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