Mathematics 9709/62 — February/March 2018

Cambridge AS Level · Probability & Statistics 1 · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Representation of Data · Permutations and Combinations · Discrete Random Variables · The Normal Distribution · Probability

Q14MRepresentation of DataFree sample

There are 900 students in a certain year-group. An identical puzzle is given to each student and the time taken, $t$ minutes, to complete the puzzle is recorded. These times are summarised in the following frequency table.

Time taken, $t$ minutes	$t \le 3$	$3 < t \le 4$	$4 < t \le 5$	$5 < t \le 6$	$6 < t \le 8$	$8 < t \le 10$	$10 < t \le 14$
Frequency	120	180	200	160	110	80	50

On the grid, draw a cumulative frequency graph to represent the data. Use your graph to estimate the median time taken by these students to complete the puzzle.

cumulative_frequency_grid

DifficultyMedium-Easy

Worked solution

Approach

Calculate the cumulative frequencies by adding the frequencies progressively. Plot these against the upper class boundaries on a coordinate grid. To estimate the median, locate the cumulative frequency corresponding to half the total number of students on the y-axis, read across to the curve, and then down to the x-axis.

Working

First, compute the cumulative frequencies:

\begin{aligned} \text{At } t = 0 &: \quad cf = 0 \\ \text{At } t = 3 &: \quad cf = 120 \\ \text{At } t = 4 &: \quad cf = 120 + 180 = 300 \\ \text{At } t = 5 &: \quad cf = 300 + 200 = 500 \\ \text{At } t = 6 &: \quad cf = 500 + 160 = 660 \\ \text{At } t = 8 &: \quad cf = 660 + 110 = 770 \\ \text{At } t = 10 &: \quad cf = 770 + 80 = 850 \\ \text{At } t = 14 &: \quad cf = 850 + 50 = 900 \end{aligned}

Plot the points $(0, 0)$ , $(3, 120)$ , $(4, 300)$ , $(5, 500)$ , $(6, 660)$ , $(8, 770)$ , $(10, 850)$ , $(14, 900)$ on the grid with linear scales, joining them with a smooth curve or straight lines.

To estimate the median:

Total number of students: $n = 900$
Median position: $\frac{n}{2} = \frac{900}{2} = 450$

Locate $cf = 450$ on the y-axis, draw a horizontal line to the cumulative frequency curve, then draw a vertical line down to the x-axis to read the median time.

From the graph, the median time is approximately $4.8$ minutes.

(As a check, linear interpolation between $(4, 300)$ and $(5, 500)$ gives $t = 4 + \frac{450 - 300}{500 - 300} \times (5 - 4) = 4 + \frac{150}{200} = 4.75$ , which is consistent with the graph reading of approximately $4.8$ .)

Answer

4.8

minutes

Final answer

4.8 minutes

Detailed explanation

Walkthrough

The question asks us to represent grouped time data using a cumulative frequency graph and then use that graph to estimate the median.

Step 1: Compute cumulative frequencies.
We start with $cf = 0$ at $t = 0$ . Then we add each frequency to the running total:

At $t = 3$ : $cf = 120$
At $t = 4$ : $cf = 120 + 180 = 300$
At $t = 5$ : $cf = 300 + 200 = 500$
At $t = 6$ : $cf = 500 + 160 = 660$
At $t = 8$ : $cf = 660 + 110 = 770$
At $t = 10$ : $cf = 770 + 80 = 850$
At $t = 14$ : $cf = 850 + 50 = 900$

These values are plotted against the upper class boundaries on the graph.

Step 2: Draw the cumulative frequency graph.
Plot the points $(0, 0)$ , $(3, 120)$ , $(4, 300)$ , $(5, 500)$ , $(6, 660)$ , $(8, 770)$ , $(10, 850)$ , $(14, 900)$ on a coordinate grid with linear scales. The x-axis represents time in minutes and the y-axis represents cumulative frequency. Join the points with a smooth curve or straight lines — both are acceptable.

Step 3: Estimate the median.
The median is the value that splits the data into two equal halves. With $n = 900$ students, the median position is $\frac{900}{2} = 450$ . On the cumulative frequency graph, find $cf = 450$ on the y-axis, draw a horizontal line to intersect the curve, then drop a vertical line down to the x-axis. The x-value at this intersection is the estimated median.

Reading from the graph, the median is approximately $4.8$ minutes. This is consistent with linear interpolation between the points $(4, 300)$ and $(5, 500)$ , which gives $4.75$ .

Key Takeaways

Cumulative frequency is found by adding frequencies progressively up to each upper class boundary.
A cumulative frequency graph is plotted with upper class boundaries on the x-axis and cumulative frequencies on the y-axis, starting at $(0, 0)$ .
The median is estimated by finding $\frac{n}{2}$ on the cumulative frequency axis, reading across to the curve, and then down to the x-axis.
Graph readings are approximate; linear interpolation between the bounding points gives a more precise value.

Common Mistakes

Forgetting to include the starting point $(0, 0)$ on the cumulative frequency graph.
Plotting cumulative frequencies against lower class boundaries instead of upper class boundaries.
Using the wrong position for the median: the median position is $\frac{n}{2}$ , not $\frac{n+1}{2}$ (which is used for discrete data).
Reading the median incorrectly from the graph by not drawing horizontal and vertical lines properly.

Things to Be Careful About

Always start the cumulative frequency graph at $(0, 0)$ .
Use linear scales on both axes.
Label both axes clearly: "TIME, IN MINUTES" on the x-axis and "CUMULATIVE FREQUENCY" on the y-axis.
The median value read from the graph is an estimate; the mark scheme accepts a range of approximately $4.7 \leq m < 4.9$ .
Ensure all points are plotted correctly — a single error in cumulative frequency will shift the entire curve.

Techniques used

calculate cumulative frequenciesplot cumulative frequency graphread median from cumulative frequency graph

The rest of this paper

7 more questions

Q2Permutations and Combinations4M
Q3Probability6M
Q4Discrete Random Variables6M
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Q6Permutations and Combinations6M
Q7The Normal Distribution8M
Q8Discrete Random Variables · The Normal Distribution10M

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