard deviation = √4.2857 = 2.0702
2006 standard deviation = √27.7143 = 5.2644
B) The primary difference is that variation is higher in the 2006 season than the 2005 season.
Also,