Statistics & regression — Mathematics Grade 12

6 exam-style question sets on Statistics & regression (Paper 2), each with a hint and a fully worked answer. The app holds 30 questions on this section in total, including variants of every set below, and lets you mark yourself part by part.

Regression & standard deviationGrouped data & estimated meanRegression & correlationOgives & the five-number summaryGrouped data, histograms & standard deviation

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Question 1

Regression & standard deviationRoutine12 marksPaper 2

An agricultural adviser in the eastern Free State studied 8 dryland maize farms. For each farm she recorded the total rainfall (in mm) during the growing season and the maize yield (in kg per hectare) that the farm obtained.

Rainfall (mm)	380	420	450	490	520	560	600	640
Yield (kg per hectare)	4 650	5 820	5 310	6 380	7 140	6 790	8 230	8 460

aDetermine the equation of the least squares regression line for this data in the form $\overset{y}{^} = a + b x$ . Round $a$ and $b$ off to TWO decimal places. (3)
bUse the regression line to predict the yield of a farm in this area that received 500 mm of rain during the growing season. (2)
cWrite down the correlation coefficient of the data, correct to TWO decimal places. (1)
dDescribe, in words, what the value of $r$ tells us about the relationship between the rainfall and the yield. (1)
eThe adviser also recorded the yields (in kg per hectare) of 7 farms in a neighbouring district: 5 840; 6 620; 4 870; 7 150; 6 330; 7 480; 4 410 Calculate the mean yield of these 7 farms. (2)
fCalculate the standard deviation of the yields of these 7 farms. (1)
gHow many of these 7 farms obtained a yield that is more than ONE standard deviation below the mean? Show your working. (2)

Hint

Enter the eight pairs in STAT mode (two variables) to read off $a$ , $b$ and $r$ ; for the seven yields use one-variable STAT mode to read $\overset{x}{ˉ}$ and $σ_{x}$ , then subtract the standard deviation from the mean and count the yields that lie below that value.

Worked answer

a) From the calculator: $a = - 661, 41$ ✓ and $b = 14, 30$ ✓
$\overset{y}{^} = - 661, 41 + 14, 30 x$ ✓

b) $\overset{y}{^} = - 661, 41 + 14, 30 (500)$ ✓ $= 6488, 59$
Predicted yield ≈ 6 489 kg per hectare ✓

c) $r = 0, 96$ ✓

d) There is a strong positive correlation: farms that received more rain during the season tended to obtain higher yields ✓

e) $\overset{x}{ˉ} = \frac{5 840 + 6 620 + 4 870 + 7 150 + 6 330 + 7 480 + 4 410}{7}$ ✓ $= \frac{42 700}{7} = 6100$ kg per hectare ✓

f) $σ = 1053, 35$ kg per hectare ✓ (population standard deviation from the calculator)

g) $\overset{x}{ˉ} - σ = 6100 - 1053, 35 = 5046, 65$ ✓
Yields below 5 046,65 kg per hectare: 4 870 and 4 410, so TWO farms ✓

Modelled on Nov 2023 Paper 2, Question 1 — same skills, new scenario.

Question 2

Grouped data & estimated meanProblem-solving8 marksPaper 2

A speed camera on the N12 near Klerksdorp recorded the speed, $v$ (in km/h), of every vehicle that passed it during one morning. The speeds are grouped in the table below.

Speed (km/h)	Frequency	Cumulative frequency
$60 \leq v < 70$	7
$70 \leq v < 80$	15
$80 \leq v < 90$	24
$90 \leq v < 100$	11
$100 \leq v < 110$	3

aComplete the cumulative frequency column of the table. (2)
bWrite down the number of vehicles that passed the camera during the morning. (1)
cHow many of these vehicles were travelling at less than 80 km/h? (1)
dLater that day the camera recorded a further $k$ vehicles. Their speeds fell in the classes $70 \leq v < 80$ and $100 \leq v < 110$ in the ratio $1 : 2$ respectively, and in no other class. The estimated mean speed of ALL the vehicles recorded that day is 85 km/h. Calculate the value of $k$ . (4)

Hint

Cumulative frequency is a running total of the frequencies. For the last part use class midpoints: the estimated mean is $\frac{\sum f \cdot m}{n}$ . Write the new total as $60 + k$ , add $\frac{1}{3} k$ vehicles at midpoint 75 and $\frac{2}{3} k$ at midpoint 105 to $\sum f \cdot m$ , set the fraction equal to 85 and solve the linear equation.

Worked answer

a) Cumulative frequencies (running totals): 7; 22; 46; 57; 60 ✓✓

b) 60 vehicles ✓

c) Fewer than 80 km/h: cumulative frequency at 80, i.e. $7 + 15 = 22$ vehicles ✓

d) Midpoints: 65; 75; 85; 95; 105
$\sum f \cdot m = 7 (65) + 15 (75) + 24 (85) + 11 (95) + 3 (105) = 4980$ ✓
The extra vehicles add $\frac{1}{3} k$ at midpoint 75 and $\frac{2}{3} k$ at midpoint 105, i.e. $25 k + 70 k = 95 k$ to the total, and the number of vehicles becomes $60 + k$ ✓
$\frac{4 980 + 95 k}{60 + k} = 85$ ✓
$4980 + 95 k = 5100 + 85 k$
$10 k = 120$
$k = 12$ ✓ (4 vehicles in $70 \leq v < 80$ and 8 in $100 \leq v < 110$ )

Modelled on Nov 2023 Paper 2, Question 2 — same skills, new scenario.

Question 3

Regression & correlationRoutine10 marksPaper 2

A second-hand electronics dealer sold 13 laptops of the same model. For each laptop he recorded its age (in months) and the price (in rand) at which it was resold. The information is given in the table below.

Age of laptop (months)	3	6	9	12	15	18	22	25	29	33	36	40	44
Resale price (R)	9 520	8 400	8 920	7 350	7 580	6 150	6 820	5 480	6 000	2 920	4 580	3 000	3 450

aDetermine the least squares regression line of the data in the form $\overset{y}{^} = a + b x$ . (3)
bWrite down the correlation coefficient of the data, correct to TWO decimal places. (1)
cPredict the resale price of a laptop of this model that is 20 months old. (2)
dWrite down the mean resale price of the 13 laptops. (1)
eThe standard deviation of the resale prices is R2 123,21. The dealer decides to increase EVERY resale price by R250. Write down the standard deviation of the increased prices. (1)
fThe dealer regards the price predicted by the regression line as the target price for a laptop of that age. Calculate the largest amount by which the actual resale price of a laptop in this data set fell below its target price. (2)

Hint

Enter the pairs in STAT mode: the calculator gives $a$ , $b$ , $r$ and $\overset{y}{ˉ}$ directly; for the last part compare each observed value with the value the line predicts at the same age.

Worked answer

a) From the calculator: $a = 9695, 69$ ✓ and $b = - 157, 10$ ✓
$\overset{y}{^} = 9695, 69 - 157, 10 x$ ✓

b) $r = - 0, 95$ ✓

c) $\overset{y}{^} = 9695, 69 - 157, 10 (20)$ ✓ $= 6553, 69$
Predicted resale price ≈ R6 553,69 ✓

d) $\overset{y}{ˉ} = 6166, 92$ , i.e. R6 166,92 ✓

e) R2 123,21 ✓ — adding the same constant to every value shifts the whole data set but does NOT change the spread, so the standard deviation stays the same.

f) The laptop aged 33 months lies furthest below the line:
$\overset{y}{^} = 9695, 69 - 157, 10 (33) = 4511, 39$ ✓
Shortfall $= 4511, 39 - 2920 = 1591, 39$ , i.e. R1 591,39 ✓

Modelled on Nov 2024 Paper 2, Question 1 — same skills, new scenario.

Question 4

Ogives & the five-number summaryRoutine10 marksPaper 2

On a busy Saturday evening a pizza restaurant delivered 60 orders. The time, in minutes, that each delivery took was recorded. The ogive (cumulative frequency curve) below represents the delivery times.

aUse the ogive to write down the median delivery time. (1)
bUse the ogive to write down the lower quartile of the delivery times. (1)
cDetermine the interquartile range of the delivery times. (2)
dThe fastest delivery took 6 minutes and the slowest took 88 minutes. Use this information and the ogive to draw a box and whisker diagram of the delivery times. (2)
eCalculate the percentage of the orders that took 75 minutes or longer to deliver. (2)
fThe restaurant gives a customer a voucher of R15 for EVERY 10 minutes, or part thereof, that a delivery takes longer than 45 minutes. Calculate the value of the voucher for the slowest delivery. (2)

This question has a diagram, shown in the app.

Hint

Read across from one quarter, one half and three quarters of 60 on the cumulative frequency axis to the curve; for the voucher, a block of 10 minutes that has only just started still counts in full.

Worked answer

a) Half the orders: read across at a cumulative frequency of 30 → median $= 40$ minutes ✓

b) Read across at a cumulative frequency of 15 → $Q_{1} = 25$ minutes ✓

c) Read across at a cumulative frequency of 45 → $Q_{3} = 60$ minutes ✓
IQR $= Q_{3} - Q_{1} = 60 - 25 = 35$ minutes ✓

d) Five-number summary: 6; 25; 40; 60; 88 ✓
Box from 25 to 60 with the median line at 40; whiskers reaching to 6 and 88 ✓

e) The ogive reads 54 at 75 minutes, so $60 - 54 = 6$ orders took 75 minutes or longer ✓
$\frac{6}{60} \times 100 = 10%$ ✓

f) Time above 45 minutes: $88 - 45 = 43$ minutes, and $43 \div 10 = 4, 3$ , so 5 blocks are started ✓
Voucher $= 5 \times 15 =$ R75 ✓

Modelled on Nov 2024 Paper 2, Question 2 — same skills, new scenario.

Question 5

Regression & correlationRoutine8 marksPaper 2

A drilling contractor maintains 10 borehole pumps of the same model on farms around Polokwane. For each pump he recorded how many years it has been in service, together with the rate, in litres per minute, at which it now delivers water.

Years in service	1	2	3	4	5	6	8	9	10	12
Delivery rate (litres per minute)	64	57	60	52	54	49	44	42	36	35

aUse your calculator to determine the least-squares regression equation of this data set, in the form $\overset{y}{^} = a + b x$ . (3)
bPredict the delivery rate of a pump of this model that has been in service for 7 years. (2)
cCan the prediction in part (b) be trusted? Motivate your answer by referring to the correlation coefficient of this data set. (2)
dInterpret the gradient of your regression equation in the context of these pumps. (1)

Hint

Enter the ten pairs in STAT mode to read off $a$ , $b$ and $r$ directly; substitute $x = 7$ for the prediction, and remember that a value of $∣ r ∣$ close to 1 signals a strong linear relationship.

Worked answer

a) From the calculator: $a = 65, 20$ ✓ and $b = - 2, 65$ ✓
$\overset{y}{^} = 65, 20 - 2, 65 x$ ✓

b) $\overset{y}{^} = 65, 20 - 2, 65 (7)$ ✓ $= 46, 65$
Predicted delivery rate ≈ 46,65 litres per minute ✓

c) $r = - 0, 98$ ✓
Yes — $∣ r ∣$ is very close to 1, so the linear association between age and delivery rate is very strong (and 7 years lies inside the range of the data), which makes the prediction dependable ✓

d) On average, the delivery rate of a pump decreases by about 2,65 litres per minute for EVERY additional year in service ✓

Modelled on Nov 2025 Paper 2, Question 1 — same skills, new scenario.

Question 6

Grouped data, histograms & standard deviationComplex12 marksPaper 2

A courier depot in Polokwane weighed every parcel that it dispatched on one morning. The masses were organised into classes, and the CUMULATIVE frequency of each class is given in the table below.

Mass, $m$ (in kg)	Cumulative frequency
$0 \leq m < 10$	6
$10 \leq m < 20$	19
$20 \leq m < 30$	37
$30 \leq m < 40$	49
$40 \leq m < 50$	56
$50 \leq m < 60$	60

aWrite down the total number of parcels dispatched that morning. (1)
bHow many parcels had a mass of at least 20 kg but less than 40 kg? (2)
cDetermine the frequency of each of the SIX mass classes and hence draw a histogram of this data. (3)
dWhat does the shape of your histogram reveal about the skewness of the mass distribution? (1)
eThe depot also stores NINE crates for a single client. The mean mass of these crates is 18 kg, but the mass of one crate, $k$ kg, is unreadable on the record sheet: 11; 14; 16; 17; 18; 19; 20; 23; $k$ (all in kg). Determine $k$ , calculate the standard deviation of the NINE masses, and state how many of the masses lie outside the interval $(\overset{x}{ˉ} - σ; \overset{x}{ˉ} + σ)$ . (5)

Hint

Differences between consecutive cumulative values give the class frequencies; in the last part find $k$ from the mean first, then let the calculator supply $σ_{n}$ and compare each mass with the interval bounds.

Worked answer

a) $n = 60$ ✓ (the final cumulative value)

b) $49 - 19$ ✓ $= 30$ parcels ✓

c) Class frequencies (differences of consecutive cumulative values): $6; 13; 18; 12; 7; 4$ ✓✓
Histogram: SIX adjacent bars of equal width over the classes from 0 kg to 60 kg, with heights 6; 13; 18; 12; 7 and 4 ✓

d) Skewed to the right (positively skewed) ✓ — the tallest bars stand over the lighter classes and the frequencies tail off towards the heavier masses.

e) $\frac{138 + k}{9} = 18$ ✓ $\Rightarrow k = 162 - 138 = 24$ ✓
$σ = \frac{( 11 - 18 ) ^{2} + ( 14 - 18 ) ^{2} + \dots + ( 24 - 18 ) ^{2}}{9} = \frac{136}{9} = 3, 89$ kg ✓✓
$(\overset{x}{ˉ} - σ; \overset{x}{ˉ} + σ) = (14, 11; 21, 89)$ : the masses 11; 14; 23 and 24 lie outside this interval, i.e. FOUR masses ✓

Modelled on Nov 2025 Paper 2, Question 2 — same skills, new scenario.

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Grade 12 Mathematics: Statistics & regression

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