Sampling, data presentation and summary statistics – IB Maths exam questions with mark scheme

1StandardCalculator7 marks

The times, in minutes, taken by 12 students to complete a puzzle are:

3, 5, 6, 6, 7, 8, 9, 10, 11, 12, 14, 31

(a) Find the median, the lower quartile, the upper quartile and the interquartile range. [4]

(b) A value is an outlier if it is more than $1.5\times$ IQR above the upper quartile or below the lower quartile. Show that 31 is an outlier. [3]

Median $=8.5$A1
$Q_1=6$A1
$Q_3=11.5$A1
IQR $=5.5$A1
Upper boundary $=Q_3+1.5\times\text{IQR}=11.5+1.5(5.5)$M1
$=19.75$A1
$31>19.75$, so 31 is an outlierR1
Answer: (a) median $8.5$, $Q_1=6$, $Q_3=11.5$, IQR $=5.5$ (b) boundary $19.75<31$
Examiner tip: Quote the boundary value and compare it with 31: just saying '31 is much bigger' earns no reasoning mark.

2StandardCalculator6 marks

The number of goals scored by a team in each of 30 matches is summarised below.

Goals $x$	1	2	3	4	5
Frequency	4	7	10	6	3

(a) Find the mean and the standard deviation of the number of goals. [3]

A new variable is defined as $y=3x+2$.

(b) Find the mean and the standard deviation of $y$. [3]

Mean $=\dfrac{\sum fx}{\sum f}=\dfrac{4+14+30+24+15}{30}$ (or uses GDC 1-var stats)M1
Mean $=2.9$ (accept $\tfrac{87}{30}=2.9$)A1
Standard deviation $=1.16$ (population value $\sigma_x=1.1647\ldots$; do not accept the sample value $1.18$)A1
Uses $\bar y=3\bar x+2$M1
Mean of $y=3(2.9)+2=10.7$A1
Standard deviation of $y=3\sigma_x=3\times1.1647\ldots=3.49$ (adding 2 does not change the spread; accept $3.48$ from $3\times1.16$)A1
Answer: (a) mean $=2.9$, $\sigma=1.16$ (b) mean $=10.7$, $\sigma=3.49$
Examiner tip: Under $y=ax+b$ the mean becomes $a\bar x+b$ but the standard deviation only scales by $|a|$.

3StandardCalculator6 marks

The times, $t$ minutes, that 50 customers wait in a queue are recorded.

Time $t$	$0\le t<10$	$10\le t<20$	$20\le t<30$	$30\le t<40$	$40\le t<60$
Frequency	5	12	18	9	6

(a) Write down the modal class. [1]

(b) Estimate the mean waiting time. [2]

Modal class $20\le t<30$A1
Uses mid-interval values $5,15,25,35,50$: $\sum fx=1270$ (allow $\sum fx/50$)M1
Mean $\approx25.4$ minutesA1
Cumulative frequencies $5,\ 17,\ 35,\ 44,\ 50$A1
Median is the 25th value, in the class $20\le t<30$: interpolates $20+\dfrac{25-17}{18}\times10$M1
Median $\approx24.4$ minutes (allow $24$ to $25$ read from a cumulative frequency graph)A1
Answer: (a) $20\le t<30$ (b) $25.4$ min (c) $24.4$ min
Examiner tip: Use midpoints (not class boundaries) for the mean, and use $n/2$ to locate the median class; the 40-60 class has width 20.

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4StandardCalculator5 marks

The box plots show the marks of two classes, A and B, on the same test.

(a) Find the interquartile range for each class. [2]

(b) Find the range for each class. [1]

(c) A student claims: 'Class B has more consistent marks because its median is lower.' Comment on this claim, using the interquartile ranges. [2]

Class A: IQR $=31-18=13$A1
Class B: IQR $=28-20=8$A1
Range A $=40-12=28$; range B $=45-8=37$A1
The median measures the average (location), not the spread, so the claim's reasoning is invalidR1
Class B is more consistent for the middle 50% since its IQR is smaller ($8<13$) (a comment that B's range is larger, $37>28$, because of extreme values may be added but is not required)R1
Answer: (a) A: 13, B: 8 (b) A: 28, B: 37 (c) claim has wrong reason; B is more consistent by IQR
Examiner tip: Consistency is about spread (IQR or standard deviation), never the median; mention the extreme values in the range as a caveat.

5CoreNon-calculator4 marks

A sixth-form college has 540 students in Year 12 and 660 students in Year 13. The principal wants to find out how long students spend travelling to college each day.

(a) State whether the daily travel time is discrete or continuous data. [1]

(b) The principal first suggests asking the first 60 students who arrive at the college car park one morning. Give one reason why this sample is likely to be biased. [1]

Instead, a stratified sample of 60 students is taken, stratified by year group.

(a) Continuous (time is measured, not counted). Do not accept 'quantitative' aloneA1
(b) Any valid reason, e.g. it is a convenience sample, so only students who drive or are driven are included / students who arrive early or by car may have different travel times / not every student has an equal chance of selectionR1
(c) Uses proportion of population, e.g. $\frac{540}{1200}\times 60$ or $\frac{60}{1200}=\frac{1}{20}$ of each year groupM1
Year 12: 27 and Year 13: 33 (both required)A1
Answer: (a) continuous (b) e.g. convenience sample: only car users, not all students equally likely to be chosen (c) 27 Year 12, 33 Year 13
Examiner tip: Stratified means proportional to each group's size, so the two sample sizes must add back to 60 and be in the ratio 540 : 660.

6CoreNon-calculator4 marks

A set of temperature readings, in degrees Celsius, has mean $24$, standard deviation $3.5$ and interquartile range $5$.

Each reading is converted using $y=2x-5$, where $x$ is the original reading.

(a) Find the mean of the converted readings. [1]

(b) Find the standard deviation of the converted readings. [1]

(d) Find the interquartile range of the converted readings. [1]

(a) $2(24)-5=43$A1
(b) $2\times 3.5=7$ (the $-5$ has no effect)A1
(c) $49$ (follow through: square of their (b))A1
(d) $2\times 5=10$A1
Answer: (a) 43 (b) 7 (c) 49 (d) 10
Examiner tip: Adding or subtracting a constant shifts the mean but leaves every measure of spread unchanged; multiplying scales spread by the multiplier (variance by its square).

7StandardCalculator6 marks

The heights, $h$ cm, of 80 tomato plants were measured. The cumulative frequency graph shows the results.

(a) Use the graph to find (i) the median height; (ii) the interquartile range. [3]

(b) Find the 90th percentile of the heights. [1]

(a)(i) Median $=40$ cm (reading at cumulative frequency 40)A1
(ii) Reads $Q_1$ at cf 20 and $Q_3$ at cf 60: $Q_1=30$, $Q_3\approx 49$ (accept $48$ to $50$)M1
IQR $\approx 19$ cm (accept $18$ to $20$)A1
(b) Reads at cf $72$: $60$ cmA1
(c) Reads cf at $h=45$: $\approx 51$ (accept $49$ to $53$)M1
$80-51=29$ plants (accept $27$ to $31$; follow through from their reading)A1
Answer: (a)(i) 40 cm (ii) about 19 cm (b) 60 cm (c) about 29 plants
Examiner tip: Many students give the cumulative frequency at 45 cm (about 51) as the answer: 'taller than' means subtract it from the total of 80.

8StandardCalculator5 marks

A survey asked households how many pets they own. The results are shown in the table.

Number of pets, $x$	0	1	2	3	4
Frequency	5	$k$	12	8	3

The mean number of pets is $1.8$.

(a) Find the value of $k$. [3]

(b) Find the standard deviation of the number of pets. [1]

(a) Correct expression for mean: $\dfrac{0(5)+1(k)+2(12)+3(8)+4(3)}{28+k}$ (allow $\frac{k+60}{28+k}$)M1
Sets equal to $1.8$: $k+60=1.8(28+k)$ or equivalentM1
$k=12$A1
(b) $\sigma=1.12$ (1.12249…) from GDC. Accept $1.12$; do not accept $1.14$ (sample sd $s_{n-1}$)A1
(c) Median $=2$ (20th and 21st values both 2)A1
Answer: (a) $k=12$ (b) 1.12 (c) 2
Examiner tip: The total frequency is $28+k$, not $28$: forgetting $k$ in the denominator is the most common error.

9HardCalculator7 marks

Two groups of runners record their 5 km times in minutes.

Group A: 20 runners, mean $15$, standard deviation $2$
Group B: 30 runners, mean $18$, standard deviation $3$

(a) Find the mean time for all 50 runners. [2]

(b) Show that, for Group A, $\sum x^2=4580$. [2]

(a) Totals: $20\times15+30\times18\,(=300+540=840)$M1
Mean $=\dfrac{840}{50}=16.8$ minA1
(b) Uses $\sigma^2=\dfrac{\sum x^2}{n}-\bar{x}^2$: $4=\dfrac{\sum x^2}{20}-225$M1
$\sum x^2=20\times229=4580$ (AG): rearrangement must be seenA1
(c) Group B: $\sum x^2=30(9+324)=9990$A1
Combined variance $=\dfrac{4580+9990}{50}-16.8^2\,(=291.4-282.24=9.16)$M1
$\sigma=3.03$ min (3.0265…)A1
Answer: (a) 16.8 min (c) 3.03 min
Examiner tip: You cannot average the two standard deviations: rebuild each group's $\sum x^2$ first, then combine.

10StretchNon-calculator7 marks

The five numbers $3,\ 5,\ 7,\ a,\ b$, where $a$ and $b$ are integers with $a<b$, have mean $6$ and variance $8$.

(a) Show that $a+b=15$ and $a^2+b^2=137$. [3]

(b) Hence find the values of $a$ and $b$. [2]

A sixth number, $6$, is added to the data.

(c) Without further calculation of the mean, explain why the standard deviation decreases, and find the new variance. [2]

(a) $\dfrac{15+a+b}{5}=6\Rightarrow a+b=15$ (AG)A1
Uses $\sigma^2=\dfrac{\sum x^2}{n}-\bar{x}^2$: $8=\dfrac{9+25+49+a^2+b^2}{5}-36$ (accept $\frac{\sum(x-6)^2}{5}=8$ form)M1
$83+a^2+b^2=220\Rightarrow a^2+b^2=137$ (AG)A1
(b) Substitutes $b=15-a$: $a^2+(15-a)^2=137\Rightarrow a^2-15a+44=0$M1
$(a-4)(a-11)=0$, so $a=4,\ b=11$ (since $a<b$)A1
(c) The added value equals the mean $6$, so the mean is unchanged and $\sum(x-\bar x)^2$ is unchanged, but it is now divided by $6$ instead of $5$ (accept equivalent: the new value is no distance from the mean)R1
New variance $=\dfrac{5\times 8}{6}=\dfrac{20}{3}$ (accept $6.67$ or $6.\dot{6}$)A1
Answer: (b) $a=4,\ b=11$ (c) the added value equals the mean, so the sum of squared deviations stays 40 but is shared over 6 values; new variance $\frac{20}{3}$
Examiner tip: Adding a value equal to the mean adds zero to the squared deviations but increases $n$, so the spread must fall.

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