For each data set, start adding the frequencies from the smallest value up until you pass the 18th and 19th positions. Which actual data values sit at those positions?

Instead of writing out all 36 numbers, multiply each value by how many times it appears, add these products to get the total sum, then divide by 36. Do this for both data sets and then compare the results.

Once you know which data set has the larger mean and which has the larger median, look for the choice whose wording matches both comparisons at the same time.

In an expression line, type the weighted sum for data set M divided by 36:

(54 + 76 + 98 + 1110 + 136 + 152) / 36

The value that Desmos returns is the mean of data set M.

In another expression line, type the corresponding expression for data set N:

(51 + 73 + 95 + 117 + 139 + 1511) / 36

Compare this output to the one from Step 1 to see which data set has the larger mean, then combine that with your hand-calculated medians to choose the answer choice that matches both comparisons.

You must compare both the mean and the median of data set M with those of data set N.

• Each data set has 36 values.
• Median: For 36 ordered numbers, the median is the average of the 18th and 19th values.
• Mean: Mean $= \dfrac{\text{sum of all values}}{36}$ for each data set.

We will use the frequencies to find where the 18th and 19th values fall (for the median) and to compute the total sum (for the mean).

Use cumulative frequencies to locate the 18th and 19th values for data set M.

For data set M:

• Value 5 occurs 4 times: positions 1–4
• Value 7 occurs 6 times: positions 5–10
• Value 9 occurs 8 times: positions 11–18
• Value 11 occurs 10 times: positions 19–28
• Value 13 occurs 6 times: positions 29–34
• Value 15 occurs 2 times: positions 35–36

The 18th value is the last 9, and the 19th value is the first 11.

So the median of M is

Now do the same for data set N.

For data set N:

• Value 5 occurs 1 time: position 1
• Value 7 occurs 3 times: positions 2–4
• Value 9 occurs 5 times: positions 5–9
• Value 11 occurs 7 times: positions 10–16
• Value 13 occurs 9 times: positions 17–25
• Value 15 occurs 11 times: positions 26–36

The 18th and 19th values both fall in the block of 13s, so both are 13.

Thus the median of N is

So the median of M is 10 and the median of N is 13, meaning the median of M is less than the median of N.

Use the frequencies to compute the total sum for each data set, then divide by 36.

Data set M:

So the mean of M is

Data set N:

So the mean of N is

Therefore, the mean of M is less than the mean of N.

From the work:

• Median: $\text{median}_M = 10$ and $\text{median}_N = 13$, so $\text{median}_M < \text{median}_N$.
• Mean: $\text{mean}_M \approx 9.78$ and $\text{mean}_N \approx 11.94$, so $\text{mean}_M < \text{mean}_N$.

So both the mean and the median of data set M are less than the corresponding values for data set N. The statement that says this is: “The mean and median of data set M are both less than the mean and median of data set N.”