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.

Diamond Ring Listed below are the amounts (dollars) it costs for marriage proposal packages at the different Major League Baseball stadiums. Five of the teams don’t allow proposals. Are there any outliers, and are they likely to have much of an effect on the measures of variation?

39 50 505055 55 75 85 100 115 175 175 200 209 250 250 350 400 450 500 500 500 500 1500 2500

The costs of the marriage proposal packages (in dollars) at 25 different stadiums are provided.

Range: It measures the distance between the maximum and the minimum values.

Mathematically, it is expressed as follows:

$\text{Range}=\text{Maximum}\text{Value}-\text{Minimum}\text{Value}$

Variance: It is the value obtained by summing the squared terms of the distance between each value from the mean and then dividing by the total sample size minus one. Its formula is as follows:

.

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

Standard deviation: It represents the square root of the variance. Its formula is as follows:

$s=\sqrt{s^2}$

The range is computed as follows:

.$$\begin{gathered} \text{Range}=\text{Maximum}\text{Value}-\text{Minimum}\text{Value} \\ =2500-39 \\ =2461.0\text{dollars} \end{gathered}$$

Hence, the range is equal to 2461.0 dollars.

The sample mean is computed as follows:

$$\begin{gathered} \overline{x}=\frac{{\sum }_{i=1}^n{x}_i}{n} \\ =\frac{39+50+....+2500}{25} \\ =365.3\text{dollars} \end{gathered}$$

Thus, the sample mean is equal to 365.3 dollars.

The variance is calculated as follows, substituting the values:

$$\begin{gathered} s^2=\frac{{\sum }_{i=1}^n{\left(x_i-\overline{x}\right)}^2}{n-1} \\ =\frac{{\left(39-365.3\right)}^2+{\left(50-365.3\right)}^2+...+{\left(2500-365.3\right)}^2}{25-1} \\ =290400.4\text{dollars}\text{square} \end{gathered}$$

Hence, the variance is equal to 290400.4 dollars square.

The standard deviation is obtained as follows, substituting the value of variance:

$$\begin{gathered} s=\sqrt{s^2} \\ =\sqrt{290400.4} \\ =538.9 \text{dollars} \end{gathered}$$

Hence, the standard deviation is equal to 538.9 dollars.

Outliers are observations that are quite different from the other observations in the dataset. By inspection, it is observed that the values1500 dollars and2500 dollars are the two outliers present in the dataset.

If the outliers are excluded, the following changes can be noted in each measure of variation.

• Highest value for range changes from 2500 to 500, leading to a smaller value of the range.
• The mean value would change as it is not resistant, which leads to a large change in the difference of observations from the mean. As a result, both variance and standard deviation would change largely.

Thus, the presence of outliers has asignificantly large effect on the measures of variation.