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O Level Elementary Mathematics Statistics Probability Quiz

Free O Level E Maths Statistics quiz, Qwen3.6 Exam version, with questions, answers, and O Level-style practice for Singapore students.

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O Level Elementary Mathematics From Real Exams Generated by Qwen3.6 Plus Updated 2026-08-17

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Answers

O-Level Elementary Mathematics Quiz - Statistics Probability (Answer Key)

1.
(a) Median:
Total n=30n = 30. Median position is 30+12=15.5\frac{30+1}{2} = 15.5th value.
15th value = 169, 16th value = 170.
Median = 169+1702=169.5\frac{169 + 170}{2} = 169.5 cm.
[1]

(b) Interquartile Range (IQR):
Q1Q_1 position = 30+14=7.75\frac{30+1}{4} = 7.75th value \approx 8th value = 165.
Q3Q_3 position = 3(30+1)4=23.25\frac{3(30+1)}{4} = 23.25th value \approx 23rd value = 176.
IQR=Q3Q1=176165=11IQR = Q_3 - Q_1 = 176 - 165 = 11 cm.
[2]

2.
(a) Cumulative Frequency Curve:
Points plotted: (1,5),(2,18),(3,35),(4,46),(5,50)(1, 5), (2, 18), (3, 35), (4, 46), (5, 50).
Smooth curve drawn through points, starting from (0,0)(0,0).
[3]

(b) Estimation:
At t=3.5t = 3.5, read from graph.
Cumulative frequency 40.5\approx 40.5 (Accept 40–41).
Students spending more than 3.5 hours = 5040.5=9.550 - 40.5 = 9.5.
Answer: 9 or 10 students.
[2]

3.
Total mass of boys = 8×65=5208 \times 65 = 520 kg.
Total mass of girls = 12×55=66012 \times 55 = 660 kg.
Total mass = 520+660=1180520 + 660 = 1180 kg.
Total students = 8+12=208 + 12 = 20.
Mean mass = 118020=59\frac{1180}{20} = 59 kg.
[2]

4.
(a) Mean:
fx=(1×4)+(2×8)+(3×12)+(4×10)+(5×6)=4+16+36+40+30=126\sum fx = (1\times4) + (2\times8) + (3\times12) + (4\times10) + (5\times6) = 4 + 16 + 36 + 40 + 30 = 126.
f=40\sum f = 40.
Mean = 12640=3.15\frac{126}{40} = 3.15.
[2]

(b) Standard Deviation:
fx2=(12×4)+(22×8)+(32×12)+(42×10)+(52×6)=4+32+108+160+150=454\sum fx^2 = (1^2\times4) + (2^2\times8) + (3^2\times12) + (4^2\times10) + (5^2\times6) = 4 + 32 + 108 + 160 + 150 = 454.
Variance = fx2f(Mean)2=45440(3.15)2=11.359.9225=1.4275\frac{\sum fx^2}{\sum f} - (\text{Mean})^2 = \frac{454}{40} - (3.15)^2 = 11.35 - 9.9225 = 1.4275.
Standard Deviation = 1.42751.19\sqrt{1.4275} \approx 1.19 (3 s.f.).
[3]

5.
(a) IQR Comparison:
IQRA=5525=30IQR_A = 55 - 25 = 30.
IQRB=5035=15IQR_B = 50 - 35 = 15.
Class A has the larger IQR.
[1]

(b) Consistency:
Class B has a smaller IQR (and smaller range), so Class B is more consistent.
OR
Class A has a larger spread, so Class A is less consistent.
[1]

6.
Total balls = 5+3+2=105 + 3 + 2 = 10.
(a) P(Red)=510=12P(\text{Red}) = \frac{5}{10} = \frac{1}{2}.
[1]
(b) P(Not Blue)=1P(Blue)=1310=710P(\text{Not Blue}) = 1 - P(\text{Blue}) = 1 - \frac{3}{10} = \frac{7}{10}.
[1]

7.
Sample space: {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\}.
(a) Primes: {2,3,5}\{2, 3, 5\}. P(Prime)=36=12P(\text{Prime}) = \frac{3}{6} = \frac{1}{2}.
[1]
(b) >4> 4: {5,6}\{5, 6\}. P(>4)=26=13P(>4) = \frac{2}{6} = \frac{1}{3}.
[1]

8.
P(Not Rain)=1P(Rain)=10.3=0.7P(\text{Not Rain}) = 1 - P(\text{Rain}) = 1 - 0.3 = 0.7.
[1]

9.
Mutually exclusive: P(AB)=P(A)+P(B)P(A \cup B) = P(A) + P(B).
P(AB)=0.4+0.25=0.65P(A \cup B) = 0.4 + 0.25 = 0.65.
[1]

10.
Possibility Diagram:
Rows/Cols labelled 1-4.
16 outcomes listed/shown in grid.
[2]

11.
(a) Sum = 5: (1,4),(2,3),(3,2),(4,1)(1,4), (2,3), (3,2), (4,1). 4 outcomes.
P(Sum=5)=416=14P(\text{Sum}=5) = \frac{4}{16} = \frac{1}{4}.
[1]
(b) Sum < 4: (1,1),(1,2),(2,1)(1,1), (1,2), (2,1). 3 outcomes.
P(Sum<4)=316P(\text{Sum}<4) = \frac{3}{16}.
[1]

12.
Sample Space: {HH,HT,TH,TT}\{HH, HT, TH, TT\}.
Exactly one Head: {HT,TH}\{HT, TH\}.
P(1 Head)=24=12P(\text{1 Head}) = \frac{2}{4} = \frac{1}{2}.
[2]

13.
n(Physics)=60n(\text{Physics}) = 60, n(Chem)=50n(\text{Chem}) = 50, n(Both)=20n(\text{Both}) = 20.
n(Physics only)=6020=40n(\text{Physics only}) = 60 - 20 = 40.
n(Chem only)=5020=30n(\text{Chem only}) = 50 - 20 = 30.
n(Neither)=100(40+30+20)=10n(\text{Neither}) = 100 - (40 + 30 + 20) = 10.

(a) P(Neither)=10100=110P(\text{Neither}) = \frac{10}{100} = \frac{1}{10} or 0.10.1.
[2]
(b) P(Physics only)=40100=25P(\text{Physics only}) = \frac{40}{100} = \frac{2}{5} or 0.40.4.
[1]

14.
Total balls = 10.
(a) P(WW)=410×39=1290=215P(WW) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}.
[2]
(b) P(WB or BW)=(410×69)+(610×49)=2490+2490=4890=815P(WB \text{ or } BW) = \left(\frac{4}{10} \times \frac{6}{9}\right) + \left(\frac{6}{10} \times \frac{4}{9}\right) = \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}.
[2]

15.
P(M)=0.8P(M) = 0.8, P(S)=0.7P(S) = 0.7. Independent.
(a) P(Both)=0.8×0.7=0.56P(\text{Both}) = 0.8 \times 0.7 = 0.56.
[1]
(b) P(Exactly One)=P(MS)+P(MS)P(\text{Exactly One}) = P(M \cap S') + P(M' \cap S).
P(MS)=0.8×(10.7)=0.8×0.3=0.24P(M \cap S') = 0.8 \times (1-0.7) = 0.8 \times 0.3 = 0.24.
P(MS)=(10.8)×0.7=0.2×0.7=0.14P(M' \cap S) = (1-0.8) \times 0.7 = 0.2 \times 0.7 = 0.14.
Total = 0.24+0.14=0.380.24 + 0.14 = 0.38.
[2]

16.
Numbers 1-20.
A={3,6,9,12,15,18}A = \{3, 6, 9, 12, 15, 18\} (6 numbers).
B={2,4,6,8,10,12,14,16,18,20}B = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20\} (10 numbers).
AB={6,12,18}A \cap B = \{6, 12, 18\} (3 numbers).

(a) P(A)=620=310P(A) = \frac{6}{20} = \frac{3}{10}.
[1]
(b) P(AB)=320P(A \cap B) = \frac{3}{20}.
[1]
(c) P(AB)=P(A)+P(B)P(AB)=620+1020320=1320P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{6}{20} + \frac{10}{20} - \frac{3}{20} = \frac{13}{20}.
[1]

17.
Total marbles = 10. With replacement means probabilities remain constant.
(a) P(RR)=310×310=9100=0.09P(RR) = \frac{3}{10} \times \frac{3}{10} = \frac{9}{100} = 0.09.
[1]
(b) P(RB)=310×710=21100=0.21P(RB) = \frac{3}{10} \times \frac{7}{10} = \frac{21}{100} = 0.21.
[1]

18.
fx=(0×5)+(1×15)+(2×20)+(3×8)+(4×2)\sum fx = (0\times5) + (1\times15) + (2\times20) + (3\times8) + (4\times2)
=0+15+40+24+8=87= 0 + 15 + 40 + 24 + 8 = 87.
Total days = 50.
Mean = 8750=1.74\frac{87}{50} = 1.74 cars.
[2]

19.
(a) For normal distribution, probability within 1 SD is approx 0.68 (or 68%).
[1]
(b) Mean + 2 SD = 30+2(5)=30+10=4030 + 2(5) = 30 + 10 = 40 cm.
[1]

20.
Total people = 10. Choosing 3.
Number of ways = (103)=10×9×83×2×1=120\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120.
[2]