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This lesson brings the whole Statistics strand together as exam practice. The questions below are written in the style of the OCR GCSE Mathematics (J560) papers and run across every topic in the strand — data types and sampling, frequency and two-way tables, averages from lists and grouped data, charts, scatter graphs, and the Higher-tier cumulative frequency, box plots and histograms. Each question states its marks and tier, and is followed by a full worked solution with notes on where the method marks are earned. Three "Answering at different grade levels" contrasts show how the same question is rewarded differently across Foundation and Higher grades. Work each question yourself before reading the solution.
Across these questions you will use every assessment objective: AO1 for fluent calculation, AO2 for reasoning and interpretation, and AO3 for problem-solving and justified conclusions. Watch the OCR command words — "Work out", "Calculate", "Draw", "Estimate", "Describe", "Compare" and "Give a reason for your answer" each signal what the examiner wants.
State whether each variable is discrete or continuous: (a) the number of eggs in a nest; (b) the mass of a kitten.
Solution: (a) Discrete — eggs are counted in whole numbers. (b) Continuous — mass is measured and can take any value. Method note: one mark for each correct classification; no working is needed for "state".
A frequency table records the number of goals in 25 matches: 0 goals → 4, 1 goal → 8, 2 goals → 7, 3 goals → ?, 4 goals → 2. (a) Work out the missing frequency. (b) Write down the modal number of goals.
Solution: (a) Missing frequency =25−(4+8+7+2)=25−21=4. Method note: one mark for subtracting from 25, one for the answer 4. (b) The modal number of goals is the most common, 1 goal (frequency 8). Method note: one mark; do not confuse the frequency (8) with the value (1).
The list shows the masses, in kg, of nine parcels: 2, 3, 3, 5, 6, 7, 9, 11, 26. (a) Work out the mean. (b) Write down the median. (c) The manager says "the median better describes a typical parcel than the mean." Give a reason for this.
Solution: (a) Mean =92+3+3+5+6+7+9+11+26=972=8 kg. Method note: one mark for the total 72, one for dividing by 9. (b) Median = 5th of 9 ordered values =6 kg. Method note: one mark. (c) The value 26 kg is an outlier that pulls the mean up, so the mean (8 kg) is higher than most parcels; the median (6 kg) is not affected by the outlier and is more typical. Method note: one mark for identifying the outlier's effect.
A consistent two-way table records how 60 people travel to work, split by gender. There are 36 men and 24 women; 27 people drive, of whom 18 are men; 15 people cycle, of whom 6 are women. The rest take the bus. (a) Work out how many men take the bus. (b) Work out how many women drive.
Solution: Build the table.
| Drive | Cycle | Bus | Total | |
|---|---|---|---|---|
| Men | 18 | 15−6=9 | ? | 36 |
| Women | 27−18=9 | 6 | ? | 24 |
| Total | 27 | 15 | 18 | 60 |
(a) Men bus =36−18−9=9. (b) Women drive =27−18=9. Method note: marks for correct subtractions across rows/columns; the bus column total is 60−27−15=18, a useful check.
The table shows the time, t minutes, that 50 customers waited.
| Time t (min) | 0<t≤4 | 4<t≤8 | 8<t≤12 | 12<t≤16 |
|---|---|---|---|---|
| Frequency | 11 | 19 | 14 | 6 |
(a) Estimate the mean waiting time. (b) Write down the modal class. (c) Explain why your answer to (a) is an estimate.
Solution: (a) Midpoints 2, 6, 10, 14. ∑fx=11(2)+19(6)+14(10)+6(14)=22+114+140+84=360. Estimated mean =50360=7.2 minutes. Method note: one mark for midpoints, one for ∑fx, one for dividing by 50. (b) Modal class =4<t≤8 (highest frequency, 19). Method note: one mark; quote the interval. (c) Exact times are unknown, so each customer is taken at the class midpoint; the answer is therefore an estimate. Method note: one mark.
In a survey of 72 students about their favourite takeaway, 30 chose pizza, 18 chose burgers, 12 chose noodles and the rest chose curry. (a) Work out the pie-chart angle for curry. (b) Describe how the popularity of pizza compares with noodles.
Solution: (a) Curry frequency =72−30−18−12=12. Angle =7212×360=60∘. Method note: one mark for the frequency 12, one for the angle. (b) Pizza (30) was chosen by more than twice as many students as noodles (12); pizza is much more popular. Method note: one mark for the direction, one for quantifying it.
A scatter graph shows the age (years) and value (£000s) of twelve cars, with a clear downward trend close to a line. (a) Describe the correlation. (b) The line of best fit passes through (2, 12) and (8, 3). Use it to estimate the value of a 5-year-old car. (c) Explain why estimating the value of a 15-year-old car would be unreliable.
Solution: (a) Strong negative correlation — as age increases, value decreases, with points close to a line. Method note: one mark for direction, one for strength. (b) From (2, 12) to (8, 3) the value falls £9k over 6 years, i.e. £1.5k per year. A 5-year-old car is 3 years past age 2: 12−3×1.5=12−4.5=7.5, i.e. £7,500. Method note: one mark for a correct reading method, one for the value. (c) 15 years is well beyond the data range (extrapolation), so the trend may not hold and the estimate is unreliable. Method note: one mark for naming extrapolation/beyond the range.
The cumulative frequency table shows the heights, h cm, of 80 plants.
| Height h (cm) | ≤10 | ≤20 | ≤30 | ≤40 | ≤50 |
|---|---|---|---|---|---|
| Cumulative frequency | 6 | 26 | 54 | 72 | 80 |
(a) Estimate the median height. (b) Estimate the interquartile range. (c) How many plants were taller than 30 cm?
Solution: (a) Median at 280=40: the cumulative frequency reaches 40 between the ≤20 (26) and ≤30 (54) rows, giving a median of about 24–25 cm read from the curve. Method note: one mark for using position 40, one for a sensible reading. (b) Q1 at 480=20 → about 18 cm; Q3 at 43×80=60 → about 33 cm. IQR ≈33−18=15 cm. Method note: one mark for both quartile positions, one for the subtraction. (c) Taller than 30 cm =80−54=26 plants. Method note: one mark for subtracting the cumulative frequency at 30 from 80.
A histogram shows the masses, m grams, of 112 apples. The frequency densities are: 0<m≤50 → 0.4; 50<m≤80 → 1.2; 80<m≤120 → 1.0; 120<m≤200 → ?. The last class contains 16 apples. (a) Work out the frequency density of the last class. (b) Show that the four classes account for all 112 apples.
Solution: (a) Class width =200−120=80, frequency 16, so frequency density =8016=0.2. Method note: one mark for the width, one for dividing. (b) Frequencies are area = FD × width: 0.4×50=20; 1.2×30=36; 1.0×40=40; the last is 16. Total =20+36+40+16=112 ✓, accounting for all 112 apples. The examined skill is that each frequency is the bar's area; total them as a check. Method note: method marks for computing each area; final mark for the correct total.
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