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Chapter 3: Problem 44

The following data give the prices of seven textbooks randomly selected from a university bookstore. \(\begin{array}{lll}\$ 89 & \$ 170 & \$ 104\end{array}\) \(\begin{array}{llll}\$ 113 & \$ 56 & \$ 161 & \$ 147\end{array}\) a. Find the mean for these data. Calculate the deviations of the data values from the mean. Is the sum of these deviations zero? b. Calculate the range, variance, and standard deviation.

Short Answer

Expert verified
The mean is \$120. The sum of deviations from the mean is zero. Range is \$114 while the variance and standard deviation can be calculated following the steps provided and using the formula provided in the step-by-step solution.

Step by step solution

01

Calculate the Mean

The mean (average) is found by adding all the data points and dividing by the total number of points.\( Mean = \frac{\$(89 + 170 + 104 + 113 + 56 + 161 + 147)}{7} = \$120\)
02

Calculate the Deviations

The deviation of each point is the absolute difference between each point and the mean. Add each of these and confirm if the sum is zero. As per the properties of the mean, the sum of the deviations will always be zero.
03

Calculate the Range

The range is the difference between the highest price (\$170) and the lowest price (\$56). Thus, the range is \$114.
04

Calculate the Variance

Variance is calculated by taking the average of the squared differences from the Mean: \[ Variance = \frac{(89-120)^2 + (170-120)^2 + (104-120)^2 + (113-120)^2 + (56-120)^2 + (161-120)^2 + (147-120)^2}{7}\]
05

Calculate the Standard Deviation

The standard deviation is the square root of the Variance. After evaluating the variance, take the square root of the result to get the Standard Deviation.

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Key Concepts

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

Mean Calculation
The mean, often called the average, represents the central value of a data set. It is the sum of all the data points divided by the total number of data points. This calculation helps us understand the overall tendency of the data. In our example with textbook prices:
  • Add all prices: \( 89 + 170 + 104 + 113 + 56 + 161 + 147 = 840 \)
  • Divide by the number of textbooks, which is 7: \( \frac{840}{7} = 120 \)
So, the mean price of the textbooks is \( \$120 \). Knowing the mean gives us a snapshot of the typical textbook price at this bookstore.
Data Deviations
Data deviations show how much each data point differs from the mean. To find deviations, subtract the mean from each value:
  • \( 89 - 120 = -31 \)
  • \( 170 - 120 = 50 \)
  • \( 104 - 120 = -16 \)
  • \( 113 - 120 = -7 \)
  • \( 56 - 120 = -64 \)
  • \( 161 - 120 = 41 \)
  • \( 147 - 120 = 27 \)
The sum of these deviations equals zero, illustrating a key property of the mean: The data points are symmetrically distributed around it. This feature validates the calculation, ensuring that the mean accurately reflects the data's center.
Range
The range is a simple way to gauge how spread out the data points are. It is calculated by subtracting the smallest value from the largest value in the set. In the textbook example:
  • Highest price: \( \\(170 \)
  • Lowest price: \( \\)56 \)
  • Thus, the Range: \( 170 - 56 = 114 \)
The range of \( \$114 \) indicates the difference between the most and least expensive textbooks. Though easy to compute, the range only provides a basic picture of variability and can be influenced by extreme values.
Variance
Variance measures the degree of spread within a data set by determining how far each number in the set is from the mean. To find it, calculate the squared deviations from the mean and then find their average:
  • \((89-120)^2 = 961\)
  • \((170-120)^2 = 2500\)
  • \((104-120)^2 = 256\)
  • \((113-120)^2 = 49\)
  • \((56-120)^2 = 4096\)
  • \((161-120)^2 = 1681\)
  • \((147-120)^2 = 729\)
Calculate the average of these values:\[\text{Variance} = \frac{961 + 2500 + 256 + 49 + 4096 + 1681 + 729}{7} \approx 1841.14\]Variance indicates how much the textbook prices fluctuate around the mean. A higher variance implies greater variability.
Standard Deviation
Standard deviation is the square root of the variance, giving a measure of spread in the same unit as the data. It is less abstract than variance and traces how much data diverge from the mean:\[\text{Standard Deviation} = \sqrt{1841.14} \approx 42.90\]This value, approximately \( 42.90 \), offers a clear depiction of the average variability of the textbook prices. The smaller the standard deviation, the closer the data points are to the mean. Conversely, a higher value reflects more spread. Standard deviation is particularly useful in comparing variability between datasets.

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Most popular questions from this chapter

Explain how the interquartile range is calculated. Give one example.

A survey of young people's shopping habits in a small city during the summer months of 2012 showed the following: Shoppers aged 12 to 14 years took an average of 8 shopping trips per month and spent an average of \(\$ 14\) per trip. Shoppers aged 15 to 17 years took an average of 11 trips per month and spent an average of \(\$ 18\) per trip. Assume that this city has 1100 shoppers aged 12 to 14 years and 900 shoppers aged 15 to 17 years. a. Find the total amount spent per month by all these 2000 shoppers in both age groups. b. Find the mean number of shopping trips per person per month for these 2000 shoppers. c, Find the mean amount spent per person per month by shoppers aged 12 to 17 years in this city.

A sample of 2000 observations has a mean of 74 and a standard deviation of 12 . Using Chebyshev's theorem, find at least what percentage of the observations fall in the intervals \(\bar{x} \pm 2 s, \bar{x} \pm 2.5 s\), and \(\bar{x} \pm 3 s\). Note that here \(\bar{x} \pm 2 \mathrm{~s}\) represents the interval \(\bar{x}-2 s\) to \(\bar{x}+2 s\), and so on.

The mean 2011 income for five families was \(\$ 99.520 .\) What was the total 2011 income of these five families?

The following data give the numbers of text messages sent by a high school student on 40 randomly selected days during 2012: \(\begin{array}{llllllllll}32 & 33 & 33 & 34 & 35 & 36 & 37 & 37 & 37 & 37 \\\ 38 & 39 & 40 & 41 & 41 & 42 & 42 & 42 & 43 & 44 \\ 44 & 45 & 45 & 45 & 47 & 47 & 47 & 47 & 47 & 48 \\ 48 & 49 & 50 & 50 & 51 & 52 & 53 & 54 & 59 & 61\end{array}\) a. Calculate the values of the three quartiles and the interquartile range. Where does the value 49 fall in relation to these quartiles? b. Determine the approximate value of the 91 st percentile. Give a brief interpretation of this percentile. c. For what percentage of the days was the number of text messages sent 40 or higher? Answer by finding the percentile rank of 40 .
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