KCSE 2025 Maths P1 Q12 — Histogram Frequencies from Bar Heights

KCSE 2025 Form 4 Statistics

Published

The Question

“A histogram represents marks scored in a test using three classes: 10 to 14, 15 to 19, and 20 to 29. The heights of the rectangular bars are 1.4, 2, and 1.5 respectively. Only the frequency of the first class, which is 7, is given. Calculate the two missing frequencies for the classes 15 to 19 and 20 to 29.”

1

Recall what the bar height means

In a histogram the height of each bar is the frequency density, not the frequency itself. Frequency density is the frequency divided by the class width. Rearranging that relationship gives a formula for the frequency, which is what we actually want to find.

Frequency density=FrequencyClass width\text{Frequency density} = \frac{\text{Frequency}}{\text{Class width}}
⇒Frequency=Height×Class width\Rightarrow \text{Frequency} = \text{Height} \times \text{Class width}
2

Find the true class widths using class boundaries

The class width comes from the class boundaries, not the stated limits. The class 10 to 14 really stretches from 9.5 to 14.5, so its width is 5. The class 15 to 19 runs from 14.5 to 19.5, also a width of 5. But 20 to 29 runs from 19.5 to 29.5, giving a width of 10. Getting this last width right is the key to the whole question.

10–14: 14.5−9.5=510\text{--}14:\ 14.5 - 9.5 = 5
15–19: 19.5−14.5=515\text{--}19:\ 19.5 - 14.5 = 5
20–29: 29.5−19.5=1020\text{--}29:\ 29.5 - 19.5 = 10
3

Check the method on the known class

Before trusting the formula, test it on the class whose frequency we already know. Multiplying the height of the first bar by its width should reproduce the given frequency of 7. It does, which confirms we are using the correct widths and the correct formula.

1.4×5=7✓1.4 \times 5 = 7 \quad \checkmark
4

Compute the two missing frequencies

Now apply the same formula to the other two classes. Multiply each bar height by its class width. The class 15 to 19 has height 2 and width 5, and the class 20 to 29 has height 1.5 and width 10.

15–19: 2×5=1015\text{--}19:\ 2 \times 5 = 10
20–29: 1.5×10=1520\text{--}29:\ 1.5 \times 10 = 15

Final Result

The missing frequencies are 10 for the class 15 to 19 and 15 for the class 20 to 29.

Why this method works

A histogram uses area, not height, to represent frequency, so bars of unequal width can still compare fairly. That is why the vertical axis measures frequency density, the frequency spread out over each unit of class width. Multiplying a bar's height by its width simply recovers the area of the bar, which equals the frequency. The class 20 to 29 is twice as wide as the others, so even with a modest height of 1.5 it holds the largest frequency once you account for that extra width.

✓

Verify each result by dividing frequency by width to recover the height: 10 divided by 5 is 2, and 15 divided by 10 is 1.5, matching the given bar heights.