Question

the table shows the battery lives, in hours, of ten brand a batteries and ten brand b batteries. battery life (hours) brand a 22.5 17.0 21.0 23.0 22.0 18.5 22.5 20.0 19.0 23.0 brand b 20.0 19.5 20.5 16.5 14.0 17.0 11.0 19.5 21.0 12.0 which would be the best measure of variability to use to compare the data? only brand a data is symmetric, so standard deviation is the best measure to compare variability. only brand b data is symmetric, so the median is the best measure to compare variability. both distributions are symmetric, so the mean is the best measure to compare variability. both distributions are skewed left, so the interquartile range is the best measure to compare variability.

Explanation:

Step1: Check data symmetry

By looking at the data values of Brand A and Brand B, we can see that neither distribution is symmetric. Brand A values range from 17.0 - 23.0 and Brand B from 11.0 - 21.0 in a non - symmetric way.

Step2: Recall measure of variability for skewed data

For skewed data distributions, the interquartile range (IQR) is a better measure of variability than the standard deviation (which is for symmetric data), and the mean and median are measures of central tendency not variability. Since both distributions seem to be skewed (Brand B has lower values pulling it left - skewed), the IQR is the best measure to compare variability.

Answer:

Both distributions are skewed left, so the interquartile range is the best measure to compare variability.