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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 c. Standard deviation b. Variance d. Coefficient of variation

Short Answer

Expert verified
Range: 22, Variance: 75.2, Standard deviation: 8.67, Coefficient of variation: 4.87%.

Step by step solution

01

Calculate the Range

The range is the difference between the highest and lowest scores.First, identify the highest score: 190Next, identify the lowest score: 168The range is found by subtracting the lowest score from the highest score:\[\text{Range} = 190 - 168 = 22\]
02

Calculate the Mean

To compute the variance and standard deviation, first find the mean of the scores.Add all the scores together:\[182 + 168 + 184 + 190 + 170 + 174 = 1068\]Divide the sum by the number of scores (6) to find the mean:\[\text{Mean} = \frac{1068}{6} = 178\]
03

Calculate the Variance

First, find the differences between each score and the mean, then square each difference:\[(182-178)^2 = 16, (168-178)^2 = 100, (184-178)^2 = 36\]\[(190-178)^2 = 144, (170-178)^2 = 64, (174-178)^2 = 16\]Sum these squared differences:\[16 + 100 + 36 + 144 + 64 + 16 = 376\]Finally, divide by the number of scores minus 1 to find the sample variance:\[\text{Variance} = \frac{376}{5} = 75.2\]
04

Calculate the Standard Deviation

The standard deviation is the square root of the variance:\[\text{Standard Deviation} = \sqrt{75.2} \approx 8.67\]
05

Calculate the Coefficient of Variation

The coefficient of variation is the standard deviation divided by the mean, expressed as a percentage:\[\text{Coefficient of Variation} = \left(\frac{8.67}{178}\right) \times 100 \approx 4.87\%\]

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Key Concepts

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

Range
The range is one of the simplest measures in descriptive statistics. It tells us the spread of a data set by showing the difference between the highest and the lowest values. In our example, we have a series of bowling scores: 182, 168, 184, 190, 170, and 174.
To calculate the range, we first need to identify the highest score, which is 190, and the lowest score, which is 168.
Subtract the smallest value from the largest:
  • The range = 190 - 168 = 22
The range gives us a basic idea of the spread of the scores, but it doesn't provide information about the distribution of scores within that range.
Variance
Variance is a measure of how spread out a set of numbers is. It gives a more detailed picture of data variability than the range. To calculate variance, we follow these steps:
  • First, find the mean (average) of the data.
  • Then, subtract the mean from each number to find the deviation of each data point.
  • Square each of these deviations.
  • Finally, average these squared deviations. Since we deal with a sample, we divide by the number of scores minus one.
For our bowling scores:
  • Mean = (182 + 168 + 184 + 190 + 170 + 174) / 6 = 178
  • Squared deviations: (182-178)^2, (168-178)^2, (184-178)^2, (190-178)^2, (170-178)^2, and (174-178)^2.
  • The sum of these squared deviations is 376.
  • The sample variance = 376 / 5 = 75.2
Variance quantifies the degree of spread in the data, providing crucial insight into data dispersion.
Standard Deviation
The standard deviation is closely linked to variance. It represents the average distance of each data point from the mean, but it is expressed in the same units as the data. Unlike variance, which is in squared units, the standard deviation is in the original units.

Calculating the standard deviation involves taking the square root of the variance. For our data set, use the variance value we calculated:
  • Standard Deviation = \( \sqrt{75.2} \approx 8.67 \)
This metric makes it easier to interpret the variability of data as it reflects dispersion in the same units as the dataset, which in this case represents scores.
Coefficient of Variation
The coefficient of variation (CV) is a standardized measure of dispersion relative to the mean and is expressed as a percentage. It is particularly useful for comparing the degree of variation between different data sets that may have different units or mean values.

To find the coefficient of variation, divide the standard deviation by the mean and multiply by 100 to get a percentage.
  • For our case: Coefficient of Variation = \( \left( \frac{8.67}{178} \right) \times 100 \approx 4.87\% \)
This tells us that the standard deviation is around 4.87% of the mean score. The CV is helpful in providing a relative measure of variability, making it advantageous for comparisons across different datasets or phenomena.

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Most popular questions from this chapter

Consider the sample data in the following frequency distribution. $$\begin{array}{ccc} \text { Class } & \text { Midpoint } & \text { Frequency } \\ 3-7 & 5 & 4 \\ 8-12 & 10 & 7 \\ 13-17 & 15 & 9 \\ 18-22 & 20 & 5 \end{array}$$ a. Compute the sample mean. b. Compute the sample variance and sample standard deviation.

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The following data show the trailing 52 -week primary share earnings and book values as reported by 10 companies (The Wall Street Journal, March 13,2000 ). $$\begin{array}{lrr} & \text { Book } & \\ \text { Company } & \text { Value } & \text { Earnings } \\ \text { Am Elec } & 25.21 & 2.69 \\ \text { Columbia En } & 23.20 & 3.01 \\ \text { Con Ed } & 25.19 & 3.13 \\ \text { Duke Energy } & 20.17 & 2.25 \\ \text { Edison Int'l } & 13.55 & 1.79 \\ \text { Enron Cp. } & 7.44 & 1.27 \\ \text { Peco } & 13.61 & 3.15 \\ \text { Pub Sv Ent } & 21.86 & 3.29 \\ \text { Southn Co. } & 8.77 & 1.86 \\ \text { Unicom } & 23.22 & 2.74 \end{array}$$ a. Develop a scatter diagram for the data with book value on the \(x\) -axis. b. What is the sample correlation coefficient, and what does it tell you about the relationship between the earnings per share and the book value?

Consumer Review posts reviews and ratings of a variety of products on the Internet. The following is a sample of 20 speaker systems and their ratings (http://www.audioreview.com). The ratings are on a scale of 1 to \(5,\) with 5 being best. $$\begin{array}{lllr} \text { Speaker } & \text { Rating } & \text { Speaker } & \text { Rating } \\\ \text { Infinity Kappa 6.1 } & 4.00 & \text { ACI Sapphire III } & 4.67 \\ \text { Allison One } & 4.12 & \text { Bose 501 Series } & 2.14 \\ \text { Cambridge Ensemble II } & 3.82 & \text { DCM KX-212 } & 4.09 \\ \text { Dynaudio Contour 1.3 } & 4.00 & \text { Eosone RSF1000 } & 4.17 \\ \text { Hsu Rsch. HRSW12V } & 4.56 & \text { Joseph Audio RM7si } & 4.88 \\ \text { Legacy Audio Focus } & 4.32 & \text { Martin Logan Aerius } & 4.26 \\ \text { Mission 73li } & 4.33 & \text { Omni Audio SA 12.3 } & 2.32 \\ \text { PSB 400i } & 4.50 & \text { Polk Audio RT12 } & 4.50 \\ \text { Snell Acoustics D IV } & 4.64 & \text { Sunfire True Subwoofer } & 4.17 \\ \text { Thiel CS1.5 } & 4.20 & \text { Yamaha NS-A636 } & 2.17\end{array}$$ a. Compute the mean and the median. b. Compute the first and third quartiles. c. Compute the standard deviation. d. The skewness of this data is -1.67 . Comment on the shape of the distribution. e. What are the \(z\) -scores associated with Allison One and Omni Audio? f. Do the data contain any outliers? Explain.
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