Question 45

The modes of A and B must be equal.

The statement that the modes of A and B must be equal is false. Since the distributions are different, particularly with A being more spread out and right-skewed while B is tightly clustered and roughly symmetric, it is entirely possible for their modes to differ.

A) The means of A and B could be equal.

This statement is true because the means of two data sets can be equal even if their distributions are different. Since both data sets have the same number of observations, it is conceivable that the average values could align despite the differing shapes of their distributions.

B) The medians of A and B could be equal.

This statement is also true. The median is a measure of central tendency that can be the same for different distributions, especially if the overall positions of the middle values align, regardless of the spread or skewness of the data.

C) The modes of A and B must be equal.

This statement is false, as the mode represents the most frequently occurring value in a data set. Given that A is more spread out and right-skewed while B is tightly clustered, their modes can differ significantly, contradicting the assertion that they must be equal.

D) The range of A is likely greater than the range of B.

This statement is likely true because A is described as more spread out, which usually results in a larger range. Since B is tightly clustered, its range would be expected to be smaller in comparison to A's.

Conclusion

The false statement is that the modes of A and B must be equal, as the different distributions can lead to differing modes. The other options correctly reflect that means and medians can be equal despite differences in distribution, and the spread of A suggests that its range is likely greater than that of B. Thus, only option C does not hold true in the context provided.