From the shape of the distribution we would expect:

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BlackTom AI · Accepted Answer

When looking at the dot plot, observe how the data pile up more heavily on the left side (lower numbers of calls) while a tail extends toward higher call counts. This pattern is characteristic of a right-skewed (positively skewed) distribution. In skewed distributions, the mean is sensitive to extreme values in the tail, so it tends to be pulled in the direction of the tail, i.e., toward higher values, more than the median.

Option a: 'the mean number of calls to be the same as the median' — This would be true for a perfectly symmetric distribution, not for a right-skewed one. The presence of a longer right tail means asymmetry, so the mean will not coincide with the median.

Option b: 'the mean number of calls to be less than the median' — This would occur if the distribution were left-skewed (tail on the left). The plot shows a longer tail to the right, so the mean exceeding the median is more plausible, not the opposite.

Option c: 'the mean number of calls to be greater than the median' — Given the right-skew nature with a few higher values pulling the average upward while the median stays more central, this relation is expected. The mean is dragged to the right by the tail, while the median is more resistant to extreme values.

Option d: 'the mean number of calls to equal the interquartile range' — The mean and the interquartile range measure different concepts (central tendency vs spread). They are not expected to be equal in any typical distribution, and this statement is not a comparison of central tendency as the other options are. It’s an apples-to-oranges claim rather than a descriptive relationship between mean and median.

In summary, the observed right-skew in the distribution suggests that the mean will be pulled toward the higher end of the scale relative to the median, making the mean larger than the median.

From the shape of the distribution we would expect:Single choice

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