Secondary 3 Elementary Mathematics Statistics Probability Quiz

Secondary 3 Elementary Mathematics Quiz - Statistics Probability (Answer Key)

Total Marks: 40

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Section A: Data Handling and Representation

Question 1 [3 marks]

(a) Mode = 2
Reasoning: The mode is the value with the highest frequency. Frequency of 2 books is 9, which is the highest.

(b) Mean = 2.1
Working:
Mean = Σ(fx) / Σf
= (0×3 + 1×7 + 2×9 + 3×6 + 4×4 + 5×1) / 30
= (0 + 7 + 18 + 18 + 16 + 5) / 30
= 64 / 30
= 2.133... ≈ 2.1 (3 s.f.)

(c) Median = 2
Reasoning: For 30 students, median is the average of the 15th and 16th values.
Cumulative frequencies: 3, 10, 19, 25, 29, 30.
Both 15th and 16th values fall in the "2 books" category.
Median = 2.

Marking notes:

• 1 mark for correct mode
• 1 mark for correct mean calculation (method + answer)
• 1 mark for correct median with reasoning
• Accept 2.13 for mean if 3 s.f. used

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Question 2 [3 marks]

(a) Stem-and-leaf diagram:

15 | 2 8
16 | 0 2 5 5 7 8 9
17 | 0 1 2 3 4 5 6 8
18 | 0 2 5

Key: 15 | 2 = 152 cm

(b) IQR = 13.5 cm
Working:
n = 20
Q1 position = (20+1)/4 = 5.25 → between 5th and 6th values
5th value = 165, 6th value = 165 → Q1 = 165
Q3 position = 3(20+1)/4 = 15.75 → between 15th and 16th values
15th value = 175, 16th value = 176 → Q3 = 175.75
IQR = Q3 - Q1 = 175.75 - 165 = 10.75 cm

Alternative method (using n/4 and 3n/4):
Q1 at 5th value = 165, Q3 at 15th value = 175 → IQR = 10 cm
Both methods accepted if clearly shown.

Marking notes:

• 1 mark for correct ordered stem-and-leaf with key
• 1 mark for correct Q1 and Q3 identification
• 1 mark for correct IQR calculation
• Common mistake: forgetting to order leaves or missing key

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Question 3 [3 marks]

(a) Modal class = 60 < t ≤ 90
Reasoning: Highest frequency is 15.

(b) Estimated mean = 68.25 minutes
Working:
Use midpoints: 15, 45, 75, 105, 135
Σ(fx) = 5(15) + 12(45) + 15(75) + 6(105) + 2(135)
= 75 + 540 + 1125 + 630 + 270 = 2640
Mean = 2640 / 40 = 66 minutes

Correction: 2640/40 = 66 exactly.
Wait, recalculating: 75+540=615, +1125=1740, +630=2370, +270=2640. 2640/40 = 66.
Estimated mean = 66 minutes

(c) Median lies in class 60 < t ≤ 90
Reasoning: n = 40, median position = 20.5th value.
Cumulative frequencies: 5, 17, 32, 38, 40.
20.5th value falls in the 3rd class (60 < t ≤ 90).

Marking notes:

• 1 mark for modal class
• 1 mark for estimated mean (midpoints + calculation)
• 1 mark for correct median class with reasoning
• Common mistake: using class boundaries instead of midpoints

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Question 4 [3 marks]

From cumulative frequency curve:

(a) Median ≈ 55 marks
Method: Median at 40th value (80/2). Read horizontally from 40 on CF axis to curve, then down to marks axis.

(b) IQR ≈ 35 marks
Method: Q1 at 20th value → ≈ 38 marks. Q3 at 60th value → ≈ 73 marks. IQR = 73 - 38 = 35 marks.

(c) Students scoring > 75 marks ≈ 12
Method: At 75 marks, CF ≈ 68. Students above = 80 - 68 = 12.

Marking notes:

• 1 mark each for reasonable estimates from graph
• Allow ±2 marks tolerance for reading off graph
• Must show evidence of reading from graph (horizontal/vertical lines)

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Question 5 [3 marks]

(a) Comparisons:

1. Central tendency: Group B has higher median (18 vs 15), indicating students in Group B generally receive more pocket money.
2. Spread: Group B has larger IQR (13 vs 12) and larger range (27 vs 25), indicating more variability in pocket money amounts.

(b) Claim is INCORRECT.
Reasoning: Median of Group B = 18. For Group A, Q3 = 22. Since Q3 (22) > 18, more than 25% of Group A students receive ≥ 22, which is > 18. In fact, 50% of Group A receive ≥ 15, and 25% receive ≥ 22. The claim says "at least 75% receive less than 18" — but 25% receive ≥ 22 (>18), and some between 15-22 also receive ≥ 18. So less than 75% receive < 18.

Marking notes:

• 1 mark for each valid comparison (max 2)
• 1 mark for correct evaluation of claim with reasoning
• Must reference specific values from box plot

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Section B: Probability

Question 6 [2 marks]

(a) P(Red) = 5/10 = 1/2 = 0.5
(b) P(Not Blue) = 1 - P(Blue) = 1 - 3/10 = 7/10 = 0.7
Or: P(Red or Green) = 5/10 + 2/10 = 7/10

Marking notes: 1 mark each. Accept fractions, decimals, or percentages.

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Question 7 [2 marks]

(a) A = {2, 4, 6}, B = {2, 3, 5} → A ∩ B = {2}
(b) A ∪ B = {2, 3, 4, 5, 6} → P(A ∪ B) = 5/6
Or: P(A) + P(B) - P(A∩B) = 3/6 + 3/6 - 1/6 = 5/6

Marking notes: 1 mark each. Must list set for (a).

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Question 8 [3 marks]

Let R = rainy, C = cloudy. P(R) = 0.4, P(C) = 0.7, P(R∩C) = 0.3

(a) P(R ∪ C) = P(R) + P(C) - P(R∩C) = 0.4 + 0.7 - 0.3 = 0.8
(b) P(C but not R) = P(C) - P(R∩C) = 0.7 - 0.3 = 0.4
(c) Check independence: P(R) × P(C) = 0.4 × 0.7 = 0.28 ≠ 0.3 = P(R∩C)
→ Not independent (since P(R∩C) ≠ P(R)P(C))

Marking notes:

• 1 mark each part
• For (c), must show calculation and conclusion

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Question 9 [4 marks]

(a) Tree diagram:

First draw:          Second draw:
W (4/10) ────────── W (3/9) → WW
        └────────── B (6/9) → WB
B (6/10) ────────── W (4/9) → BW
        └────────── B (5/9) → BB

Probabilities on branches: 4/10, 6/10, then conditional probabilities.

(b) P(same colour) = P(WW) + P(BB) = (4/10)(3/9) + (6/10)(5/9) = 12/90 + 30/90 = 42/90 = 7/15 ≈ 0.467

(c) P(2nd is White) = P(WW) + P(BW) = (4/10)(3/9) + (6/10)(4/9) = 12/90 + 24/90 = 36/90 = 2/5 = 0.4
Note: Same as P(1st is White) due to symmetry.

Marking notes:

• 1 mark for correct tree diagram with probabilities
• 1 mark for P(same colour) with working
• 1 mark for P(2nd White) with working
• 1 mark for correct final answers
• Common mistake: using replacement probabilities

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Question 10 [4 marks]

(a) Venn diagram:
Two overlapping circles B (Biology) and C (Chemistry).
Intersection = 8.
B only = 18 - 8 = 10.
C only = 15 - 8 = 7.
Neither = 30 - (10+8+7) = 5.

(b) P(Biology but not Chemistry) = 10/30 = 1/3

(c) P(Biology | Chemistry) = P(B∩C) / P(C) = 8/15

Marking notes:

• 1 mark for correct Venn diagram with all values
• 1 mark for (b)
• 1 mark for (c) with conditional probability formula
• 1 mark for correct answers

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Question 11 [3 marks]

(a) Multiples of 3: 3, 6 → 2 outcomes. P(win) = 2/8 = 1/4 = 0.25
(b) P(exactly 1 win in 2 spins) = 2 × (1/4) × (3/4) = 6/16 = 3/8 = 0.375
Binomial: n=2, p=1/4, P(X=1) = 2C1 × (1/4)¹ × (3/4)¹
(c) Expected winners = 120 × 1/4 = 30

Marking notes:

• 1 mark each part
• For (b), must consider both orders (WL and LW)

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Question 12 [3 marks]

Let M = pass Math, E = pass English. P(M)=0.75, P(E)=0.8, P(M∩E)=0.65

(a) P(at least one) = P(M∪E) = 0.75 + 0.8 - 0.65 = 0.9
(b) P(exactly one) = P(M∪E) - P(M∩E) = 0.9 - 0.65 = 0.25
Or: P(M only) + P(E only) = (0.75-0.65) + (0.8-0.65) = 0.1 + 0.15 = 0.25
(c) P(E | M) = P(M∩E) / P(M) = 0.65 / 0.75 = 13/15 ≈ 0.867

Marking notes:

• 1 mark each part
• For (b), accept either method

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Question 13 [4 marks]

Total = 10 marbles (6R, 4B). Draw 3 without replacement.

(a) P(3 Red) = (6/10)(5/9)(4/8) = 120/720 = 1/6 ≈ 0.167
(b) P(exactly 2 Red) = P(RRB) + P(RBR) + P(BRR)
= 3 × (6/10)(5/9)(4/8) = 3 × 120/720 = 360/720 = 1/2 = 0.5
Wait: For exactly 2 red, the blue can be in any of 3 positions. Each sequence: (6/10)(5/9)(4/8) = 120/720. But careful: after 2 reds drawn, 4 blue remain out of 8. Yes, 4/8. So 3 × 120/720 = 360/720 = 1/2.
(c) P(at least 1 Blue) = 1 - P(no Blue) = 1 - P(3 Red) = 1 - 1/6 = 5/6 ≈ 0.833

Marking notes:

• 1 mark for (a)
• 1 mark for (b) with correct combination count (3 ways)
• 1 mark for (c) using complement
• 1 mark for correct final answers

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Question 14 [3 marks]

P(X)=0.5, P(Y)=0.4, P(X∪Y)=0.7

(a) P(X∩Y) = P(X) + P(Y) - P(X∪Y) = 0.5 + 0.4 - 0.7 = 0.2
(b) P(X|Y) = P(X∩Y)/P(Y) = 0.2/0.4 = 0.5
(c) Mutually exclusive means P(X∩Y) = 0. Here P(X∩Y) = 0.2 ≠ 0 → NOT mutually exclusive.

Marking notes:

• 1 mark each part
• For (c), must state definition and compare

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Question 15 [3 marks]

P(defective) = 0.02, P(not defective) = 0.98. n = 5 independent trials.

(a) P(none defective) = (0.98)⁵ ≈ 0.9039
(b) P(at least 1 defective) = 1 - P(none) = 1 - 0.9039 = 0.0961
(c) Expected defective in 1000 = 1000 × 0.02 = 20

Marking notes:

• 1 mark each part
• For (a), accept (0.98)^5 or calculated value
• For (c), must be whole number

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Section C: Combined Problems

Question 16 [3 marks]

(a) Mean = Σ(fx)/Σf = (1×4 + 2×8 + 3×15 + 4×12 + 5×11) / 50
= (4 + 16 + 45 + 48 + 55) / 50 = 168 / 50 = 3.36

(b) P(both scored 5) = (11/50) × (10/49) = 110/2450 = 11/245 ≈ 0.0449

Marking notes:

• 1 mark for mean
• 1 mark for correct probability setup (without replacement)
• 1 mark for correct answer

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Question 17 [3 marks]

(a) Mean = (0×15 + 1×45 + 2×25 + 3×10 + 4×5) / 100
= (0 + 45 + 50 + 30 + 20) / 100 = 145/100 = 1.45 cars

(b) P(exactly 2 of 3 own 2 cars):
Households with 2 cars = 25. Without replacement.
P = 3C2 × (25/100) × (24/99) × (75/98)
= 3 × (25/100) × (24/99) × (75/98)
= 3 × (1/4) × (24/99) × (75/98)
= 3 × (1/4) × (8/33) × (75/98)
= 3 × 600 / (4×33×98) = 1800 / 12936 = 75/539 ≈ 0.139

Marking notes:

• 1 mark for mean
• 1 mark for correct binomial-like setup with without-replacement probabilities
• 1 mark for correct answer

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Question 18 [4 marks]

P(on time) = 0.85, P(late) = 0.15. n = 5 independent days. Binomial distribution.

(a) P(5 on time) = (0.85)⁵ ≈ 0.4437
(b) P(exactly 4 on time) = 5C1 × (0.85)⁴ × (0.15)¹ = 5 × 0.5220 × 0.15 ≈ 0.3915
(c) P(late ≥ 2) = 1 - P(late 0) - P(late 1)
P(late 0) = (0.85)⁵ ≈ 0.4437
P(late 1) = 5 × (0.85)⁴ × (0.15) ≈ 0.3915
P(late ≥ 2) = 1 - 0.4437 - 0.3915 = 0.1648

Marking notes:

• 1 mark each part
• For (c), accept complement method or direct sum of P(2)+P(3)+P(4)+P(5)
• Answers to 4 d.p. or 3 s.f.

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Question 19 [4 marks]

3 Gold (G), 7 Silver (S). Total 10. Draw 2 without replacement.

(a) P(G) = 3/10 = 0.3
(b) P(S | G) = 7/9 (after 1 gold removed, 9 left with 7 silver)
(c) P(G ∩ S) = P(G) × P(S|G) = (3/10) × (7/9) = 21/90 = 7/30 ≈ 0.233
(d) Check independence: P(G) × P(S) = (3/10) × (7/10) = 21/100 = 0.21
But P(G∩S) = 7/30 ≈ 0.233 ≠ 0.21 → NOT independent
Reason: Drawing without replacement makes events dependent.

Marking notes:

• 1 mark each part
• For (d), must show calculation and conclusion

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Question 20 [4 marks]

P(Dog) = 0.6, P(Cat) = 0.4, P(Dog∩Cat) = 0.15

(a) Venn diagram:
Two overlapping circles. Intersection = 0.15.
Dog only = 0.6 - 0.15 = 0.45.
Cat only = 0.4 - 0.15 = 0.25.
Neither = 1 - (0.45+0.15+0.25) = 0.15.

(b) P(neither) = 0.15
(c) P(Cat | Dog) = P(Dog∩Cat) / P(Dog) = 0.15 / 0.6 = 0.25
(d) Check independence: P(Dog) × P(Cat) = 0.6 × 0.4 = 0.24
P(Dog∩Cat) = 0.15 ≠ 0.24 → NOT independent
The claim is false. Having a dog and having a cat are negatively associated (households with dogs are less likely to have cats than if independent).

Marking notes:

• 1 mark for Venn diagram with correct probabilities
• 1 mark for (b)
• 1 mark for (c) with conditional probability
• 1 mark for (d) with calculation and conclusion

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End of Answer Key