Statistics: regression, correlation, data presentation – 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.

4StandardCalculator7 marks

Six students recorded the number of hours of revision, $x$, and their test score, $y$. The data are shown in the table and on the scatter diagram.

$x$	2	4	5	7	8	10
$y$	14	19	21	28	29	36

(a) Use your graphic display calculator to find the Pearson correlation coefficient $r$ and the equation of the regression line of $y$ on $x$. [3]

(b) Describe the correlation and interpret the gradient of the line in context. [2]

(d) Explain why the line should not be used to predict the score of a student who revises for 20 hours. [1]

$r=0.996$ (accept $0.99586\ldots$ or any more accurate value)A1
Gradient $a=2.74$ ($2.73809\ldots$); accept $\frac{115}{42}$A1
Intercept $b=8.07$ ($8.07142\ldots$), so $y=2.74x+8.07$ (award only if written as an equation in $y$ and $x$)A1
Strong positive (linear) correlation (accept 'very strong positive'; do not accept 'positive' alone)A1
Each extra hour of revision is associated with about $2.74$ (accept $2.7$) more marks on average, in context (do not accept a causal claim such as 'revision causes')A1
$y\approx24.5$ (accept $24.51$ from the 3 s.f. coefficients)A1
$x=20$ is outside the data range ($2$ to $10$ hours), so this is extrapolation and the prediction is unreliableR1
Answer: (a) $r=0.996$, $y=2.74x+8.07$ (b) very strong positive; $\approx2.7$ marks per hour (c) $\approx24.5$ (d) extrapolation
Examiner tip: Interpretation must be in context and say 'on average' or 'associated with'; do not claim causation.

5StandardCalculator5 marks

A café owner records the midday temperature, $x\,^\circ$C, and the sales of cold drinks, $y$ (in pounds), on 25 days with temperatures between $14\,^\circ$C and $30\,^\circ$C. The regression line of $y$ on $x$ is $y=12.5x-90$ and $r=0.94$.

(a) Interpret the value $12.5$ in context. [1]

(b) Describe the correlation. [1]

(d) Explain why the model should not be used at $5\,^\circ$C. [1]

(e) The owner concludes that higher temperatures cause higher sales. Comment on this claim. [1]

For each extra $1\,^\circ$C, sales increase by about £12.50 on average (must be in context)A1
Strong positive (linear) correlation, as $r=0.94$ is close to 1 (do not accept 'positive' alone)A1
$12.5(25)-90=$ £222.50 (accept $222.5$ or £223 to 3 s.f.)A1
$5\,^\circ$C is outside the range of the data ($14$ to $30\,^\circ$C), so this would be extrapolation and unreliable (also accept: the model gives negative sales, $12.5(5)-90=-27.5$, which is impossible; either reason scores)R1
Correlation does not imply causation: the data show only an association (naming a possible third factor, e.g. time of year or sunshine, is not required but may be credited instead)R1
Answer: (a) +£12.50 per °C (b) strong positive (c) £222.50 (d) extrapolation (e) correlation is not causation
Examiner tip: 'Positive correlation' alone is not enough for the description mark: say how strong it is (e.g. strong positive).

6HardCalculator6 marks

The table shows five pairs of values. One value, $k$, is missing. The regression line of $y$ on $x$ is $y=3x+5$.

$x$	2	4	6	8	10
$y$	12	15	$k$	27	36

(a) Write down the coordinates of the mean point $(\bar x,\bar y)$ and find the value of $k$. [4]

(b) Find the residual (actual $-$ predicted) at $x=6$. [2]

$\bar x=6$A1
The regression line passes through $(\bar x,\bar y)$: $\bar y=3(6)+5$M1
$\bar y=23$A1
$\dfrac{12+15+k+27+36}{5}=23\Rightarrow k=25$A1
Predicted value at $x=6$ is $23$M1
Residual $=25-23=2$A1
Answer: (a) $(6,23)$, $k=25$ (b) $+2$
Examiner tip: Remember that the least-squares line always passes through $(\bar x,\bar y)$: this lets you find a missing value without calculating a gradient.

7StandardCalculator5 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.

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