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Scores turned in by an amateur golfer at the Bonita Fairways Golf Course in Bonita Springs, Florida, during 2005 and 2006 are as follows: $$\begin{array}{llllllll} 2005 \text { Season: } & 74 & 78 & 79 & 77 & 75 & 73 & 75 & 77 \\ 2006 \text { Season: } & 71 & 70 & 75 & 77 & 85 & 80 & 71 & 79 \end{array}$$ 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
The mean score is 76 for both years, but 2006 has a higher variability in scores. Improvement in some low scores is seen in 2006.

Step by step solution

01

Calculate the Mean for 2005

To find the mean score for the 2005 season, add all the scores together and divide by the number of scores. \[\text{Mean}_{2005} = \frac{74 + 78 + 79 + 77 + 75 + 73 + 75 + 77}{8} = \frac{608}{8} = 76\]
02

Calculate the Mean for 2006

Similarly, calculate the mean score for the 2006 season:\[\text{Mean}_{2006} = \frac{71 + 70 + 75 + 77 + 85 + 80 + 71 + 79}{8} = \frac{608}{8} = 76\]
03

Calculate the Standard Deviation for 2005

Standard deviation measures the amount of variation or dispersion in a set of values. Use the formula:\[\text{SD}_{2005} = \sqrt{\frac{\sum (x_i - \text{Mean}_{2005})^2}{n}}\]Where \(x_i\) represent each score, \(\text{Mean}_{2005} = 76\), and \(n = 8\). 1. Calculate deviations: - \(74 - 76 = -2\), \(78 - 76 = 2\), \(79 - 76 = 3\), \(77 - 76 = 1\), \(75 - 76 = -1\), \(73 - 76 = -3\), \(75 - 76 = -1\), \(77 - 76 = 1\).2. Square deviations: - \(4, 4, 9, 1, 1, 9, 1, 1\).3. Sum of squared deviations: - \(4 + 4 + 9 + 1 + 1 + 9 + 1 + 1 = 30\).4. Divide by \(n\) and take the square root:\[\text{SD}_{2005} = \sqrt{\frac{30}{8}} \approx 1.94\]
04

Calculate the Standard Deviation for 2006

Use the same process as Step 3 for the 2006 scores:\[\text{SD}_{2006} = \sqrt{\frac{\sum (x_i - \text{Mean}_{2006})^2}{n}}\]1. Calculate deviations: - \(71 - 76 = -5\), \(70 - 76 = -6\), \(75 - 76 = -1\), \(77 - 76 = 1\), \(85 - 76 = 9\), \(80 - 76 = 4\), \(71 - 76 = -5\), \(79 - 76 = 3\).2. Square deviations: - \(25, 36, 1, 1, 81, 16, 25, 9\).3. Sum of squared deviations:- \(25 + 36 + 1 + 1 + 81 + 16 + 25 + 9 = 194\).4. Divide by \(n\) and take the square root:\[\text{SD}_{2006} = \sqrt{\frac{194}{8}} \approx 4.93\]
05

Compare 2005 and 2006 Performance

Evaluate the two seasons using the mean and standard deviation:- The mean scores for both years are the same, at \(76\). However, the standard deviation for 2006 (\(4.93\)) is higher than that of 2005 (\(1.94\)), indicating greater variability and inconsistency in the 2006 scores.- The primary difference in performance is the variability; less consistency is observed in 2006. Although the average score did not change, the performance improved in terms of achieving lower scores like 71 and 70, but at the cost of some very high scores such as 85.

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Key Concepts

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

Mean Calculation
When it comes to analyzing any set of data, mean calculation is one of the fundamental steps. The mean, often simply referred to as the average, gives us a quick snapshot of the overall performance. To calculate it, we sum up all data points and then divide by the number of data points. For the golfer's scores, there are eight scores in each season. The calculation for each year involves adding up all the scores and dividing the total by eight. In our derived example, both the 2005 and 2006 seasons have a mean score of 76. Despite experiencing different playing conditions or scoring highs and lows, the golfer's average across the years remained consistent. This calculation helps to see that, on average, the golfer performed with parity over the two years. However, the mean alone does not tell the full story, which is where measures like standard deviation come into play.
Performance Analysis
Analyzing performance is critical for identifying trends, improvements, or declines over a period. For the amateur golfer, the performance analysis involves comparing not just the mean scores but also the standard deviation, which highlights score consistency. In the case of this golfer, both seasons had the same mean score, which might suggest a similar level of performance both years. However, by examining the standard deviations, a
  • Consistent 2005: A lower standard deviation in 2005 indicates that the golfer's performance was more consistent that year.
  • Variable 2006: A higher standard deviation in 2006 points to more variability, with scores ranging more widely from low to high.
The perfect illustration: though the average stayed the same, 2006 had instances of both exceptional and poor scores, suggesting inconsistent performance. By coupling the mean with this measure of variability, we get a deeper understanding of the overall performance.
Data Variability
Data variability refers to how spread out or clustered data points are. It's a crucial piece of the puzzle in understanding data patterns or behaviors. For the golfer in question, the variability can be seen in terms of how tightly or loosely the scores bunch around the mean. The standard deviation is the statistical measure used to quantify this variability. In the context of the golfer's scores:
  • In 2005, with a standard deviation of approximately 1.94, the scores were closely grouped around the mean. This low variability shows more predictable performance.
  • In 2006, the standard deviation jumped to about 4.93, indicating scores varied more widely from the mean. This higher variability depicts less predictability in performance.
Understanding data variability helps the golfer and coaches to identify patterns or anomalies. It shines a light on areas needing improvement, particularly in achieving consistent performance. While an average alone might look similar across periods, examining variability shows us the true performance dynamics at play.

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Most popular questions from this chapter

Consumer Reports provided overall customer satisfaction scores for AT\&T, Sprint, T-Mobile, and Verizon cell-phone services in major metropolitan areas throughout the United States. The rating for each service reflects the overall customer satisfaction considering a variety of factors such as cost, connectivity problems, dropped calls, static interference, and customer support. A satisfaction scale from 0 to 100 was used with 0 indicating completely dissatisfied and 100 indicating completely satisfied. The ratings for the four cellphone services in 20 metropolitan areas are as shown (Consumer Reports, January 2009 ). $$\begin{array}{lcccc} \text { Metropolitan Area } & \text { AT\&T } & \text { Sprint } & \text { T-Mobile } & \text { Verizon } \\ \text { Atlanta } & 70 & 66 & 71 & 79 \\ \text { Boston } & 69 & 64 & 74 & 76 \\ \text { Chicago } & 71 & 65 & 70 & 77 \\ \text { Dallas } & 75 & 65 & 74 & 78 \\ \text { Denver } & 71 & 67 & 73 & 77 \\ \text { Detroit } & 73 & 65 & 77 & 79 \\ \text { Jacksonville } & 73 & 64 & 75 & 81 \\ \text { Las Vegas } & 72 & 68 & 74 & 81 \\ \text { Los Angeles } & 66 & 65 & 68 & 78 \\ \text { Miami } & 68 & 69 & 73 & 80 \\ \text { Minneapolis } & 68 & 66 & 75 & 77 \\ \text { Philadelphia } & 72 & 66 & 71 & 78 \\ \text { Phoenix } & 68 & 66 & 76 & 81 \\ \text { San Antonio } & 75 & 65 & 75 & 80 \\ \text { San Diego } & 69 & 68 & 72 & 79 \\ \text { San Francisco } & 66 & 69 & 73 & 75 \\ \text { Seattle } & 68 & 67 & 74 & 77 \\ \text { St. Louis } & 74 & 66 & 74 & 79 \\ \text { Tampa } & 73 & 63 & 73 & 79 \\ \text { Washington } & 72 & 68 & 71 & 76 \end{array}$$ a. Consider T-Mobile first. What is the median rating? b. Develop a five-number summary for the T-Mobile service. c. Are there outliers for T-Mobile? Explain. d. Repeat parts (b) and (c) for the other three cell-phone services. e. Show the box plots for the four cell-phone services on one graph. Discuss what a comparison of the box plots tells about the four services. Which service did Consumer Reports recommend as being best in terms of overall customer satisfaction?

Does a major league baseball team's record during spring training indicate how the team will play during the regular season? Over the last six years, the correlation coefficient between a team's winning percentage in spring training and its winning percentage in the regular season is .18 (The Wall Street Journal, March 30,2009 ). Shown are the winning percentages for the 14 American League teams during the 2008 season. a. What is the correlation coefficient between the spring training and the regular season winning percentages? b. What is your conclusion about a team's record during spring training indicating how the team will play during the regular season? What are some of the reasons why this occurs? Discuss.

Consider a sample with data values of \(10,20,12,17,\) and \(16 .\) Compute the variance and standard deviation.

Consider a sample with data values of \(10,20,12,17,\) and \(16 .\) Compute the range and interquartile range.

How do grocery costs compare across the country? Using a market basket of 10 items including meat, milk, bread, eggs, coffee, potatoes, cereal, and orange juice, Where to Retire magazine calculated the cost of the market basket in six cities and in six retirement areas across the country (Where to Retire, November/December 2003 ). The data with market basket cost to the nearest dollar are as follows: $$\begin{array}{lclr} \text { City } & \text { cost } & \text { Retirement Area } & \text { cost } \\\ \text { Buffalo, NY } & \$ 33 & \text { Biloxi-Gulfport, MS } & \$ 29 \\ \text { Des Moines, IA } & 27 & \text { Asheville, NC } & 32 \\ \text { Hartford, CT } & 32 & \text { Flagstaff, AZ } & 32 \\ \text { Los Angeles, CA } & 38 & \text { Hilton Head, SC } & 34 \\ \text { Miami, FL } & 36 & \text { Fort Myers, FL } & 34 \\ \text { Pittsburgh, PA } & 32 & \text { Santa Fe, NM } & 31 \end{array}$$ a. Compute the mean, variance, and standard deviation for the sample of cities and the sample of retirement areas. b. What observations can be made based on the two samples?

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