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Chapter 3: Problem 47

The following data give the numbers of pieces of junk mail received by 10 families during the past month. \(\begin{array}{llllllllll}41 & 33 & 28 & 21 & 29 & 19 & 14 & 31 & 39 & 36\end{array}\) Find the range, variance, and standard deviation.

Short Answer

Expert verified
The range of number of pieces of junk mail received by 10 families during the past month is 27; the variance is 53.69 and the standard deviation is 7.33.

Step by step solution

01

Compute the Range

Range is calculated by subtracting the lowest data point from the highest data point. Here, the lowest data point is 14 and the highest is 41. So the range is 41-14=27.
02

Compute the Mean

Before calculating variance and standard deviation, it is necessary to calculate the mean or average. This is obtained by adding all data points and dividing the sum by the number of data points. So the mean is \( \frac{41+33+28+21+29+19+14+31+39+36}{10} = 29.1 \)
03

Compute Variance

Variance requires summing the squared difference from the mean for each data point and dividing that by the number of data points. The variance will be \( \frac{(41-29.1)^2+(33-29.1)^2+(28-29.1)^2+(21-29.1)^2+(29-29.1)^2+(19-29.1)^2+(14-29.1)^2+(31-29.1)^2+(39-29.1)^2+(36-29.1)^2}{10} = 53.69 \)
04

Compute Standard Deviation

Standard deviation is the square root of variance. The standard deviation will be \( \sqrt{53.69} = 7.33 \)

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Key Concepts

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

Understanding the Range
The range is one of the simplest ways to measure the spread or distribution of a data set.
In essence, the range helps us understand how much the numbers in our data set vary.
To find the range, you simply subtract the smallest number in your data set from the largest number.

For example, with our data set of junk mail received by families, we have values ranging from 14 pieces to 41 pieces.
This means that the range is calculated as follows:
  • Find the highest value (41)
  • Find the lowest value (14)
  • Subtract the lowest from the highest: 41 - 14 = 27
So, the range of this data set is 27, indicating a wide spread between the fewest and most junk mail pieces received in a month.
The range provides a quick snapshot but can be influenced significantly by outliers or extreme values.
Delving into Variance
Variance gives us a deeper understanding of data distribution by quantifying the average squared differences from the mean.
This means it tells us, on average, how far each number in our data set is from the mean.

To calculate variance, follow these steps:
  • Start by calculating the mean. For our data: \( \frac{41+33+28+21+29+19+14+31+39+36}{10} = 29.1 \)
  • For each data point, subtract the mean and square the result.
  • Sum up all the squared results.
  • Finally, divide by the number of data points (10 in this case).
As calculated, the variance is 53.69.
This high variance indicates that there is a significant spread around the mean of 29.1.
Grasping Standard Deviation
Standard deviation is a nifty tool that tells us the spread of a data set around its mean.
It is essentially the square root of the variance and provides insights into the average distance each data point is from the mean without the requirement to square the distances.

Our calculated variance was 53.69. Thus, the standard deviation is:
  • Take the square root of the variance
  • \( \sqrt{53.69} = 7.33 \)
The standard deviation of 7.33 indicates how much variation exists from the average amount of junk mail received.
A smaller standard deviation would mean that data points tend to be closer to the mean, while a larger standard deviation indicates more variation in the data.
It is especially useful in comparing the spread between different data sets.

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Most popular questions from this chapter

In a study of distances traveled to a college by commuting students, data from 100 commuters yielded a mean of \(8.73\) miles. After the mean was calculated, data came in late from three students, with respective distances of \(11.5,7.6\), and \(10.0\) miles. Calculate the mean distance for all 103 students.

The 2009 gross sales of all companies in a large city have a mean of \(\$ 2.3\) million and a standard deviation of \(\$ .6\) million. Using Chebyshev's theorem, find at least what percentage of companies in this city had 2009 gross sales of a. \(\$ 1.1\) to \(\$ 3.5\) million b. \(\$ .8\) to \(\$ 3.8\) million c. \(\$ .5\) to \(\$ 4,1\) million

The mean monthly mortgage paid by all home owners in a town is \(\$ 2365\) with a standard deviation of \(\$ 340\) a. Using Chebyshev's theorem, find at least what percentage of all home owners in this town pay a monthly mortgage of i. \(\$ 1685\) to \(\$ 3045\) ii. \(\$ 1345\) to \(\$ 3385\) \({ }^{*} \mathbf{b}\). Using Chebyshev's theorem, find the interval that contains the monthly mortgage payments of at least \(84 \%\) of all home owners.

One property of the mean is that if we know the means and sample sizes of two (or more) data sets, we can calculate the combined mean of both (or all) data sets. The combined mean for two data sets is calculated by using the formula $$ \text { Combined mean }=\bar{x}=\frac{n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}}{n_{1}+n_{2}} $$ where \(n_{1}\) and \(n_{2}\) are the sample sizes of the two data sets and \(\bar{x}_{1}\) and \(\bar{x}_{2}\) are the means of the two data sets, respectively. Suppose a sample of 10 statistics books gave a mean price of \(\$ 140\) and a sample of 8 mathematics books gave a mean price of \(\$ 160\). Find the combined mean. (Hint: For this example: \(\left.n_{1}=10, n_{2}=8, \bar{x}_{1}=\$ 140, \bar{x}_{2}=\$ 160 .\right)\)

Briefly explain what summary measures are used to construct a box-and-whisker plot.
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