Which of the following sets of data
has a mode of 48 and a median of 20. Is it A) 48, 21, 11, 48, 20,
17? B) 21, 48, 19, 48, 17, 11? C) 47, 47, 11, 48, 20, 17? D) 10, 16, 19, 21, 47, 47? Or E) 20, 48, 48, 11, 11, 19?
We recall that the mode is the most
frequently occurring value. Therefore, we need to find a set of
data where 48 is the most common or most frequently occurring number. 48 does not occur in set D. Therefore, this cannot be the
correct answer. Whilst there is a 48 in set C,
there are two 47s. Therefore, the mode of set C is
47. We can, therefore, rule this out as
the correct answer.
Set E has two 48s, but it also has
two 11s. This means that it has two modes,
or it’s bimodal. It has a mode of 11 and 48. This means that option E is also
incorrect. Both option A and option B have two
48s. This is the most frequently
occurring value in both data sets. This means that the mode of both of
these is 48.
Let’s now consider our second piece
of information. The median of the data set has to
be 20. We know that the median is the
middle value once the numbers are in ascending or descending order. Once we have put both of these data
sets in order, we notice that they have an even number of values, in this case,
six. This means that there are two
middle members, in set A, 20 and 21 and in set B, 19 and 21.
The median can be calculated by
finding the mean of these two values. This is the same as finding the
midpoint of the two values. The mean of 20 and 21 is 20.5. And the mean of 19 and 21 is
20. This means that set A has a median
of 20.5 and set B has a median of 20. We can, therefore, rule out set
A. The set of data that has a mode of
48 and a median of 20 is set B, 21, 48, 19, 48, 17, and 11.