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Chapter 3: Problem 16

A bowler's scores for six games were \(182,168,184,190,170,\) and \(174 .\) Using these data as a sample, compute the following descriptive statistics: a. Range b. Variance c. Standard deviation d. Coefficient of variation

Short Answer

Expert verified
Range: 22, Variance: 62.67, Standard deviation: 7.91, CV: 4.44%

Step by step solution

01

Compute the Range

To find the range of the scores, we subtract the lowest score from the highest score. From the given data, the highest score is 190 and the lowest score is 168. Thus, the range is:\[\text{Range} = 190 - 168 = 22\]
02

Compute the Mean

The mean (average) score is calculated by summing up all the scores and dividing by the number of scores. The scores are 182, 168, 184, 190, 170, and 174.\[\text{Mean} = \frac{182 + 168 + 184 + 190 + 170 + 174}{6} = \frac{1068}{6} = 178\]
03

Compute the Variance

Variance is calculated by taking the mean of the squared differences between each score and the mean.\[\text{Variance} = \frac{(182-178)^2 + (168-178)^2 + (184-178)^2 + (190-178)^2 + (170-178)^2 + (174-178)^2}{6}\]Calculate each squared difference:\[(182-178)^2 = 16, \quad (168-178)^2 = 100, \quad (184-178)^2 = 36, \quad (190-178)^2 = 144, \quad (170-178)^2 = 64, \quad (174-178)^2 = 16\]Sum of squared differences:\[16 + 100 + 36 + 144 + 64 + 16 = 376\]Now divide by 6 to find the variance:\[\text{Variance} = \frac{376}{6} \approx 62.67\]
04

Compute the Standard Deviation

The standard deviation is the square root of the variance. From Step 3, we have:\[\text{Standard Deviation} = \sqrt{62.67} \approx 7.91\]
05

Compute the Coefficient of Variation

The coefficient of variation (CV) is the standard deviation divided by the mean, expressed as a percentage. Thus:\[\text{CV} = \left(\frac{7.91}{178}\right) \times 100 \approx 4.44\%\]

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Key Concepts

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

Range
The range is a basic measure of variation in a data set. It indicates the spread between the highest and lowest values. To calculate the range, you simply subtract the smallest number from the largest number in the data set. For our bowling scores, with a maximum score of 190 and a minimum score of 168, the calculation is straightforward:
  • Range = 190 - 168 = 22
This result tells us that there is a 22-point difference between the highest and lowest scores bowled.
Variance
Variance measures how far each number in the set is from the mean and thus from every other number in the set. First, calculate the mean score by adding all scores and dividing by the number of scores:
  • Mean = \(\frac{182 + 168 + 184 + 190 + 170 + 174}{6} = 178\)
Next, subtract the mean from each score to find the difference, square each of these differences, and then find the average of these squared differences:
  • Variances: \((182 - 178)^2, (168 - 178)^2, \ldots\)
  • Sum of squares = 376
  • Variance = \(\frac{376}{6} \approx 62.67\)
A higher variance indicates a wider spread in the data. Here, a variance of approximately 62.67 means the scores vary moderately from the mean.
Standard Deviation
Standard deviation is another measure of spread, showing how much variation or dispersion exists from the average. It is the square root of the variance. In our example, the variance is 62.67, so the standard deviation is:
  • Standard Deviation = \(\sqrt{62.67} \approx 7.91\)
This result implies that on average, each bowler's score deviates about 7.91 points from the mean. A smaller standard deviation would indicate that more scores are closer to the mean, while a larger standard deviation means more scores are spread out from the mean.
Coefficient of Variation
The coefficient of variation (CV) is a standardized measure of dispersion of a probability distribution or frequency distribution. It is expressed as a percentage, indicating the ratio of the standard deviation to the mean:
  • CV = \((\frac{7.91}{178}) \times 100 \approx 4.44\%\)
This percentage shows the extent of variability in relation to the mean. A CV of 4.44% means that the standard deviation is a small fraction of the mean, indicating relatively less variability in bowling scores compared to their average.

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Most popular questions from this chapter

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