The speeds, $v\,\text{km/h}$, of 80 vehicles travelling along a road were recorded. The results are shown in the table. Speed ($v\,\text{km/h}$) Frequency --- --- $30 < v \leq 40$ 10 $40 < v \leq 50$ 18 $50 < v \leq 60$ 27 $60 < v \leq 70$ 19 $70 < v \leq 80$ 6 (a) Calculate an estimate of the mean speed of the vehicles. \_\_\_\_\_\_ $\text{km/h}$ (b) Draw the cumulative frequency diagram. (c) Use your cumulative frequency diagram to find an estimate for (i) the median,…

Mathematics (Syllabus D) 4024/21 — May/June 2020

Cambridge O-Level · Calculator · worked solutions for every part, with the mark scheme

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Topics Algebra and Graphs · Number · Mensuration · Trigonometry · Geometry · Statistics · +2 more

Q1StatisticsFree sample

The speeds, $v\,\text{km/h}$ , of 80 vehicles travelling along a road were recorded.
The results are shown in the table.

Speed ( $v\,\text{km/h}$ )	Frequency
$30 < v \leq 40$	10
$40 < v \leq 50$	18
$50 < v \leq 60$	27
$60 < v \leq 70$	19
$70 < v \leq 80$	6

(a)

Calculate an estimate of the mean speed of the vehicles.

______ $\text{km/h}$

DifficultyMedium-Easy

Worked solution

Approach

For grouped data, estimate the mean by using the midpoint of each class interval as the representative value, multiplying by the frequency, summing, and dividing by the total frequency.

Working

The midpoints of the class intervals are:

Speed (km/h)	Midpoint ( $x$ )	Frequency ( $f$ )	$f \times x$
$30 < v \leq 40$	$35$	$10$	$350$
$40 < v \leq 50$	$45$	$18$	$810$
$50 < v \leq 60$	$55$	$27$	$1485$
$60 < v \leq 70$	$65$	$19$	$1235$
$70 < v \leq 80$	$75$	$6$	$450$

\begin{aligned} \sum f x &= 35 \times 10 + 45 \times 18 + 55 \times 27 + 65 \times 19 + 75 \times 6 \\ &= 350 + 810 + 1485 + 1235 + 450 \\ &= 4330 \end{aligned}

The total frequency is:

\sum f = 10 + 18 + 27 + 19 + 6 = 80

The estimated mean speed is:

\text{Mean} = \frac{\sum f x}{\sum f} = \frac{4330}{80} = 54.125

Answer

54.1 \text{ km/h} \quad (\text{accept } 54.125 \text{ or } 54.12 \text{ to } 54.13)

Final answer

54.1

Detailed explanation

Walkthrough

When data is given in class intervals rather than individual values, we cannot find the exact mean. Instead, we estimate it by assuming every value in a class is equal to the midpoint of that class. The midpoint of $30 < v \leq 40$ is $\frac{30+40}{2} = 35$ , and similarly for the other classes. Multiplying each midpoint by its frequency gives an estimated total for that class. Adding all these gives $4330$ , and dividing by the total number of vehicles ( $80$ ) gives the estimated mean speed of $54.125$ km/h. The mark scheme accepts answers rounded to 1 or 3 significant figures: $54.1$ , $54.12$ , or $54.13$ .

Key Takeaways

For grouped data, the estimated mean uses class midpoints.
Always verify that the sum of frequencies equals the stated total before dividing.
Roundings to 1 or 3 significant figures are accepted unless an exact form is required.

Common Mistakes

Using the lower or upper class boundary instead of the midpoint.
Forgetting to divide by the total frequency ( $80$ ) and instead dividing by the number of classes ( $5$ ).
Arithmetic errors in multiplying midpoints by frequencies or in summing.
Not showing working ("nfww" means no follow-through without working shown).

Things to Be Careful About

The mark scheme explicitly states "nfww" (no follow-through without working), so all steps must be shown.
The answer $54.125$ is exact; $54.1$ , $54.12$ , and $54.13$ are all accepted rounded forms.
Always double-check the sum of frequencies equals $80$ ; if it does not, the mean calculation is invalid.

Techniques used

find the midpoint of each class intervalmultiply each midpoint by its frequencydivide the total by the sum of frequencies

(b)

Draw the cumulative frequency diagram.

Fig. 1

DifficultyMedium-Easy

Worked solution

Approach

Cumulative frequency is found by adding each frequency to the running total. Plot the cumulative frequency against the upper class boundary of each interval, then join the points with a smooth curve (an ogive). The curve should start at the lower boundary of the first class with cumulative frequency $0$ .

Working

Upper class boundary ( $v$ )	Frequency	Cumulative frequency
$40$	$10$	$10$
$50$	$18$	$28$
$60$	$27$	$55$
$70$	$19$	$74$
$80$	$6$	$80$

Also include the starting point $(30, 0)$ .

Answer

Plot the points $(30, 0)$ , $(40, 10)$ , $(50, 28)$ , $(60, 55)$ , $(70, 74)$ , $(80, 80)$ and draw a smooth S-shaped curve through them, starting at $(30, 0)$ and ending at $(80, 80)$ .

Final answer

Points plotted: (30, 0), (40, 10), (50, 28), (60, 55), (70, 74), (80, 80) with a smooth S-curve through them

Detailed explanation

Walkthrough

Cumulative frequency at a given upper boundary is the total number of vehicles with speed less than or equal to that boundary. Starting from the first class $30 < v \leq 40$ with frequency $10$ , the cumulative frequency at $v = 40$ is $10$ . Adding the next frequency $18$ gives $28$ at $v = 50$ . Continuing: $28 + 27 = 55$ at $v = 60$ ; $55 + 19 = 74$ at $v = 70$ ; $74 + 6 = 80$ at $v = 80$ . We also include $(30, 0)$ as the starting point since no vehicles have speed less than $30$ km/h. Plot these six points on the grid in Fig. 1 and join them with a smooth curve. The curve should be S-shaped (ogive), rising slowly at first, then steeply in the middle, then flattening out as it approaches the total frequency of $80$ .

Key Takeaways

Cumulative frequency is a running total of frequencies.
Plot cumulative frequency against the upper class boundary, not the midpoint.
Always include the starting point at the lower boundary of the first class with cumulative frequency $0$ .
The curve must be smooth, not a series of straight line segments.

Common Mistakes

Plotting against the midpoint of each class instead of the upper class boundary.
Forgetting the starting point $(30, 0)$ .
Drawing straight lines between points instead of a smooth curve.
Miscalculating the running total (e.g., $10 + 18 = 27$ instead of $28$ ).
Not reaching a final cumulative frequency of $80$ at $v = 80$ .

Things to Be Careful About

The mark scheme awards B2 for 4 or 5 correct points plotted, or B1 for correct cumulative frequencies stated. All points should be accurate to within half a grid square.
The curve must be smooth; jagged or angular curves may lose marks.
Axes are already labelled in Fig. 1; do not relabel them.

Techniques used

compute cumulative frequencies at upper class boundariesplot points on the given griddraw a smooth cumulative frequency curve

(c)

Use your cumulative frequency diagram to find an estimate for

(i)

the median,

______ $\text{km/h}$

DifficultyMedium-Easy

Worked solution

Approach

The median is the value at the $\frac{n}{2}$ position in the ordered data. For $n = 80$ , the median position is $40$ . Draw a horizontal line from cumulative frequency $40$ to the curve, then drop a vertical line to the speed axis to read the median.

Working

Median position $= \frac{80}{2} = 40$ .

From the cumulative frequency diagram, at cumulative frequency $40$ , the corresponding speed is approximately $54$ km/h.

Answer

54 \text{ km/h} \quad (\text{accept } 53 \text{ to } 55)

Final answer

Detailed explanation

Walkthrough

The median divides the data into two equal halves. With $80$ vehicles, the median is the average of the $40$ th and $41$ st values, which we approximate by reading the value at cumulative frequency $40$ from the diagram. On the cumulative frequency curve, locate $40$ on the vertical axis, draw a horizontal line to the curve, then draw a vertical line down to the horizontal axis. The reading is approximately $54$ km/h. The mark scheme accepts any value from $53$ to $55$ km/h, allowing for normal reading error from a hand-drawn diagram.

Key Takeaways

The median position for $n$ values is $\frac{n}{2}$ .
Reading from a cumulative frequency diagram involves horizontal then vertical lines from the position to the curve to the axis.
Allow for reading error; the mark scheme gives a tolerance range.

Common Mistakes

Using the wrong position (e.g., $\frac{n}{4}$ for the median instead of $\frac{n}{2}$ ).
Reading the cumulative frequency value instead of the speed value from the horizontal axis.
Drawing horizontal/vertical lines in the wrong direction.
Not allowing for reading error from a hand-drawn curve.

Things to Be Careful About

The answer is read from the student's own diagram in part (b), so follow-through marks are awarded based on their diagram.
The mark scheme accepts $53$ to $55$ km/h, reflecting normal reading tolerance.
Always read the correct axis: speed is on the horizontal axis, cumulative frequency on the vertical.

Techniques used

find the median position from the total frequencyread the median value from the cumulative frequency diagram

(ii)

the interquartile range.

______ $\text{km/h}$

DifficultyMedium

Worked solution

Approach

The lower quartile (LQ) is at the $\frac{n}{4}$ position and the upper quartile (UQ) is at the $\frac{3n}{4}$ position. Read both values from the cumulative frequency diagram and subtract: $\text{IQR} = \text{UQ} - \text{LQ}$ .

Working

Lower quartile position $= \frac{80}{4} = 20$ .

From the cumulative frequency diagram, at cumulative frequency $20$ , the corresponding speed is approximately $46$ km/h.

Upper quartile position $= \frac{3 \times 80}{4} = 60$ .

From the cumulative frequency diagram, at cumulative frequency $60$ , the corresponding speed is approximately $63$ km/h.

\text{Interquartile range} = \text{UQ} - \text{LQ} = 63 - 46 = 17 \text{ km/h}

Answer

17 \text{ km/h} \quad (\text{accept } 14 \text{ to } 18)

Final answer

Detailed explanation

Walkthrough

The interquartile range (IQR) measures the spread of the middle $50\%$ of the data. It is calculated as $\text{UQ} - \text{LQ}$ . For $n = 80$ :

Lower quartile position: $\frac{80}{4} = 20$ . Reading from the diagram at cumulative frequency $20$ , the speed is approximately $46$ km/h (between $v = 40$ where CF = $10$ and $v = 50$ where CF = $28$ ).
Upper quartile position: $\frac{3 \times 80}{4} = 60$ . Reading from the diagram at cumulative frequency $60$ , the speed is approximately $63$ km/h (between $v = 60$ where CF = $55$ and $v = 70$ where CF = $74$ ).

The IQR is $63 - 46 = 17$ km/h. The mark scheme accepts any answer from $14$ to $18$ km/h, allowing for reading error from the hand-drawn diagram. One mark is awarded for correctly identifying LQ (45 to 47) or UQ (61 to 63), and the second mark for the correct IQR calculation.

Key Takeaways

Lower quartile is at position $\frac{n}{4}$ , upper quartile at $\frac{3n}{4}$ .
IQR = UQ - LQ; always subtract the smaller from the larger.
Reading from a cumulative frequency diagram requires careful horizontal and vertical line construction.
Allow for reading tolerance; the mark scheme gives a range.

Common Mistakes

Using the wrong positions (e.g., $\frac{n}{2}$ for both quartiles).
Subtracting in the wrong order (LQ - UQ gives a negative value).
Reading the cumulative frequency value instead of the speed value.
Not allowing for reading error from a hand-drawn curve.
Forgetting that the mark scheme awards partial credit for correct quartile readings.

Things to Be Careful About

The answer is read from the student's own diagram in part (b), so follow-through marks are awarded based on their diagram.
The mark scheme accepts IQR from $14$ to $18$ km/h, reflecting normal reading tolerance.
One mark is given for a correct LQ (45 to 47) or UQ (61 to 63), even if the other is wrong.
Always show the subtraction: IQR = UQ - LQ.

Techniques used

find the lower quartile positionfind the upper quartile positionread quartile values from the cumulative frequency diagramsubtract lower quartile from upper quartile

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