There are 51 data points. For an odd number of data points, which position (1st, 2nd, 3rd, ...) is the median? Use the cumulative frequencies to find which temperature is at that position.

When you add a new value of 0°C, it will go at the very beginning when the data are ordered. How does this change the positions of the existing values? Which original positions now become the 26th and 27th values in the new list of 52 data points?

Once you know which temperatures are at the middle positions before and after adding 0, decide whether the median changes. Then compare the old and new means to see whether the mean increased, decreased, or stayed the same.

In Desmos, define the temperature and frequency lists:

• Type T = [18,19,20,21,22,23,24,25]
• Type F = [4,5,6,9,10,8,5,4] Then on a new line, type mean0 = sum(T*F) / sum(F) and look at the decimal value of mean0 for the original mean.

On a new line, type mean1 = sum(T*F) / (sum(F) + 1). This uses the same total sum but increases the number of data points by 1. Compare the values of mean0 and mean1 in Desmos to see whether the mean increased or decreased.

If you want to check the median using Desmos, you can build the full list of 51 temperatures, for example: L0 = [18,18,18,18,19,19,19,19,19,20,20,20,20,20,20,...] continuing according to the frequencies, and then type median(L0) to see the original median. For the new data set, define L1 = [0, L0] or manually add 0 at the start, and then type median(L1) to see the new median; compare the two medians.

To find the original mean, first compute the total of all 51 temperatures.

Multiply each temperature by its frequency and add:

There are 51 days, so the original mean is

Keep this value in mind for comparison later.

In the new data set, you add one more value, 0°C.

• The new sum is $1100 + 0 = 1100$ (the total does not change).
• The new number of values is $51 + 1 = 52$.

So the new mean is

You will compare this value to the original mean in the final step.

First, find the original median.

With 51 data points (an odd number), the median is the value in position

when the data are listed in order.

Use cumulative frequencies to locate the 26th value:

• 18°C: positions 1–4
• 19°C: positions 5–9
• 20°C: positions 10–15
• 21°C: positions 16–24
• 22°C: positions 25–34

So the 26th value is 22°C. That is the original median.

Now, add one more value of 0°C. This becomes the very first value when the data are ordered.

• The new data set has 52 values (an even number).
• The median is the average of the 26th and 27th values.

Because the 0°C reading is inserted at the very beginning, every original value moves one position to the right:

• New position 2 = old position 1
• New position 3 = old position 2
• ...
• New position 26 = old position 25
• New position 27 = old position 26

From the original data, both the 25th and 26th values are 22°C (they are in the 22°C group, positions 25–34). So, in the new data set, the 26th and 27th values are both 22°C, and their average is also 22°C.

So the original median is 22°C, and the new median is also 22°C.

From the calculations:

• Original mean: $\dfrac{1100}{51} \approx 21.6$
• New mean: $\dfrac{1100}{52} \approx 21.2$

Since the total stayed the same but there are more data points, the new mean is less than the original mean.

For the median, we found that both the original and new medians are 22°C, so the medians are equal.

Therefore, the correct comparison is: The mean of the new data set is less than the mean of the original data set, and the medians of the two data sets are equal.