9.
Identify the three measures of central tendency in the following distributions:
a. A1 and B1= mode, A2 and B2 = median, A3 and B3 = mean
b. A1 and B3= mode, A2 and B2 = median, A3 and B1 = mean
c. A3 and B1= mode, A2 and B2 = median, A1 and B3 = mean
d. A1 and B1= median, A2 and B2 = mode, A3 and B3 = mean

The correct option is (a):
A1 and B1 = mode, A2 and B2 = median, A3 and B3 = mean. In a skewed distribution, the mode lies at the peak, the mean is pulled toward the long tail, and the median lies between them.​

Introduction

Measures of central tendency—mean, median and mode—occupy different positions in skewed distributions, unlike in a normal distribution where all three coincide. In right (positively) and left (negatively) skewed curves, the peak, the middle point and the tail together reveal which point is mode, median or mean. This MCQ uses two schematic curves (A and B) to test that conceptual relationship.​

Understanding the two curves

• In a positively skewed (right‑skewed) distribution, the long tail is on the right; the order from left to right is mode → median → mean.​

• In a negatively skewed (left‑skewed) distribution, the long tail is on the left; the order from left to right is mean → median → mode.​

In both cases, the mode is at the highest point (peak), the mean is pulled furthest toward the tail, and the median lies between mean and mode.​

For graph A (left panel), the tail is to the right (positive skew), so A1 is at the peak, A2 in the middle, and A3 closest to the tail.
For graph B (right panel), the tail is to the left (negative skew), so B1 is at the peak, B2 in the middle, and B3 closest to the tail.

Thus:

• A1, B1: highest point → mode

• A2, B2: central point between the other two → median

• A3, B3: closest to tail → mean​

Option‑by‑option explanation

Option (a)

A1 and B1 = mode, A2 and B2 = median, A3 and B3 = mean

• A1 and B1 lie exactly at the peak of each curve, matching the definition of mode as the most frequent value located at the highest point of the distribution.​

• A2 and B2 lie between the other two markers, which fits the median, always between mean and mode in skewed distributions.​

• A3 and B3 are displaced toward the long tail (right in A, left in B), consistent with the mean, which is most affected by extreme values and pulled toward the tail.​

Therefore, option (a) is correct.

Option (b)

A1 and B3 = mode, A2 and B2 = median, A3 and B1 = mean

• A1 as mode is fine, but B3 is near the left tail, not at the peak; a mode cannot lie in the tail of a unimodal distribution.​

• Assigning B1 (the peak) as mean contradicts the rule that mean in a skewed distribution is shifted toward the tail, not at the maximum.​

So option (b) misplaces mean and mode for curve B and is incorrect.

Option (c)

A3 and B1 = mode, A2 and B2 = median, A1 and B3 = mean

• A3 is near the right tail in curve A and cannot be the mode, which must sit at the peak; similarly, B3 is near the left tail and cannot be the mean’s partner for A1 as mean.​

• Assigning A1 as mean ignores that, in a right‑skewed distribution, the mean lies to the right of both median and mode, not at the peak.​

Hence option (c) reverses the roles of mean and mode in graph A and is wrong.

Option (d)

A1 and B1 = median, A2 and B2 = mode, A3 and B3 = mean

• A2 and B2 lie at the peaks visually, so labeling them as mode would fit, but the figure clearly labels A1 and B1 at the top; the option swaps the labels relative to the true peaks shown.​

• Making A1 and B1 the median places the median at the highest frequency, which only occurs in a perfectly symmetric distribution, not in clearly skewed ones.​

Thus option (d) conflicts with both the diagram and the known ordering of central tendency in skewed distributions and is incorrect.

Key takeaway

In skewed distributions, remember the visual rule:

• Right‑skewed: peak (mode) on the left, tail to the right → mode < median < mean.​

• Left‑skewed: tail to the left → mean < median < mode.​

Using the peak–middle–tail logic quickly identifies A1/B1 as mode, A2/B2 as median and A3/B3 as mean, confirming option (a) as the correct answer.