Question

:In Exercises 5–20, find the range, variance, and standard deviation for the given sample data. Include appropriate units (such as “minutes”) in your results. (The same data were used in Section 3-1, where we found measures of center. Here we find measures of variation.) Then answer the given questions.

Most Expensive Colleges Listed below in dollars are the annual costs of tuition and fees at the 10 most expensive colleges in the United States for a recent year (based on data from U.S. News & World Report). The colleges listed in order are Columbia, Vassar, Trinity, George Washington, Carnegie Mellon, Wesleyan, Tulane, Bucknell, Oberlin, and Union. What does this “top 10” list tell us about the variation among costs for the population of all U.S. college tuitions?

49,138 47,890 47,510 47,343 46,962 46,944 46,930 46,902 46,870 46,785

Short answer

The values of the calculated statistics are as follows:

The range is equal to 2353.0 dollars.

Variance is equal to 527910.5 dollars square.

Thestandarddeviation is equal to 726.6 dollars.

The sample values consist of the costs of only the top 10 colleges in the US. Thus, the variation among these costs does not tell anything about the entire population of all US colleges.

Step-by-step solution

Given information

The given data shows the annual costs of tuition and fees of the top 10 colleges of the United States.

Formulae for measures of variation 

Therange measures the overall variation in a dataset. It is computed as follows:

$$\begin{gathered} \text{Range}=\text{Maximum}\text{Value}-\text{Minimum}\text{Value} \\ =49138-46785 \\ =2353.0\text{dollars} \end{gathered}$$

Hence, the value of the range is 2353.0 dollars.

Thevariance measures the square of the deviation between the values from the mean value. It is computed as follows:

$s^2=\frac{{\sum }_{i=1}^n{\left(x_i-\overline{x}\right)}^2}{n-1}$

Thestandard deviation measures the deviation between the values from the mean value. It is computed as follows:

$s=\sqrt{s^2}$

Compute sample standard deviation and variance 

The sample mean is computed as follows:

$$\begin{gathered} \overline{x}=\frac{{\sum }_{i=1}^n{x}_i}{n} \\ =\frac{49+138+....+785}{10} \\ =47327.4\text{dollars} \end{gathered}$$

Thus, the sample mean is equal to 47327.4 dollars.

The variance is calculated as follows, substituting the values:

$$\begin{gathered} s^2=\frac{{\sum }_{i=1}^n{\left(x_i-\overline{x}\right)}^2}{n-1} \\ =\frac{{\left(49-47327.4\right)}^2+{\left(138-47327.4\right)}^2+...+{\left(785-47327.4\right)}^2}{10-1} \\ =527910.5\text{dollars} \end{gathered}$$

Hence, the value of variance is 527910.5 dollars square.

The standard deviation is calculated as follows:

$$\begin{gathered} s=\sqrt{s^2} \\ =\sqrt{527910.5} \\ =726.6\text{dollars} \end{gathered}$$

Hence, the value of standard deviation is 726.6 dollars.

Evaluate the quality of sampled values

The sample fees are taken from the top 10 colleges in the US. They are randomly selected samples taken from the population of all fees concerning all US colleges.

Thus, the values of the variation from the sample are not appropriate representative of the costs of the entire population of US colleges.