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. Herewe find measures of variation.) Then answer the given questions.

What Happens in Vegas . . . Listed below are prices in dollars for one night at different hotels located on Las Vegas Boulevard (the “Strip”). How useful are the measures of variation for someone searching for a room?

212 77 121 104 153 264 195 244

Short answer

The value of the range of the prices is equal to 187.0 dollars.

The variance of the prices is equal to 4626.2 dollars squared.

The standard deviation of the prices is equal to 68.0 dollars.

The sample has a collection of eight different hotels with different price ranges. Thus, the values can be useful in searching for a room on Las Vegas Boulevard as it provides a fair idea of the available hotels.

Step-by-step solution

Given information

The given data shows the prices for one night at eight different hotels in Las Vegas.

The number of values (n) is 8.

Computation of the measures of variation

The measures of dispersion are used for analyzing the spread of a given set of values.

The following are the three highly used measures of variation:

The rangeis the value obtained when the minimum value is subtracted from the maximum value.

$$\begin{gathered} \text{Range}=\text{Maximum}\text{Value}-\text{Minimum}\text{Value} \\ =264-77 \\ =187.0\text{dollars} \end{gathered}$$

.

Therefore, for the given sample of prices, the range is equal to 187.0 dollars.

Sample variance$\left(s^2\right)$is used to determine the inherent variation in the sample. The square of the unit of the data forms the unit of variance. It is calculated using the following formula:

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

Here,

x represents the sampled values, and

$\overline{x}$is the sample mean.

The sample mean is calculated as

$$\begin{gathered} \overline{x}=\frac{{\sum }_{1=1}^n{x}_i}{n} \\ =\frac{212+77+...+244}{8} \\ =\frac{1370}{8} \\ \approx 171.3 \end{gathered}$$

.

Thus, the sample mean is 171.3 million.

The variance of the sample is calculated as

$$\begin{gathered} s^2=\frac{{\sum }_{i=1}^n{\left(x_i-\overline{x}\right)}^2}{n-1} \\ =\frac{{\left(212-171.3\right)}^3+{\left(77-171.3\right)}^3+...+{\left(244-171.3\right)}^3}{8-1} \\ =\frac{32383.5}{7} \\ =4626.2 \end{gathered}$$

.

Therefore, the sample variance of the prices for a one-night stay is equal to 4626.2${\left(\text{dollars}\right)}^2$.

Thestandard deviation of the sample also shows the variation in the data. The units of the values in the data are the units of standard deviation. It is calculated using the following formula as

$$\begin{gathered} s=\sqrt{s^2} \\ =\sqrt{4626.2} \\ \approx 68.0\text{dollars} \end{gathered}$$

Therefore, the sample standard deviation of the prices for a one-night stay is equal to 68.0 dollars.

Interpretation

As the sample consists of a range of eight different hotels with different prices, it can be said that the values of the measures of variation are useful for a person looking for a room on Las Vegas Boulevard.