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:

• First, check the symmetry of the data. For Brand A, arrange the data: \(17.0,18.5,19.0,20.0,21.0,22.0,22.5,22.5,23.0,23.0\). The data is relatively symmetric. For Brand B, arrange the data: \(11.0,12.0,14.0,16.5,17.0,19.5,19.5,20.0,20.5,21.0\). There is an out - lier (if we assume \(120\) is a typo, but even without considering it as a typo, the non - outlier part is not symmetric).
• When data is symmetric, standard deviation (a measure related to the mean) is a good measure of variability. When data is skewed (Brand B has a potential outlier which can cause skewness), inter - quartile range is better. But since Brand A is symmetric and Brand B is not (and we are comparing), for symmetric data (Brand A), standard deviation is appropriate.

Answer:

Only Brand A data is symmetric, so standard deviation is the best measure to compare variability.