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91Ó°ÊÓ

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 \(74 \quad 78 \quad 79 \quad 77 \quad 75 \quad 73 \quad 75 \quad 77\) 2006 Season \(71 \quad 70 \quad 75 \quad 77 \quad 85 \quad 80 \quad 71 \quad 79\) a. Use the mean and standard deviation to evaluate the golfer's performance over the two-year period. b. What is the primary difference in performance between 2005 and \(2006 ?\) What improvement, if any, can be seen in the 2006 scores?

Short Answer

Expert verified
In 2006, the mean score remained 76, but greater score variability indicates inconsistency, though some lower individual scores suggest potential improvement.

Step by step solution

01

Calculate the Mean for 2005

To find the mean for the 2005 scores, sum all the scores and divide by the number of scores. The scores for 2005 are: 74, 78, 79, 77, 75, 73, 75, 77. The sum is \(74 + 78 + 79 + 77 + 75 + 73 + 75 + 77 = 608\). There are 8 scores, so the mean is \(\frac{608}{8} = 76\).
02

Calculate the Mean for 2006

For 2006, the scores are: 71, 70, 75, 77, 85, 80, 71, 79. The sum is \(71 + 70 + 75 + 77 + 85 + 80 + 71 + 79 = 608\). Again, there are 8 scores, so the mean is \(\frac{608}{8} = 76\).
03

Calculate the Standard Deviation for 2005

First, find the variance for the 2005 scores by calculating the average of the squared differences from the mean. The squared differences from 76 are: \((-2)^2,\ 2^2,\ 3^2,\ 1^2,\ (-1)^2,\ (-3)^2,\ (-1)^2,\ 1^2\) which are: 4, 4, 9, 1, 1, 9, 1, 1. The sum of these squared differences is 30. The variance is \(\frac{30}{8} = 3.75\), so the standard deviation is \(\sqrt{3.75} \approx 1.94\).
04

Calculate the Standard Deviation for 2006

For the 2006 scores, find the squared differences from the mean (76): \((-5)^2,\ (-6)^2,\ (-1)^2,\ 1^2,\ 9^2,\ 4^2,\ (-5)^2,\ 3^2\), which are: 25, 36, 1, 1, 81, 16, 25, 9. The sum is 194. The variance is \(\frac{194}{8} = 24.25\), so the standard deviation is \(\sqrt{24.25} \approx 4.92\).
05

Compare Performance

Comparing the means, both years have the same average score (76). However, the standard deviation for 2006 (4.92) is much higher than for 2005 (1.94), indicating more variability in the 2006 scores. The primary difference is the increased inconsistency in performance in 2006.
06

Analyze Improvements

Although there is more variability in 2006, some improvement can be observed through the lower individual scores achieved, such as 70 and 71, which were not present in 2005, suggesting the golfer did occasionally perform better than any score in 2005.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean Calculation
The mean is a measure of the average score, offering insight into overall performance across a dataset. In this exercise, we have two separate sets of scores for an amateur golfer, spanning the years 2005 and 2006. For each year, we first sum the scores, and then divide by the total number of scores.

For 2005, the golfer had scores of: 74, 78, 79, 77, 75, 73, 75, and 77.
  • Sum of scores: 608
  • Number of scores: 8
  • Mean: \( \frac{608}{8} = 76 \)
Similarly, for 2006, the scores were: 71, 70, 75, 77, 85, 80, 71, and 79.
  • Sum of scores: 608
  • Number of scores: 8
  • Mean: \( \frac{608}{8} = 76 \)
Both years exhibit the same mean score of 76, which indicates that, on average, the golfer's performance was consistent across these two years.
Standard Deviation
Standard deviation is a crucial statistical measure used to understand the variability or spread of a set of values in relation to the mean. A higher standard deviation indicates a wider spread of scores from the mean, reflecting inconsistent performance.

To calculate the standard deviation, we first compute the variance, which is the average of the squared differences between each score and the mean.

For 2005:
  • Mean: 76
  • Squared differences: 4, 4, 9, 1, 1, 9, 1, 1
  • Variance: \( \frac{30}{8} = 3.75 \)
  • Standard Deviation: \( \sqrt{3.75} \approx 1.94 \)

For 2006:
  • Mean: 76
  • Squared differences: 25, 36, 1, 1, 81, 16, 25, 9
  • Variance: \( \frac{194}{8} = 24.25 \)
  • Standard Deviation: \( \sqrt{24.25} \approx 4.92 \)
While both years have the same mean, the standard deviation is dramatically higher in 2006. This indicates that the golfer’s scores were much more varied in that year, showcasing inconsistency.
Variance Analysis
Variance analysis provides insight into the degree of fluctuation in scores over time. It is an essential concept in descriptive statistics that highlights how much scores diverge from the mean. Here, we are using variance to understand the differences in the golfer's performance between 2005 and 2006.

For 2005, the variance is 3.75. This showcases smaller discrepancies between individual scores and the mean score of 76. The golfer's performance was steady, with minimal fluctuation between high and low scores.

In contrast, the 2006 variance of 24.25 reflects greater fluctuation in scores. The large variance indicates that the golfer had rounds of both excellent and subpar performances, as evidenced by scores like 70 and 85 in 2006. This high variance points to moments of both significant improvement and inconsistency when compared to 2005.

By examining variance alongside the mean and standard deviation, it is clear that while both years shared the same average performance, 2006 featured a notable increase in performance variability.

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