Cumulative Frequency Worked ExamplesIGCSE Maths 0580

(a) Add the frequencies as you go. Each cumulative value belongs to the upper boundary of its class.

(b) With $100$ values the median is the $50$th. Locate it: the CF passes $50$ between $m = 60$ (CF $48$) and $m = 80$ (CF $78$).

Interpolate: we need $2$ more values out of the $30$ in that class, across a class of width $20$.

Evaluate.

(c) The lower quartile is the $25$th value, which lies between CF $18$ (at $40$) and CF $48$ (at $60$).

The upper quartile is the $75$th value, between CF $48$ (at $60$) and CF $78$ (at $80$).

Subtract for the interquartile range.

Find the cumulative frequency at $m = 30$ first. It lies halfway between $20$ (CF $5$) and $40$ (CF $18$).

Evaluate — half of the $13$ students in that class.

Now the cumulative frequency at $m = 70$, halfway between $60$ (CF $48$) and $80$ (CF $78$).

Evaluate — half of the $30$ students in that class.

Subtract to count only those between the two marks. This is the key step: a difference of cumulative frequencies.

The answer must be a whole number of students, so round.

The top $20\%$ of $100$ students is $20$ students. Convert this into a position from the bottom, which is what a cumulative frequency curve measures.

So you need the mark at which the cumulative frequency is $80$ — the $80$th percentile.

Locate it: CF is $78$ at $m = 80$ and $100$ at $m = 100$, so the answer is just above $80$.

Interpolate: we need $2$ more students out of the $22$ in that class, across a width of $20$.

Evaluate.

(a) Each cumulative frequency is plotted at the upper boundary of its class — the "$m \le$" value in the table.

The first class starts at $0$, and no student scored $0$ or less, so the curve also starts at the point $(0, 0)$.

Join all six points with a smooth, increasing curve. Do not draw bars or a single straight line of best fit.

(b) With $100$ students, read across from a cumulative frequency of $50$ to the curve, then down to the mark axis.

This agrees with the value $61.3$ found by interpolation in the first example — a reading within about half a small square is acceptable.

(c) Read up from a mark of $50$ to the curve, then across. This counts the students who scored $50$ marks or less.

Subtract from the total for the number who scored more than $50$.

(a) Build the cumulative frequencies.

The median is the $50$th value, lying between CF $30$ (at $40$) and CF $70$ (at $60$).

Evaluate — exactly halfway through the class.

The lower quartile is the $25$th value, between CF $8$ (at $20$) and CF $30$ (at $40$).

The upper quartile is the $75$th value, between CF $70$ (at $60$) and CF $92$ (at $80$).

Subtract for the interquartile range.

(b) Compare the averages first: the second group's median is lower.

Then compare the spreads: the second group's IQR is smaller.

Build cumulative frequencies against the upper boundaries. The unequal widths do not affect this step.

The median is the $50$th value, between CF $40$ (at $t = 10$) and CF $80$ (at $t = 20$). This class has width $10$.

Evaluate.

The lower quartile is the $25$th value, between CF $12$ (at $5$) and CF $40$ (at $10$). This class has width $5$.

Evaluate.

The upper quartile is the $75$th value, back in the class of width $10$.

Subtract for the interquartile range.

(a) Add each frequency to the running total as you move along the table.

Write each against the upper boundary of its class.

(b) Read the cumulative frequency at $h = 30$ directly from the table.

(c) With $40$ values the median is between the $20$th and $21$st. The CF reaches $13$ at $20\text{ cm}$ and $28$ at $30\text{ cm}$, so both fall in the third class.

(a) Each cross is plotted at the upper boundary of a class. Read the cumulative frequency at each boundary.

The frequency of a class is the difference between the cumulative frequencies at its two ends.

Check against the values already given: the last class is $80 - 70 = 10$, as printed, and the frequencies total $80$.

(b) For an estimated mean, use the midpoint of each class.

Multiply each midpoint by its frequency. This is the method mark.

Add the products.

Divide by the total frequency. Without a calculator, divide by $10$ and then by $8$.

(a) Each club has $60$ members, so read across at these cumulative frequencies. On a curve, use $\frac{n}{2}$ and $\frac{n}{4}$ — not $\frac{n+1}{2}$.

Club A (solid curve): across from $30$ to the curve and down gives the median.

Across from $15$ and from $45$ for the quartiles, then subtract.

Club B (dashed curve): the same three readings.

Its quartiles, and the interquartile range.

(b) Compare the averages, remembering that in a race a lower time is better.

Then compare the spreads.

(c) Go up from Club B's median, $33.2$ minutes, to the Club A curve and read across. This is the number of Club A members who took less time.

Write this as a percentage of the $60$ members of Club A.

(a) Read up from $t = 40$ to the curve and across. The curve gives the number who take $40$ minutes or less.

Subtract from the total for the number who take more than $40$ minutes.

(b) The first student chosen is one of the $24$ out of $120$. The same student cannot be picked twice, so for the second there are $23$ left out of $119$.

Multiply and simplify.

As a decimal, correct to $3$ significant figures:

(c) Half of the students is $60$. Read up from $t = 25$ to the curve and across.

Only about $47$ students travel for less than $25$ minutes, which is fewer than $60$, so the claim is false. (Equivalently, the median, read across from $60$, is about $28$ minutes — more than $25$.)

The shortfall is the number of extra students needed.

(a) Keep a running total of the frequencies.

(b) Plot each cumulative frequency at the upper boundary of its class. The last class is twice as wide, so its point goes at $t = 40$, not $t = 35$.

The first class starts at $t = 10$ and nobody took $10$ minutes or less, so the cumulative frequency at $t = 10$ is $0$. Plot $(10, 0)$ — the curve starts there, not at the origin.

Join the six points with a smooth, increasing curve.

(c)(i) With $60$ people, the median is read across from a cumulative frequency of $\frac{60}{2} = 30$. On the time axis each small square is $1$ minute.

(c)(ii) The lower quartile is read across from $\frac{60}{4} = 15$. The answer is a time, read from the horizontal axis.

Check the scales first. Across, $5$ small squares make $1$ hour, so each is $0.2$ hours. Up, $5$ small squares make $20$, so each is $4$.

(a) The quartiles are at a quarter and three quarters of the total.

Read across from $40$ and from $120$ to the curve, then down to the time axis.

Subtract the two times. Subtracting the cumulative frequencies, $120 - 40 = 80$, is not the interquartile range.

(b) Convert $65\%$ into a number of teenagers. Without a calculator: $10\%$ of $160$ is $16$, so $60\%$ is $96$ and $5\%$ is $8$.

Read across from a cumulative frequency of $104$ to the curve, then down.