Question

determine the best measure of variability. justify your answer. key: 10|2 represents 102. 10: 2 3 5 6 8; 11:1 2 5; 12:3 5 6; 13:0 2 5 9 9. options: iqr, because the data distribution is skewed; standard deviation, because the data distribution is skewed; standard deviation, because the data distribution is symmetric; iqr, because the data distribution is symmetric.

Explanation:

Step1: Analyze data distribution

First, we reconstruct the data from the stem - and - leaf plot. The stem - and - leaf plot has stems 10, 11, 12, 13.

• For stem 10: leaves are 2, 3, 3, 5, 6, 8. So the data points are 102, 103, 103, 105, 106, 108.
• For stem 11: leaves are 1, 2, 5. So the data points are 111, 112, 115.
• For stem 12: leaves are 3, 5, 6. So the data points are 123, 125, 126.
• For stem 13: leaves are 0, 2, 5, 9, 9, 9. So the data points are 130, 132, 135, 139, 139, 139.

To check for skewness, we can look at the shape of the data. A skewed distribution has a longer tail on one side. Let's check the number of data points in each stem:

• Stem 10: 6 data points
• Stem 11: 3 data points
• Stem 12: 3 data points
• Stem 13: 6 data points

The data is symmetric around the middle (the number of data points in the lower stems (10, 11) and upper stems (12, 13) are balanced). When the data distribution is symmetric, the mean and standard deviation are good measures of center and variability respectively. When the data is skewed, the median and inter - quartile range (IQR) are better because the mean is affected by outliers or skewness. Since our data is symmetric, we should use standard deviation as a measure of variability.

Step2: Recall measures of variability

• The inter - quartile range (IQR) is the difference between the third quartile ($Q_3$) and the first quartile ($Q_1$), $IQR = Q_3-Q_1$. It is a resistant measure of variability, more suitable for skewed data.
• The standard deviation ($s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}$) measures the average distance of data points from the mean. It is more suitable for symmetric data because the mean is a good measure of center for symmetric data, and the standard deviation depends on the mean.

Since our data is symmetric (as we saw from the stem - and - leaf plot, the number of data points in the lower and upper parts are balanced), the standard deviation is the best measure of variability.

Answer:

Standard deviation, because the data distribution is symmetric