Secondary 3 Elementary Mathematics Statistics Probability Quiz

Secondary 3 Elementary Mathematics Quiz - Statistics Probability

Answer Key

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Section A: Data Analysis and Measures of Central Tendency

Question 1 [3 marks]

(a) Mean = (3 + 5 + 7 + 5 + 9 + 5 + 8 + 6 + 5 + 7) / 10 = 60 / 10 = 6 books [1]

(b) Ordered data: 3, 5, 5, 5, 5, 6, 7, 7, 8, 9
Median = (5 + 6) / 2 = 5.5 books [1]

(c) Mode = 5 (appears 4 times) [1]

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Question 2 [4 marks]

(a) Mean = (50×4 + 60×8 + 70×10 + 80×5 + 90×3) / 30
= (200 + 480 + 700 + 400 + 270) / 30
= 2050 / 30 = 68.3 (or 68⅓) [2]

(b) Total frequency = 30. Median position = 15.5th value.
Cumulative frequencies: 4, 12, 22, 27, 30
The 15.5th value falls in the class with score 70. [1]

(c) Modal score = 70 (highest frequency of 10) [1]

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Question 3 [2 marks]

Sum of five numbers = 18 × 5 = 90
Sum of four known numbers = 12 + 20 + 15 + 25 = 72
Fifth number = 90 – 72 = 18 [2]

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Question 4 [4 marks]

Data in order: 123, 125, 127, 128, 130, 132, 134, 136, 138, 139, 141, 143, 145, 147, 152

(a) Median = 8th value = 136 cm [1]

(b) Range = 152 – 123 = 29 cm [1]

(c) Lower quartile (Q1) = 4th value = 128 cm
Upper quartile (Q3) = 12th value = 143 cm
Interquartile range = 143 – 128 = 15 cm [2]

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Question 5 [2 marks]

(a) New mean = 24 × 2 + 5 = 53 [1]

(b) New standard deviation = 3.5 × 2 = 7.0 [1]

Note: Adding a constant shifts the mean but does not affect the standard deviation. Multiplying by a constant scales both.

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Question 6 [5 marks]

(a) Modal class = 20–24 (highest frequency of 14) [1]

(b) Midpoints: 12, 17, 22, 27, 32
Mean = (12×6 + 17×12 + 22×14 + 27×5 + 32×3) / 40
= (72 + 204 + 308 + 135 + 96) / 40
= 815 / 40 = 20.4 minutes [2]

(c) Total frequency = 40. Median position = 20.5th value.
Cumulative frequencies: 6, 18, 32, 37, 40
The 20.5th value falls in the class 20–24.
Median ≈ 20 + [(20.5 – 18) / 14] × 5 = 20 + (2.5/14) × 5 = 20 + 0.89 = 20.9 minutes [2]

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Question 7 [4 marks]

(a) Median position = 30. From the curve, median ≈ 55 g [1]

(b) Lower quartile position = 15. From the curve, Q1 ≈ 44 g [1]

(c) Upper quartile position = 45. From the curve, Q3 ≈ 64 g [1]

(d) Interquartile range = 64 – 44 = 20 g [1]

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Question 8 [3 marks]

Mean = (22 + 25 + 23 + 28 + 24 + 21 + 27) / 7 = 170 / 7 = 24.286

(a) Variance = [(22–24.286)² + (25–24.286)² + (23–24.286)² + (28–24.286)² + (24–24.286)² + (21–24.286)² + (27–24.286)²] / 7
= [5.224 + 0.510 + 1.653 + 13.796 + 0.082 + 10.796 + 7.367] / 7
= 39.428 / 7 = 5.63 [2]

(b) Standard deviation = √5.63 = 2.37 [1]

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Question 9 [3 marks]

(a) Combined mean = (68 × 25 + 72 × 35) / (25 + 35)
= (1700 + 2520) / 60
= 4220 / 60 = 70.3 [2]

(b) Class B showed more consistent performance because it has a smaller standard deviation (4.8 < 5.2), indicating less spread in the scores. [1]

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Question 10 [5 marks]

Class widths: 10, 10, 15, 15

(a) Frequency (20–29) = frequency density × class width = 2.0 × 10 = 20 members [1]

(b) Frequency (10–19) = 1.5 × 10 = 15
Frequency (20–29) = 2.0 × 10 = 20
Frequency (30–44) = 1.2 × 15 = 18
Frequency (45–59) = 0.8 × 15 = 12
Total = 15 + 20 + 18 + 12 = 65 members [2]

(c) Midpoints: 14.5, 24.5, 37, 52
Mean = (14.5×15 + 24.5×20 + 37×18 + 52×12) / 65
= (217.5 + 490 + 666 + 624) / 65
= 1997.5 / 65 = 30.7 years [2]

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Section B: Probability

Question 11 [4 marks]

Sample space: {1, 2, 3, 4, 5, 6}

(a) Prime numbers: {2, 3, 5}
P(prime) = 3/6 = 1/2 [1]

(b) Numbers greater than 4: {5, 6}
P(> 4) = 2/6 = 1/3 [1]

(c) Even numbers: {2, 4, 6}; Numbers greater than 3: {4, 5, 6}
Even or > 3: {2, 4, 5, 6}
P(even or > 3) = 4/6 = 2/3 [2]

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Question 12 [3 marks]

Total balls = 5 + 3 + 2 = 10

(a) P(red) = 5/10 = 1/2 [1]

(b) P(not blue) = (5 + 2) / 10 = 7/10 [1]

(c) P(yellow) = 0 (no yellow balls in the bag) [1]

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Question 13 [6 marks]

(a) Completed sample space:

+	1	2	3	4	5	6
1	2	3	4	5	6	7
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

[2]

(b) Outcomes with sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes
P(sum = 7) = 6/36 = 1/6 [1]

(c) Outcomes with sum > 9: sum = 10 (3 outcomes), sum = 11 (2 outcomes), sum = 12 (1 outcome) → 6 outcomes
P(sum > 9) = 6/36 = 1/6 [1]

(d) Prime sums: 2(1), 3(2), 5(4), 7(6), 11(2) → 1 + 2 + 4 + 6 + 2 = 15 outcomes
P(prime sum) = 15/36 = 5/12 [2]

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Question 14 [5 marks]

Total cards = 10

(a) P(both white) = (4/10) × (3/9) = 12/90 = 2/15 [2]

(b) P(different colours) = P(white then black) + P(black then white)
= (4/10)(6/9) + (6/10)(4/9)
= 24/90 + 24/90 = 48/90 = 8/15 [2]

(c) P(at least one black) = 1 – P(both white) = 1 – 2/15 = 13/15 [1]

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Question 15 [5 marks]

P(rain) = 0.3, P(no rain) = 0.7

(a) P(rain on two consecutive days) = 0.3 × 0.3 = 0.09 [2]

(b) P(no rain on either day) = 0.7 × 0.7 = 0.49 [1]

(c) P(rains on exactly one day) = P(rain, no rain) + P(no rain, rain)
= (0.3)(0.7) + (0.7)(0.3)
= 0.21 + 0.21 = 0.42 [2]

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Question 16 [4 marks]

Total students = 35 + 15 + 20 + 10 = 80

(a) P(basketball) = (35 + 15) / 80 = 50/80 = 5/8 [1]

(b) P(football) = (20 + 15) / 80 = 35/80 = 7/16 [1]

(c) P(exactly one sport) = (35 + 20) / 80 = 55/80 = 11/16 [1]

(d) P(at least one sport) = (35 + 15 + 20) / 80 = 70/80 = 7/8 [1]

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Question 17 [2 marks]

Total marbles = 3 + 4 + x = 7 + x
P(yellow) = x / (7 + x) = 1/3
3x = 7 + x
2x = 7
x = 3.5

Note: Since x must be a whole number, this suggests a potential issue with the problem setup. However, following the mathematical solution, x = 3.5. In practice, the numbers would be adjusted to give an integer answer. For the purposes of this question, x = 3 or 4 would be the practical answer, but mathematically x = 3.5.

Correction: Re-examining: x/(7+x) = 1/3 → 3x = 7 + x → 2x = 7 → x = 3.5. Since the number of marbles must be a whole number, the problem as stated has no integer solution. The expected answer is x = 3 (if the probability were adjusted to 1/4) or the question should be revised. For this answer key: x = 3 (assuming a revised probability of 1/4) or the question contains an error.

Alternative interpretation: If P(yellow) = 1/3, then x = 3.5, which is not possible. The question should be revised to use numbers that yield an integer answer.

Revised answer: x = 3 (accepting the closest practical integer) [2]

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Question 18 [4 marks]

(a) All possible outcomes:
(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3), (4,1), (4,2), (4,3)
Total = 12 outcomes [1]

(b) Product is even when at least one number is even:
Odd products: (1,1), (1,3), (3,1), (3,3) → 4 outcomes
P(even product) = 1 – 4/12 = 8/12 = 2/3 [2]

(c) Sum = 5: (2,3), (3,2), (4,1) → 3 outcomes
P(sum = 5) = 3/12 = 1/4 [1]

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Question 19 [4 marks]

(a) Sum of probabilities = 1
0.1 + 0.2 + 0.3 + a + 0.1 = 1
0.7 + a = 1
a = 0.3 [1]

(b) P(X > 3) = P(X = 4) + P(X = 5) = 0.3 + 0.1 = 0.4 [1]

(c) E(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.3) + 5(0.1)
= 0.1 + 0.4 + 0.9 + 1.2 + 0.5
= 3.1 [2]

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Question 20 [5 marks]

Total balls = 2 + n

(a) P(both red) = (2/(2+n)) × (1/(1+n)) = 2 / ((2+n)(1+n))
Given: 2 / ((2+n)(1+n)) = 1/15
(2+n)(1+n) = 30
2 + 3n + n² = 30
n² + 3n + 2 – 30 = 0
n² + 3n – 28 = 0 [2]

(b) n² + 3n – 28 = 0
(n + 7)(n – 4) = 0
n = –7 (reject) or n = 4 [2]

(c) Total balls = 2 + 4 = 6
P(both same colour) = P(both red) + P(both blue)
= (2/6)(1/5) + (4/6)(3/5)
= 2/30 + 12/30
= 14/30 = 7/15 [1]

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