Statistical Charts and Diagrams Worked ExamplesIGCSE Maths 0580

(a) Frequency density is frequency divided by class width — this is what the height of each histogram bar represents.

The first two classes each have width $10$.

The third class has width $20$ and the fourth has width $40$.

(b) The interval $25 < x \le 35$ sits entirely inside the class $20 < x \le 40$, which holds $36$ parcels across a width of $20$.

Assume the parcels are spread evenly through the class, and take the matching fraction.

Alternatively, multiply the frequency density by the width — the same calculation.

(a) A full circle of $360^{\circ}$ represents all $720$ people. Take the matching fraction of the total.

Simplify first: $\frac{720}{360} = 2$, so each degree represents $2$ people.

(b) Now go the other way — take the fraction of people and convert it to a fraction of the circle.

Simplify: $\frac{90}{720} = \frac{1}{8}$.

(c) A pie chart shows only the proportions in each category. The same chart would serve for $72$ people or $7200$.

(a) Use the tens digit as the stem and the units digit as the leaf. The data is already in order, so the leaves go straight in.

A key is compulsory — without it the diagram has no meaning and a mark is lost.

Check the leaf count matches the number of data values: $2 + 4 + 3 + 4 + 1$.

(b) The mode is the value appearing most often — the repeated leaf $4$ on stem $3$.

The range is largest minus smallest, both read off the ends of the diagram.

(c) With $14$ values the median is the mean of the $7$th and $8$th. Count leaves: the $7$th is $42$ and the $8$th is $45$.

(a) Sales rise consistently as the temperature rises, and the points lie very close to a straight line.

(b) Substitute $x = 23$ into the equation of the line of best fit.

Evaluate.

(c) Compare the ninth day with what the line predicts for $21^{\circ}\text{C}$.

The actual figure of $95$ is far below the predicted $180$ — the point lies well away from the trend.

Give a sensible reason in context: something other than temperature affected sales that day — rain, a closure, or a recording error.

(a) Reverse the frequency-density formula: frequency $=$ density $\times$ class width.

Apply it to each class, reading the widths from the class limits.

Add for the total frequency.

(b) Find the midpoint of each class. The last runs from $20$ to $50$, so its midpoint is $35$.

Multiply each midpoint by its frequency.

Add these products.

Divide by the total frequency.

(c) The median is the $56$th value. Cumulative frequencies are $10$, $46$, $81$, $111$, so the $56$th falls in the third class.

Build the cumulative frequencies against the upper class boundaries.

The lower quartile is at position $\frac{111}{4} = 27.75$.

That lies between CF $10$ (at $x = 4$) and CF $46$ (at $x = 10$) — a class of width $6$ containing $36$ values.

Evaluate.

The upper quartile is at position $\frac{3 \times 111}{4} = 83.25$, which lies between CF $81$ (at $20$) and CF $111$ (at $50$) — a class of width $30$ containing $30$ values.

Evaluate.

Subtract for the interquartile range.

Find how many degrees represent one student by dividing the full circle by the total.

Multiply each frequency by $6$. Football first.

Then cricket and tennis.

And swimming.

Check the four angles total $360^{\circ}$.

Check the scale using a class you already know: $0 < t \le 20$ has frequency $12$ and width $20$, and its bar is at $0.6$.

(a) Frequency $=$ frequency density $\times$ class width. The bar for $30 < t \le 40$ is at $2.2$ and the class is $10$ wide.

The bar for $60 < t \le 100$ is at $0.2$, and this class is $40$ wide.

(b) The missing bar is for $40 < t \le 60$, which has frequency $18$ and width $20$.

So draw a bar from $t = 40$ to $t = 60$ with height $0.9$, touching the bars on each side — a histogram has no gaps.

(c) "More than $50$" covers the top half of the $40 < t \le 60$ class and all of the last class. Assume the $18$ visitors are spread evenly through their class.

Add the whole of the last class.

(a) In a composite bar chart each section starts where the one below ends. For Year $10$, the bus section runs from $25$ up to $60$, so subtract.

(b) Year $11$ has $120$ students in total (the top of the bar). The car section runs from $90$ to $120$.

Take the matching fraction of $360^{\circ}$.

(c) Read the walk sections: $30$ out of $90$ in Year $9$, and $35$ out of $120$ in Year $11$. Leila is right that $35 > 30$, but the year groups are different sizes, so compare proportions.

Write both over a common denominator to compare without a calculator.

A greater proportion of Year $9$ walk, so walking is less popular in Year $11$, even though more Year $11$ students walk. Leila is not correct.

(a) Use the known class to fix the scale. The class $15 < a \le 20$ has width $5$ and frequency $35$.

That bar is $7$ large squares tall, so each large square on the vertical axis is $1$ unit of frequency density (and each small square is $0.25$). Now read the other heights.

Multiply each density by its class width: $5$, $5$, $10$, $20$ and $20$.

Add the frequencies to show the total.

(b) From $18$ to $20$ is $2$ years of the class with density $7$; from $20$ to $25$ is $5$ years of the class with density $5$. Area gives frequency.

(c) Use the class midpoints.

Multiply each midpoint by its frequency and add.

Divide by the total frequency and give the answer to $3$ significant figures.

(a) Convert the fraction to a percentage by making the denominator $100$.

Every bar in a percentage composite bar chart reaches $100\%$. The rest of the April sales were at the box office.

So draw the April bar from $0$ to $68$ shaded as online, and continue it from $68$ to $100$ shaded as box office.

(b) In February the online section ends at $72$, so the box office section runs from $72$ to $100$.

Find $28\%$ of $3 500 000 without a calculator: $1\%$ is $35 000, then multiply by $28$.

Check with the online share: $72\%$ is $72 \times 35\,000 = 2\,520\,000$, and the two parts add back to the total.

Bar heights in a histogram are proportional to frequency density, so find it for every class. The widths are $10$, $5$, $15$ and $30$.

The bar with frequency density $6$ is $12\text{ cm}$ tall, which fixes the scale.

Multiply each of the other frequency densities by $2$.

(a) Count how many masses are still missing.

The lightest dog is $16\text{ kg}$, so add the leaf $6$ to stem $1$, before the $8$.

The range fixes the heaviest dog. Add the leaf $3$ to stem $4$.

Three masses are left. $24$ already appears twice, so for $25$ to be the only mode it must appear more than twice — that uses all three.

Write the completed diagram with the leaves in order, and check the count: $2 + 6 + 3 + 2 = 13$.

(b) With $13$ dogs the median is the $\frac{13 + 1}{2} = 7$th value. Counting leaves: $16, 18, 24, 24, 25, 25, 25$.