Data presentation, measures of location and spread, outliers – A-level Maths exam questions with mark scheme

1StandardCalculator6 marks

The table summarises the times, $t$ minutes, that 50 customers spent in a shop.

Time, $t$ (minutes)	Frequency
$0\le t<10$	5
$10\le t<20$	12
$20\le t<30$	18
$30\le t<40$	10
$40\le t<60$	5

(a) Use linear interpolation to estimate the median time. [2]

(b) Estimate the mean and the standard deviation of the times. [4]

(a) $20+\frac{25-17}{18}\times 10$ (allow use of 25.5: $20+\frac{8.5}{18}\times10$)M1
(a) awrt 24.4 (awrt 24.7 if 25.5 used)A1
(b) uses midpoints 5, 15, 25, 35, 50 with $\frac{\sum fx}{\sum f}=\frac{1255}{50}$M1
(b) mean $=25.1$A1
(b) $\sqrt{\frac{\sum fx^2}{50}-25.1^2}$ with $\sum fx^2=38825$M1
(b) awrt 12.1 (allow awrt 12.2 using $n-1$)A1
Answer: median $\approx 24.4$ min; mean $=25.1$ min; s.d. $\approx 12.1$ min
Examiner tip: The last class is 20 minutes wide, so its midpoint is 50, not 45 – check every class width before using midpoints.

2StandardCalculator4 marks

The histogram shows the times taken by a group of people to complete a puzzle. The vertical axis is not labelled with values.

The bar for the class $10\le t<20$ represents 24 people.

(a) Find the number of people who took between 5 and 10 minutes. [2]

(b) Estimate the number of people who took between 25 and 40 minutes. [2]

(a) area $\propto$ frequency: e.g. $10\times 6$ squares $\leftrightarrow 24$, so 1 (minute × square) $\leftrightarrow 0.4$ people, or frequency density of $10\le t<20$ bar $=2.4$ giving scale 0.4 per square; applied to $5\times 8$M1
(a) 16A1
(b) $\frac{1}{2}\times 12+\frac{10}{20}\times 8$ (or equivalent area from 25 to 40 using a correct scale)M1
(b) 10A1
Answer: (a) 16; (b) 10
Examiner tip: Frequency is the area of a bar, not its height – find what one square of area is worth first.

3StandardCalculator6 marks

Sam uses the large data set to study the daily mean temperature, $x$ °C, at one UK weather station for the 31 days of a month in 2015. He finds $$\sum x = 405.3 \qquad \sum x^2 = 5392.15$$

(a) Calculate the mean and the standard deviation of the daily mean temperatures. [4]

The lower quartile is 11.8 °C and the upper quartile is 14.3 °C. An outlier is defined as a value more than $1.5\times$ the interquartile range below the lower quartile or above the upper quartile.

(b) The lowest temperature recorded was 8.6 °C and the highest was 18.4 °C. Determine whether either of these values is an outlier. Show your working. [2]

(a) $\frac{405.3}{31}$M1
(a) awrt 13.1 °CA1
(a) $\sqrt{\frac{5392.15}{31}-\left(\frac{405.3}{31}\right)^2}$M1
(a) awrt 1.73 °C (allow awrt 1.76 using $s$)A1
(b) $11.8-1.5\times2.5$ or $14.3+1.5\times 2.5$ (i.e. 8.05 or 18.05)M1
(b) both limits 8.05 and 18.05 and conclusion: 18.4 is an outlier, 8.6 is notA1
Answer: (a) mean $\approx 13.1$ °C, s.d. $\approx 1.73$ °C; (b) 18.4 °C is an outlier, 8.6 °C is not
Examiner tip: Keep the unrounded mean (13.0742…) when you square it – using 13.1 gives a noticeably wrong standard deviation.

Stuck on Data presentation, measures of location and spread, outliers?

Get it sorted in a free 30-minute session

We work through your weakest questions on this topic on Zoom. I reply the same day.

Free 30-min session on this topic Send me 5 questions a day

4StandardCalculator5 marks

Using data from the large data set for one UK weather station, Priya finds the equation of the regression line of daily total sunshine, $s$ hours, on daily mean cloud cover, $c$ oktas, to be $$s = 13.2 - 1.35c$$

(a) Give an interpretation of the value $-1.35$ in this context. [1]

(b) Use the regression line to estimate the daily total sunshine on a day with a mean cloud cover of 4 oktas. [1]

(c) Explain why it would be unreliable to use this line to estimate the daily total sunshine at this station on a day in December. [1]

In the large data set some values of daily total rainfall are recorded as 'tr'.

(d) State what 'tr' means and explain how Priya should treat these values when calculating the mean daily rainfall. [2]

(a) for each extra okta of cloud cover, daily sunshine falls by 1.35 hours (on average) – must be in contextB1
(b) $13.2-1.35\times4=7.8$ hoursB1
(c) the large data set only covers May to October, so December is outside the range of the data (extrapolation) / winter days are much shorterB1
(d) 'tr' means trace: an amount of rainfall less than 0.05 mmB1
(d) treat as 0 (accept a small value between 0 and 0.05, e.g. 0.025) so that these days are still includedB1
Answer: (a) sunshine falls 1.35 h per okta; (b) 7.8 h; (c) extrapolation – data only May–Oct; (d) trace (<0.05 mm), treat as 0
Examiner tip: Interpretations must be in context with units – 'as $c$ goes up, $s$ goes down' alone does not get the mark.

5HardCalculator7 marks

The masses, $x$ grams, of 20 apples were recorded. The data were coded using $y=\dfrac{x-150}{5}$ and the following summary statistics were obtained: $$\sum y = -12 \qquad \sum y^2 = 86$$

(a) Find the mean mass of the apples. [2]

(b) Find the standard deviation of the masses of the apples. [3]

It was later found that one apple had been recorded as 162 g when its true mass was 147 g.

(a) $\bar y=-0.6$ and attempt $\bar x=150+5\bar y$M1
(a) 147 gA1
(b) $\sigma_y=\sqrt{\frac{86}{20}-(-0.6)^2}$ $(=\sqrt{3.94})$M1
(b) $\sigma_x=5\sigma_y$ (multiply by 5 only – no adding 150)M1
(b) awrt 9.92 g (allow awrt 10.2 using $n-1$)A1
(c) $\frac{20\times147-162+147}{20}$ or $147-\frac{15}{20}$M1
(c) 146.25 g (accept 146 g)A1
Answer: (a) 147 g; (b) 9.92 g; (c) 146.25 g
Examiner tip: Coding shifts the mean but not the spread: the standard deviation is multiplied by 5 only – adding 150 to it is a common lost mark.

6StretchCalculator6 marks

Two groups of students timed how long, in minutes, they took to run a cross-country course.

	Number of students	Mean	Standard deviation
Group A	12	24.5	3.2
Group B	18	28.0	4.1

The standard deviations were calculated using $\sigma=\sqrt{\dfrac{\sum x^2}{n}-\bar{x}^2}$.

Find the mean and the standard deviation of the times of all 30 students. [6]

$\sum x_A=12\times24.5=294$ and $\sum x_B=18\times28=504$M1
combined mean $=\frac{798}{30}=26.6$A1
uses $\sum x^2=n(\sigma^2+\bar x^2)$ for at least one group, e.g. $12(3.2^2+24.5^2)$M1
$\sum x_A^2=7325.88$ and $\sum x_B^2=14414.58$ (or total 21740.46)A1
$\sqrt{\frac{21740.46}{30}-26.6^2}$M1
awrt 4.14 minutesA1
Answer: mean $=26.6$ minutes, s.d. $\approx 4.14$ minutes
Examiner tip: You cannot average the two standard deviations – rebuild $\sum x$ and $\sum x^2$ for each group first.

Stuck on this topic?

Book a free 30-minute session on WhatsApp and we'll find exactly which step is costing you the marks. Or build a timed paper on it.

Free 30-minute session