Elite Mathematics / Linear / Topics / Statistics IGCSE · 4MA1 · Topic Statistics: past paper questions & worked solutions
Every Statistics question from the 4MA1 papers in our bank, 2020–2026, sorted from easiest to hardest. Try each one first, then open the worked solution.
64 questions
215 marks in total
74% of papers (40 of 54) include it
13/41/10 easy / medium / hard
Questions, easiest first Level is estimated from marks: 1–2 easy, 3–4 medium, 5+ hard.
1. Jun 2026 4WM2HR · Q15 Easy 2 marksShow worked solution Positions of the quartiles \(n=11\): LQ is the \(\tfrac{11+1}{4}=3\)rd value, UQ is the \(9\)th value.
Read from the ordered list \(2,2,\mathbf{3},6,7,8,9,12,\mathbf{13},17,20\): LQ \(=3\), UQ \(=13\).
Interquartile range \[13-3=10\]
Final answer \(10\) minutes
2. Jun 2026 P2HR · Q15 Easy 2 marksShow worked solution Solution | Step 1 The overall median is the sixth value, 9. The lower five values are \(2,2,3,6,7\), and the upper five are \(9,12,13,17,20\).
Exclude the overall median when forming the two halves.
Solution | Step 2 \[Q_1=3,\quad Q_3=13,\quad \mathrm{IQR}=13-3=10\text{ minutes}.\]
Take each half-median, then subtract the lower quartile from the upper.
Final answer \(10\text{ minutes}\).
3. Nov 2025 4WM2H · Q16 Easy 2 marksShow worked solution Calculate value **Data (sorted):** 14, 15, 16, 18, 19, 20, 22, 25, 28, 40, 50. Number of data points *n* = 11.
Find median **Step 1: Find the Median (Q2).** The median is the $\\left(\\frac{11+1}{2}\\right) = 6^{\\text{th}}$ value. Median = 20
Find median **Step 2: Find Lower Quartile (Q1) and Upper Quartile (Q3).** Q1 is the median of the lower 5 values (14, 15, **16**, 18, 19). *Q*1 = 16 Q3 is the median of the upper 5 values (22, 25, **28**, 40, 50). *Q*3 = 28
Calculate statistic IQR = *Q*3 − *Q*1 = 28 − 16 = 12
4. Nov 2025 P1H · Q16 Easy 2 marksShow worked solution Use Statistics Toolkit correct The data are already in order:
\[
1,\ 1,\ 3,\ 4,\ 6,\ 7,\ 8,\ 9,\ 10,\ 12,\ 15,\ 25,\ 30,\ 37,\ 50
\]
There are 15 values There are 15 values.
\[
Q_1=4,\qquad Q_3=25
\]
\[
\text{IQR}=25-4=21
\]
5. Jun 2025 P1HR · Q13 Easy 2 marksShow worked solution Put the numbers in order Put the numbers in order:
\[
9,\ 10,\ 12,\ 15,\ 17,\ 18,\ 19,\ 19,\ 20,\ 25,\ 26
\]
\[
Q_1=12,\qquad Q_3=20
\]
\[
\text{IQR}=20-12=8
\]
6. May 2025 4WM2H · Q16 Easy 2 marksShow worked solution There are values. Excluding the There are \(11\) values. Excluding the median \(14\):
Calculate value Lower half: \(7,8,10,11,13\), so \(Q_1=10\).
Calculate value Upper half: \(16,17,19,21,24\), so \(Q_3=19\).
\[
\text{IQR}=19-10=9
\]
7. Jun 2024 P2HR · Q3 Easy 2 marksShow worked solution The range is The range is \(4\):
\[
15-x=4
\]
\[
x=11
\]
The median is The median is \(12.5\):
\[
\frac{x+y}{2}=12.5
\]
\[
x+y=25
\]
\[
y=25-11=14
\]
Final answer \(x=11,\ y=14\).
8. May 2024 P1HR · Q10 Easy 2 marksShow worked solution The goals in order are \[
0,\ 1,\ 2,\ 2,\ 3,\ 4,\ 4,\ 6,\ 7,\ 9,\ 11
\]
\[
Q_1=2,\qquad Q_3=7
\]
\[
\text{IQR}=7-2=5
\]
9. Nov 2023 P2H · Q13 Easy 2 marksShow worked solution The results in order are \[
1,\ 2,\ 4,\ 6,\ 6,\ 8,\ 11,\ 12,\ 13,\ 14,\ 17
\]
There are 11 values There are 11 values.
\[
Q_1=4,\qquad Q_3=13
\]
\[
\text{IQR}=13-4=9
\]
10. May 2023 P1H · Q10 Easy 2 marksShow worked solution The marks in order are \[
7,\ 10,\ 14,\ 15,\ 16,\ 17,\ 18,\ 19,\ 20,\ 20,\ 23,\ 25,\ 30,\ 39,\ 40
\]
\[
Q_1=15,\qquad Q_3=25
\]
\[
\text{IQR}=25-15=10
\]
11. Jun 2022 P1HR · Q13 Easy 2 marksShow worked solution Use Statistics Toolkit correct The runs are already in order:
\[
0,\ 1,\ 1,\ 3,\ 5,\ 6,\ 7,\ 7,\ 8,\ 9,\ 9,\ 12,\ 12,\ 15,\ 16
\]
Calculate statistic There are 15 values, so ignore the median when finding the quartiles.
\[
Q_1=3,\qquad Q_3=12
\]
\[
\text{IQR}=12-3=9
\]
12. May 2021 P1H · Q14 Easy 2 marksShow worked solution The goals in order are \[
5,\ 8,\ 9,\ 11,\ 13,\ 13,\ 14,\ 15,\ 16,\ 17,\ 20
\]
\[
Q_1=9,\qquad Q_3=16
\]
\[
\text{IQR}=16-9=7
\]
13. Jan 2021 P2H · Q12 Easy 2 marksShow worked solution Use Statistics Toolkit correct The data is already in order:
\[
2,3,3,4,5,6,6,8,9,10,11,13,14,15,18
\]
Calculate value For 15 values,
\[
Q_1=4,\qquad Q_3=13
\]
\[
\text{IQR}=13-4=9
\]
Final answer \(9\text{ minutes}\).
14. Jun 2026 P1HR · Q9 Medium 3 marksShow worked solution Solution | Step 1 \[W_{11}=11(77)=847\text{ kg},\qquad W_8=8(75.2)=601.6\text{ kg}.\]
Total weight equals number of sheep multiplied by their mean weight.
Solution | Step 2 \[W_3=847-601.6=245.4\text{ kg},\qquad \overline w_3=\frac{245.4}{3}=81.8\text{ kg}.\]
Subtract the known subgroup total and divide the remaining total by 3.
Final answer \(81.8\text{ kg}\)
15. Nov 2025 P1H · Q4 Medium 3 marksShow worked solution The range is The range is \(15\):
\[
17-e=15
\]
\[
e=2
\]
Calculate statistic The mean is \(9\), so the total is
\[
6 \times 9=54
\]
\[
2+f+7+g+12+17=54
\]
\[
f+g=16
\]
The median is The median is \(8.5\):
\[
\frac{7+g}{2}=8.5
\]
\[
g=10
\]
\[
f=16-10=6
\]
Final answer \(e=2,\ f=6,\ g=10\).
16. Nov 2025 4WM2H · Q10 Medium 3 marksShow worked solution Calculate statistic Total<sub>5</sub> = Mean × Number = 217 × 5 = 1085 grams
Solve inequality Total<sub>6</sub> = 1085 + 175 = 1260 grams
Calculate statistic $$\\text{New Mean} = \\frac{1260}{6} = 210 \\text{ grams}$$
17. Jun 2025 P2H · Q1 Medium 3 marksShow worked solution Calculate statistic The range is \(10\), and the greatest card is \(14\):
\[
14-\text{smallest}=10
\]
Use Statistics Toolkit tag correct So the smallest card is \(4\).
Calculate statistic The median is \(7.5\). For 6 cards, the median is the mean of the 3rd and 4th cards:
\[
\frac{6+x}{2}=7.5
\]
\[
x=9
\]
Calculate statistic The mode is \(6\), so the other missing card must be \(6\).
Final answer \(4,\ 6,\ 9\).
18. Jun 2025 P2HR · Q8 Medium 3 marksShow worked solution The total of the 7 The total of the 7 numbers is
\[
7 \times 60=420
\]
The total of the 3 The total of the 3 numbers is
\[
3 \times 46=138
\]
The total of the other The total of the other 4 numbers is
\[
420-138=282
\]
Their mean is \[
\frac{282}{4}=70.5
\]
19. May 2025 4WM2H · Q1 Medium 3 marksShow worked solution Calculate statistic The range is \(10\), so the first number is
\[
14-10=4
\]
Calculate statistic For six numbers, the median is the mean of the third and fourth values.
\[
\frac{6+\text{fourth value}}{2}=7.5
\]
Calculate statistic So the fourth value is \(9\). To make \(6\) the mode, the second value must also be \(6\).
Final answer the hidden numbers are \(4,6,9\).
20. May 2025 4WM2HR · Q7 Medium 3 marksShow worked solution Required Mean of 7 numbers = 60; mean of 3 of them = 46. Find the mean of the other 4.
Sum of all 7 numbers Sum of all 7 numbers: 7 × 60 = 420
Sum of the 3 numbers Sum of the 3 numbers: 3 × 46 = 138
Sum of the remaining 4 Sum of the remaining 4: 420 − 138 = 282
Mean of the 4 numbers Mean of the 4 numbers: $$\\frac{282}{4} = 70.5$$
Final answer Mean of the other 4 numbers = 70.5
21. Nov 2024 P2H · Q8 Medium 3 marksShow worked solution The total weight of the The total weight of the first 4 bananas is
\[
4 \times 145=580
\]
The total weight of the The total weight of the 5 bananas is
\[
5 \times 142=710
\]
Weight of the banana Andy Weight of the banana Andy added:
\[
710-580=130
\]
Final answer \(130\text{ g}\).
22. Jun 2024 P2H · Q1 Medium 3 marksShow worked solution The median is The median is \(10\):
\[
\frac{8+j}{2}=10
\]
\[
j=12
\]
Calculate statistic The mode is \(18\), so
\[
k=18
\]
The range is The range is \(13\):
\[
18-h=13
\]
\[
h=5
\]
Final answer \(h=5,\ j=12,\ k=18\).
23. Jun 2024 P2H · Q12 Medium 3 marksShow worked solution Total for the first 4 Total for the first 4 days:
\[
4 \times 11800=47200
\]
Total for the next 3 Total for the next 3 days:
\[
3 \times 13207=39621
\]
Total for all 7 days Total for all 7 days:
\[
47200+39621=86821
\]
Mean for the 7 days Mean for the 7 days:
\[
\frac{86821}{7}=12403
\]
24. May 2024 P1HR · Q1 Medium 3 marksShow worked solution The total of the 6 The total of the 6 cards is
\[
6 \times 11=66
\]
The known total is \[
16+15+3+2+9=45
\]
\[
x=66-45=21
\]
25. Nov 2023 P2H · Q8 Medium 3 marksShow worked solution The total for the first The total for the first 8 matches is
\[
8 \times 6=48
\]
The total for all 10 The total for all 10 matches is
\[
10 \times 7=70
\]
Compare density So the last two matches have total
\[
70-48=22
\]
The last two scores are The last two scores are \(k\) and \(k+10\):
\[
k+(k+10)=22
\]
\[
2k+10=22
\]
\[
k=6
\]
26. Jun 2023 P2HR · Q10 Medium 3 marksShow worked solution Total for Monday to Friday Total for Monday to Friday:
\[
5 \times 104=520
\]
Total for Monday to Sunday Total for Monday to Sunday:
\[
7 \times 127=889
\]
Calculate value So the total for Saturday and Sunday is
\[
889-520=369
\]
Calculate value Saturday was \(132\text{ km}\), so
\[
369-132=237
\]
Final answer \(237\text{ km}\).
27. May 2023 P1H · Q2 Medium 3 marksShow worked solution Use midpoints Use midpoints \(15,45,75,105,135\).
\[
15(5)+45(6)+75(8)+105(9)+135(2)=2160
\]
Final answer \(2160\) minutes.
28. Jun 2022 P2H · Q1 Medium 3 marksShow worked solution Calculate statistic The mode is \(9\), so
\[
d=9
\]
The range is The range is \(4\):
\[
9-a=4
\]
\[
a=5
\]
Calculate statistic The median is \(8\), so
\[
\frac{c+d}{2}=8
\]
\[
\frac{c+9}{2}=8
\]
\[
c=7
\]
The integers are in order The integers are in order and \(a<b<c\), so
\[
b=6
\]
Final answer \(a=5,\ b=6,\ c=7,\ d=9\).
29. Jun 2022 P1HR · Q3 Medium 3 marksShow worked solution The mean of the five The mean of the five cards is \(12\), so the total is
\[
5 \times 12=60
\]
The known total is \[
15+7-2+23=43
\]
\[
x=60-43=17
\]
30. Jun 2022 P2HR · Q3 Medium 3 marksShow worked solution The median is . Since The median is \(160\). Since the ordered heights are
\[
\text{Diana},\ 158,\ \text{Candela},\ 175
\]
\[
\frac{158+\text{Candela}}{2}=160
\]
\[
\text{Candela}=162
\]
The range is The range is \(21\):
\[
175-\text{Diana}=21
\]
\[
\text{Diana}=154
\]
Final answer Candela \(=162\text{ cm}\), Diana \(=154\text{ cm}\).
31. Jun 2022 P2H · Q12 Medium 3 marksShow worked solution Total mass of the 7 Total mass of the 7 parcels:
\[
7 \times 2.7=18.9
\]
Total mass of the 4 Total mass of the 4 remaining parcels:
\[
4 \times 3.3=13.2
\]
Total mass of the 3 Total mass of the 3 delivered parcels:
\[
18.9-13.2=5.7
\]
Each delivered parcel has mass Each delivered parcel has mass \(W\):
\[
3W=5.7
\]
\[
W=1.9
\]
Final answer \(1.9\text{ kg}\).
32. Jan 2022 P2HR · Q8 Medium 3 marksShow worked solution Total mass of all 50 Total mass of all 50 tigers:
\[
50 \times 218=10900
\]
Total mass of the 22 Total mass of the 22 Siberian tigers:
\[
22 \times 260=5720
\]
Total mass of the 28 Total mass of the 28 Bengal tigers:
\[
10900-5720=5180
\]
Mean mass of the Bengal Mean mass of the Bengal tigers:
\[
\frac{5180}{28}=185
\]
Final answer \(185\text{ kg}\).
33. Jan 2022 P2H · Q15 Medium 3 marksShow worked solution Put the distances in order Put the distances in order:
\[
5,\ 5,\ 7,\ 8,\ 10,\ 12,\ 13,\ 14,\ 16,\ 21,\ 23
\]
Calculate statistic There are 11 values, so the median is the 6th value.
The lower half is \[
5,\ 5,\ 7,\ 8,\ 10
\]
Calculate value so
\[
Q_1=7
\]
The upper half is \[
13,\ 14,\ 16,\ 21,\ 23
\]
Calculate value so
\[
Q_3=16
\]
\[
\text{IQR}=16-7=9
\]
Final answer \(9\text{ km}\).
34. May 2021 P1H · Q10 Medium 3 marksShow worked solution The total for the first The total for the first 5 tests is
\[
5 \times 74=370
\]
Calculate value For an average of at least \(77\) after 6 tests, the total must be at least
\[
6 \times 77=462
\]
Calculate value So the last test score must be at least
\[
462-370=92
\]
Final answer at least \(92\).
35. Jan 2021 P2H · Q2 Medium 3 marksShow worked solution Calculate statistic The median is \(7\), so the middle number is \(7\).
Calculate statistic The mode is \(8\), so use two 8s.
Calculate statistic The range is \(5\), so choose the smallest and largest numbers to differ by \(5\). One possible set is
\[
3,\ 6,\ 7,\ 8,\ 8
\]
Range Range:
\[
8-3=5
\]
Mode Mode:
\[
8
\]
Median Median:
\[
7
\]
Final answer \(3,\ 6,\ 7,\ 8,\ 8\).
36. Jan 2021 P2H · Q7 Medium 3 marksShow worked solution Total weight of 5 cocoa Total weight of 5 cocoa pods:
\[
5\times 398=1990\text{ g}
\]
Total weight of 6 cocoa Total weight of 6 cocoa pods:
\[
6\times 401=2406\text{ g}
\]
Weight of the sixth cocoa Weight of the sixth cocoa pod:
\[
2406-1990=416
\]
Final answer \(416\text{ grams}\).
37. Jan 2020 P2HR · Q2 Medium 3 marksShow worked solution Use midpoints Use midpoints \(10,30,50,70,90\).
\[
10(5)+30(11)+50(8)+70(19)+90(9)
\]
\[
=50+330+400+1330+810
=2920
\]
Final answer \(\pounds2920\).
38. Jan 2020 P1HR · Q8 Medium 3 marksShow worked solution Total weight of 5 children Total weight of 5 children:
\[
5\times 28=140\text{ kg}
\]
Total weight of the 2 Total weight of the 2 children who get off:
\[
2\times 26.5=53\text{ kg}
\]
Total weight of the 3 Total weight of the 3 children left:
\[
140-53=87\text{ kg}
\]
Mean weight Mean weight:
\[
\frac{87}{3}=29
\]
Final answer \(29\text{ kg}\).
39. Jan 2020 P2HR · Q14 Medium 3 marksShow worked solution Put the marks in order Put the marks in order:
\[
5,\ 7,\ 11,\ 12,\ 13,\ 14,\ 15,\ 16,\ 17,\ 18,\ 18
\]
\[
Q_1=11,\qquad Q_3=17
\]
\[
\text{IQR}=17-11=6
\]
40. Specimen 4WM2H · Q1 Medium 3 marksShow worked solution Calculate statistic The mean is \(12\), so the total of the five numbers is
\[
5 \times 12 = 60
\]
Calculate value The known numbers add to
\[
15+7-2+23=43
\]
Calculate value Therefore
\[
43+x=60
\]
\[
x=17
\]
41. Specimen 4WM2H · Q6 Medium 3 marksShow worked solution Calculate statistic The mode is \(9\), and \(d\) appears three times, so
\[
d=9
\]
Calculate statistic The range is \(4\), so
\[
d-a=4
\]
\[
9-a=4
\]
\[
a=5
\]
Calculate statistic There are six values, so the median is the mean of the 3rd and 4th values:
\[
\frac{c+d}{2}=8
\]
\[
\frac{c+9}{2}=8
\]
\[
c=7
\]
Solve inequality Since \(a<b<c<d\), the remaining integer is \(b=6\).
Final answer \(a=5,\ b=6,\ c=7,\ d=9\).
42. Jun 2026 P2HR · Q6 Medium 4 marksShow worked solution Solution | Step 1 The largest card is 8, so the smallest is \(8-7=1\).
Range is largest minus smallest.
Solution | Step 2 The fifth card must be 8, since 8 is the mode.
Only the fifth position can supply another 8 in the ordered cards.
Solution | Step 3 If the third card is \(z\), then \[1+3+z+7+8+8=31\quad\Rightarrow\quad z=4.\]
Use the total of all six cards.
Solution | Step 4 The ordered cards are \(1,3,4,7,8,8\), so \[\text{median}=\frac{4+7}{2}=5.5.\]
With six values, average the third and fourth.
Final answer Median \(=5.5\).
43. Jun 2026 P2H · Q13 Medium 4 marksShow worked solution Solution | Step 1 Tree A contributes \(12\times150=1800\) grams.
Total weight equals count times mean weight.
Solution | Step 2 Tree C has \(2610\div145=18\) apples.
Number of items equals total divided by mean.
Solution | Step 3 The total weight is \(1800+2700+2610=7110\) grams and the total number of apples is \(12+20+18=50\).
Add weights and counts separately; do not average the three group means.
Solution | Step 4 \[\text{Combined mean}=\frac{7110}{50}=142.2\text{ grams}.\]
Divide the total weight by the total count.
Final answer \(142.2\text{ grams}\)
44. Nov 2025 P1H · Q1 Medium 4 marksShow worked solution Use midpoints Use midpoints \(2.5,7.5,12.5,17.5,22.5\).
\[
2.5(8)+7.5(7)+12.5(3)+17.5(10)+22.5(2)=330
\]
\[
\text{estimated mean}=\frac{330}{30}=11
\]
Final answer \(11\text{ km}\).
45. Nov 2025 4WM2H · Q2 Medium 4 marksShow worked solution Find class midpoints **Step 1: Find the midpoints (*x*) for each class interval.**
0 h 10 x 5 0 \< *h* ≤ 10 ⟹ *x* = 5
10 h 20 x 15 10 \< *h* ≤ 20 ⟹ *x* = 15
20 h 30 x 25 20 \< *h* ≤ 30 ⟹ *x* = 25
30 h 40 x 35 30 \< *h* ≤ 40 ⟹ *x* = 35
40 h 50 x 45 40 \< *h* ≤ 50 ⟹ *x* = 45
Find the midpoint **Step 2: Multiply frequency (*f*) by midpoint (*x*) to find *f**x*.**
Calculate statistic $$\\text{Mean} = \\frac{\\sum fx}{\\sum f} = \\frac{1290}{60} = 21.5$$
46. May 2025 4WM2H · Q8 Medium 4 marksShow worked solution Use class midpoints Use class midpoints \(5,15,25,35,45\).
\[
\frac{5x+15(3)+25(2)+35(4)+45(2)}{x+3+2+4+2}=20
\]
\[
\frac{5x+325}{x+11}=20
\]
\[
5x+325=20x+220
\]
\[
105=15x
\]
\[
x=7
\]
47. Nov 2023 P2H · Q1 Medium 4 marksShow worked solution The modal class is the The modal class is the class with the greatest frequency:
\[
25<m\le 30
\]
Use midpoints Use midpoints \(2.5,7.5,12.5,17.5,22.5,27.5\).
\[
2.5(8)+7.5(2)+12.5(6)+17.5(4)+22.5(12)+27.5(18)=945
\]
Final answer modal class \(25<m\le 30\), estimated total \(945\) minutes.
48. Jun 2023 P2H · Q10 Medium 4 marksShow worked solution Team A Team A:
\[
11 \times 17=187
\]
Estimate the value After the next round:
\[
12 \times 18.5=222
\]
\[
\text{next score}=222-187=35
\]
Team B Team B:
\[
9 \times 18=162
\]
Estimate the value After the next round:
\[
10 \times 18.5=185
\]
\[
\text{next score}=185-162=23
\]
Difference Difference:
\[
35-23=12
\]
49. Jun 2023 P2HR · Q14 Medium 4 marksShow worked solution Akari s scores in order Akari's scores in order are
\[
20,\ 20,\ 22,\ 23,\ 25,\ 26,\ 26,\ 27,\ 28,\ 29,\ 29
\]
\[
Q_1=22,\qquad Q_3=28
\]
\[
\text{IQR}=28-22=6
\]
Machi s interquartile range is Machi's interquartile range is \(9\).
Akari is more consistent because Akari is more consistent because her interquartile range is smaller.
Final answer Akari's IQR is \(6\), and Akari is more consistent.
50. Jan 2022 P1HR · Q5 Medium 4 marksShow worked solution The range is . Since The range is \(16\). Since the greatest known mark is \(80\),
\[
80-16=64
\]
Use Statistics Toolkit tag correct So one missing mark is \(64\).
Calculate statistic The mode is \(80\), so another missing mark is \(80\).
Use Statistics Toolkit tag correct Now the marks in order are
\[
64,\ 68,\ 72,\ x,\ 75,\ 77,\ 80,\ 80
\]
Calculate statistic The median is \(74\), so
\[
\frac{x+75}{2}=74
\]
\[
x=73
\]
Final answer \(64,\ 73,\ 80\).
51. Jan 2022 P2H · Q9 Medium 4 marksShow worked solution The total number of points The total number of points is
\[
0(13)+1(17)+2(8)+3x+4(11)=77+3x
\]
The total frequency is \[
13+17+8+x+11=49+x
\]
Calculate statistic The mean is \(2\), so
\[
\frac{77+3x}{49+x}=2
\]
\[
77+3x=98+2x
\]
\[
x=21
\]
52. Jan 2021 P1HR · Q2 Medium 4 marksShow worked solution Calculate statistic The mode is \(7\), and the numbers are \(a,a,b,c\), so
\[
a=7
\]
The median of four numbers The median of four numbers is
\[
\frac{a+b}{2}
\]
Evaluate fraction So
\[
\frac{7+b}{2}=8.5
\]
\[
7+b=17
\]
\[
b=10
\]
Calculate statistic The mean is \(9\), so the total is
\[
4\times 9=36
\]
\[
7+7+10+c=36
\]
\[
c=12
\]
Final answer \(a=7,\ b=10,\ c=12\).
53. Jan 2021 P2HR · Q10 Medium 4 marksShow worked solution Use class midpoints Use class midpoints \(1,3,5,7,9\).
The total frequency is \[
x+11+8+6+5=x+30
\]
The total estimated distance is \[
1x+3(11)+5(8)+7(6)+9(5)=x+160
\]
The estimated mean is The estimated mean is \(4.25\):
\[
\frac{x+160}{x+30}=4.25
\]
\[
x+160=4.25x+127.5
\]
\[
32.5=3.25x
\]
\[
x=10
\]
54. May-Nov 2020 P1H · Q4 Medium 4 marksShow worked solution The median is The median is \(9\):
\[
\frac{x+10}{2}=9
\]
\[
x=8
\]
Calculate statistic The mean is \(11\), so the total is
\[
6 \times 11=66
\]
\[
4+7+8+10+y+y=66
\]
\[
29+2y=66
\]
\[
2y=37
\]
\[
y=18.5
\]
Final answer \(x=8,\ y=18.5\).
55. Jun 2026 P2H · Q3 Hard 5 marksShow worked solution Part (a) | Step 1 The greatest frequency is \(15\), in the interval \(27<h\le31\).
The modal class is the class with the highest frequency.
Part (b) | Step 1 The class midpoints are \(17,21,25,29,33\) hours.
Use the midpoint of each class as a representative value.
Part (b) | Step 2 \[\sum fm=2(17)+5(21)+12(25)+15(29)+6(33)\]
\[=34+105+300+435+198=1072.\]
Multiply each midpoint by its frequency and add.
Part (b) | Step 3 \[\text{Estimated mean}=\frac{1072}{2+5+12+15+6}=\frac{1072}{40}=26.8\text{ hours}.\]
Divide the estimated total by the number of candles.
Final answer (a) \(27<h\le31\)
(b) \(26.8\text{ hours}\)
56. Jun 2025 P1H · Q1 Hard 5 marksShow worked solution The modal class is the The modal class is the class with the greatest frequency.
Modal class **Modal class:** \(5<d\le 10\).
Use midpoints Use midpoints \(2.5,7.5,12.5,17.5,22.5\).
\[
2.5(26)+7.5(40)+12.5(16)+17.5(10)+22.5(8)=920
\]
\[
\text{estimated mean}=\frac{920}{100}=9.2
\]
Final answer modal class \(5<d\le 10\), estimated mean \(9.2\text{ km}\).
57. May 2025 4WM2HR · Q1 Hard 5 marksShow worked solution Required \(a\) Write down the modal class. (b) Estimate the mean weight of the 50 dogs.
Part (a) The **modal class** is the interval with the highest frequency. From the table, the highest frequency is **17**, which belongs to:
Answer Modal class: 8 \< *w* ≤ 12 kg
Find the midpoint of each Find the midpoint of each class interval:
Apply the estimated mean formula Apply the estimated mean formula: $$\\bar{x} = \\frac{\\sum f\\!x}{\\sum f} = \\frac{660}{50} = 13.2 \\text{ kg}$$
Final answer Estimated mean weight = 13.2 kg
58. Nov 2024 P1H · Q1 Hard 5 marksShow worked solution The modal class is the The modal class is the class with the greatest frequency:
\[
10<p\le 15
\]
Use midpoints Use midpoints \(12.5,17.5,22.5,27.5,32.5\).
\[
12.5(18)+17.5(16)+22.5(14)+27.5(8)+32.5(4)=1170
\]
\[
\text{estimated mean}=\frac{1170}{60}=19.5
\]
Final answer modal class \(10<p\le 15\), estimated mean \(19.5\) dollars.
59. Nov 2024 P2H · Q10 Hard 5 marksShow worked solution Evaluate fraction For year 10, Biology is \(126^\circ\) out of \(360^\circ\).
\[
\frac{126}{360}\times 300=105
\]
Calculate value For year 11:
\[
(3x+6)+(5x+8)+(7x-9)=320
\]
\[
15x+5=320
\]
\[
x=21
\]
Year 11 Biology Year 11 Biology:
\[
3(21)+6=69
\]
Difference Difference:
\[
105-69=36
\]
Final answer \(36\) more students.
60. Nov 2021 P2H · Q3 Hard 5 marksShow worked solution More than euros More than \(24\) euros:
\[
25+6=31
\]
Total frequency Total frequency:
\[
6+14+19+25+6=70
\]
\[
P(S>24)=\frac{31}{70}
\]
Use midpoints Use midpoints \(4,12,20,28,36\).
\[
4(6)+12(14)+20(19)+28(25)+36(6)=1488
\]
\[
\text{estimated mean}=\frac{1488}{70}=21.257\ldots
\]
Final answer \(\frac{31}{70}\), estimated mean \(21.3\) euros.
61. Jan 2021 P1H · Q2 Hard 5 marksShow worked solution (a) The greatest frequency is (a) The greatest frequency is \(23\), so the modal class is
\[
70<s\le 80
\]
(b) Use midpoints (b) Use midpoints \(45,55,65,75,85\):
\[
\frac{45(10)+55(16)+65(19)+75(23)+85(12)}{80}
=66.375
\]
To 3 significant figures To 3 significant figures:
\[
66.4
\]
Final answer (a) \(70<s\le80\), (b) \(66.4\text{ km/h}\).
62. May-Nov 2020 P1HR · Q4 Hard 5 marksShow worked solution The modal class is the The modal class is the class with the greatest frequency:
\[
30<t\le 40
\]
Use midpoints Use midpoints \(5,15,25,35,45\).
\[
5(4)+15(10)+25(15)+35(25)+45(6)=1690
\]
\[
\text{estimated mean}=\frac{1690}{60}=28.166\ldots
\]
Final answer modal class \(30<t\le 40\), estimated mean \(28.2\) minutes.
63. Jan 2022 P1H · Q5 Hard 6 marksShow worked solution One possible ordered set is \[
5,\ 8,\ 8,\ 20,\ 21,\ 24
\]
Check Check:
\[
\text{median}=\frac{8+20}{2}=14
\]
Calculate statistic The mode is \(8\), the smallest number is \(5\), and the largest number is \(24\).
Calculate value For the basketball scores:
\[
5 \times 21=105
\]
\[
6 \times 23=138
\]
\[
138-105=33
\]
Final answer \(5,8,8,20,21,24\) and \(33\).
64. Jan 2020 P1H · Q4 Hard 7 marksShow worked solution (a) The greatest frequency is (a) The greatest frequency is \(16\), so the modal class is
\[
3<w\le 4
\]
(b) Use class midpoints (b) Use class midpoints:
\[
\frac{2.5(12)+3.5(16)+4.5(9)+5.5(2)+6.5(1)}{40}
=3.6
\]
(c) Weights greater than kg (c) Weights greater than \(4\) kg:
\[
9+2+1=12
\]
\[
P(w>4)=\frac{12}{40}=\frac{3}{10}
\]
Final answer (a) \(3<w\le4\), (b) \(3.6\text{ kg}\), (c) \(\frac{3}{10}\).
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