Solution

This kind of question should be attacked with an upfront work. Students can be misleading by frequency or else. But dive into the problem and see the real gist of it

List Z is the sum of list X and Y. So the average of Z in the average of the single averages of X and Y : \(\frac{2.7+7.1}{2}=4.9\)

To compare the Average and the median, we have to calculate the median of Z. Z is composed of 120 numbers, so the median is the average of the two middle numbers.
X have numbers from 1 to 5, and Y from 6 to 9. As such, the two middle numbers of Z are 5 and 6. the median is \(\frac{6+5}{2}=5.5\)

So the correct answer is \(B\)
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Median is the exact mid point when a list of numbers are arranged in increasing order.

So in 1, 2, 3, 5, 7 median is 3.

But in 1, 2, 3, 5 median is \(\frac{2+3}{2}=2.5\)
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Sandy
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Hi Carcass,
Shouldn't the median be (6+6)/2= 6?
the median should be measure (order60+order61)/2
I apologized because I don't know how to explain better

The answer should be B, I believe its keyed in incorrectly.
Thank you again for putting this together!!

That is a clever solution, Carcass. I would just add for anyone who missed it, we can use the "average of the single averages" because each set has an equal number of values; in other words, their weight is equal. If the sets had unequal numbers of values, for example if the first set had 59 values and the second had 60, then we couldn't average the averages.

Having said this, when we come across a problem like this, where we can use your method it will save a lot of time, versus calculating a weighted mean, or just calculating a total sum for both sets using n and the mean (ex. 60x2.7 + 60x7.1) and then dividing to find the aggregate mean ( [60x2.7 + 60x7.1]/120 = 4.9).