KJSEA 2025 Maths Q39 — Statistics: Frequency Table, Mean & Median

KJSEA 2025 Grade 9 Statistics

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The Question

“The marks scored by 20 learners in a test were recorded. (a) Prepare a frequency table for the marks. (b) Using the table, find the mean mark and the median mark. The marks are: 15 (twice), 17 (five times), 18 (three times), 19 (six times), 22 (twice), 29 (once) and 33 (once).”

1

Tally the marks into a frequency table

Go through the recorded marks and count how many learners scored each distinct mark. Writing this as a frequency table organises the raw scores so they are easy to work with. Always check that the frequencies add up to the total number of learners, which is 20 here, so nothing is missed or double-counted.

Mark (x)Frequency (f)152175183196222291331Total20\begin{array}{c|c} \text{Mark } (x) & \text{Frequency } (f) \\ \hline 15 & 2 \\ 17 & 5 \\ 18 & 3 \\ 19 & 6 \\ 22 & 2 \\ 29 & 1 \\ 33 & 1 \\ \hline \text{Total} & 20 \end{array}

The frequencies 2 + 5 + 3 + 6 + 2 + 1 + 1 give 20, confirming all learners are accounted for.

2

Multiply each mark by its frequency

To find the mean you need the total of all the marks. Instead of adding every score one by one, multiply each mark by how many learners scored it. This gives the combined marks for each row, and adding those row totals gives the overall sum of marks.

15×2=3015 \times 2 = 30
17×5=8517 \times 5 = 85
18×3=5418 \times 3 = 54
19×6=11419 \times 6 = 114
22×2=4422 \times 2 = 44
29×1=2929 \times 1 = 29
33×1=3333 \times 1 = 33
3

Find the mean

Add all the row totals to get the sum of every mark, then divide by the number of learners. The mean is the sum of the marks divided by the frequency total, which spreads the total evenly across all 20 learners.

∑fx=30+85+54+114+44+29+33=389\sum fx = 30 + 85 + 54 + 114 + 44 + 29 + 33 = 389
Mean=∑fx∑f=38920=19.45\text{Mean} = \frac{\sum fx}{\sum f} = \frac{389}{20} = 19.45
4

Locate the median position

The median is the middle value when the marks are arranged in order. Because there are 20 values, an even number, there is no single middle score, so the median is the average of the 10th and 11th values. Use the running (cumulative) frequencies to find which marks sit in those positions.

Positions of median=202 and 202+1=10th and 11th\text{Positions of median} = \frac{20}{2} \text{ and } \frac{20}{2}+1 = 10^{\text{th}} \text{ and } 11^{\text{th}}
Cumulative: 15 ⁣:2,  17 ⁣:7,  18 ⁣:10,  19 ⁣:16\text{Cumulative: } 15\!:2,\; 17\!:7,\; 18\!:10,\; 19\!:16

The 8th to 10th values are 18, and the 11th to 16th values are 19.

5

Find the median

The 10th value is 18 and the 11th value is 19. The median is the average of these two middle marks.

Median=18+192=372=18.5\text{Median} = \frac{18 + 19}{2} = \frac{37}{2} = 18.5

Final Result

The mean mark is 19.45 and the median mark is 18.5.

Why this method works

The mean uses every mark, so multiplying each score by its frequency and dividing by the total learners is just a quick way of adding all 20 scores and sharing them equally. The median, on the other hand, depends only on position in the ordered list. With an even count of 20 values the true middle lies between the 10th and 11th scores, so averaging those two gives the central value. The cumulative frequencies tell you exactly where each mark falls in the ordered list without writing out all 20 numbers.

✓

Verify the mean: 389 divided by 20 equals 19.45, and the frequencies sum to 20, so both the total and the median positions are consistent.