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.

TV Prices Listed below are selling prices in dollars of TVs that are 60 inches or larger and rated as a “best buy” by Consumer Reports magazine. Are the measures of variation likely to be typical for all TVs that are 60 inches or larger?

1800 1500 1200 1500 1400 1600 1500 950 1600 1150 1500 1750

Short answer

The measures of variation have the following values:

• The sample range is equal to $850.0.
• The sample variance is equal to 61117.4 ${\left(\text{dollars}\right)}^2$.
• The sample standard is equal to $247.2 d.

The sample consists of prices of only those TVs that are considered the ‘best buy’. As this sample does not represent the entire population of TVs that are 60 inches or larger, the measures of variation are not typical for all the TVs that are equal to or greater than 60 inches.

Step-by-step solution

Given information

A sample of prices of ‘best buy’ TVs, which are equal to or greater than 60 inches, is given.

The sample size (n) is 12.

Computation of the measures of variation

The measures of variation are computed to gain an insight into thedegree of variation present in a sample. The values are used to analyze and infer other important statistical characteristics of the data.

Range,variance$\left(s^2\right),$ andstandard deviation$\left(s\right)$ are the three basic measures of dispersion/variation that are frequently reported in statistical research.

Calculations done for the given dataset are shown below:

The range is calculated as

$$\begin{gathered} \text{Range}=\text{Maximum}\text{Value}-\text{Minimum}\text{Value} \\ =1800-950 \\ =850.0 \end{gathered}$$

Therefore, the sample range for the prices of TVs is equal to $850.0.

The formula of the sample variance is

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

Here,

x represents the values in the sample, and

$\overline{x}$represents the mean of the sample.

The sample mean is calculated as

$$\begin{gathered} \overline{x}=\frac{{\sum }_{1=1}^n{x}_i}{n} \\ =\frac{1800+1500+...+1750}{12} \\ =\frac{17450}{12} \\ \approx 1454.2 \end{gathered}$$

.

Thus, the sample mean is $1454.2.

The sample variance is calculated as

$$\begin{gathered} s^2=\frac{{\sum }_{i=1}^n{\left(x_i-\overline{x}\right)}^2}{n-1} \\ =\frac{{\left(1800-1454.2\right)}^2+{\left(1500-1454.2\right)}^2+...{\left(1750-1454.2\right)}^2}{12-1} \\ =\frac{672291.67}{11} \\ \approx 61117.4 \end{gathered}$$

Therefore, the sample variance for the prices of TVs is equal to 61117.4${\left(\text{dollars}\right)}^2$.

The sample standard deviation is calculated as

$$\begin{gathered} s=\sqrt{s^2} \\ =\sqrt{61117.4} \\ \approx 247.2 \end{gathered}$$

Therefore, the sample standard deviation for the prices of TVs is equal to $247.2.

Interpretation

The sample contains prices of only the ‘best buy’ TVs. Thus, it cannot be considered a representative of the entire population of TVs equal to or greater than 60 inches.

Therefore, the measures of variation are not typical for all the TVs that are equal to or greater than 60 inches.