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 centre. Here we find measures of variation.) Then answer the given questions.

Sales of LP Vinyl Record Albums Listed below are annual U.S. sales of vinyl record albums (millions of units). The numbers of albums sold are listed in chronological order, and the last entry represents the most recent year. Do the measures of variation give us any information about a changing trend over time?

0.3 0.6 0.8 1.1 1.1 1.4 1.4 1.5 1.2 1.3 1.4 1.2 0.9 0.9 1.0 1.9 2.5 2.8 3.9 4.6 6.1

Short answer

The measures of variation are as follows:

The range of albums is 5.80 million.

The variance of albums is 2.09 million square.

The standard deviation of albums is 1.44 million.

No, the measures of variation do not give an insight into the variation or trend that the values follow.

Step-by-step solution

Given information

The data shows the number of vinyl record albums sold for 21 years, listed in chronological order.

The measures of variation 

The three measures of variation that are most commonly calculated are as follows:

• Range: The difference of the lowest value from the highest.
• Variance: The average squared differences from mean.
• Standard deviation: The square root of variance.

Compute the measures of variation 

The range of the given sample of 21 years is computed as follows:

$$\begin{gathered} \text{Range}=\text{Highest}\text{Value}-\text{Lowest}\text{Value} \\ =6.1-0.3 \\ =5.8\text{million} \end{gathered}$$

Thus, the range of the number of albums sold is equal to5.80 million.

The sample mean is computed as follows:

$$\begin{gathered} \overline{x}=\frac{{\sum }_{i=1}^n{x}_i}{n} \\ =\frac{0.3+0.6+....+6.1}{21} \\ =1.80 \text{million} \end{gathered}$$

Thus, the sample mean is equal to 1.80 million.

The variance of the given sample of 21 years is computed as follows:

$$\begin{gathered} s^2=\frac{{\sum }_{i=1}^n{\left(x_i-\overline{x}\right)}^2}{n-1} \\ =\frac{{\left(0.3-1.80\right)}^2+{\left(0.6-1.80\right)}^2+...+{\left(6.1-1.80\right)}^2}{21-1} \\ =2.09{\left(\text{million}\right)}^2 \end{gathered}$$

Thus, the variance of the number of albums sold is equal to2.09 million square.

The standard deviation of the given sample of 21 years is computed as follows:

$$\begin{gathered} s=\sqrt{s^2} \\ =\sqrt{2.09} \\ =1.44\text{million} \end{gathered}$$

Thus, the standard deviation of the number of albums sold is equal to 1.44 million.

Analyze the results for the time-based data

The given sample depictstime-series datarecorded for 21 years in chronological order. The measures of variation express the extent to which data is spread out. They do not provide any meaningful interpretation for the changes that take place in patterns over time.

Thus, the measures of variationdo not tell much about any particular trend in the values over the years.