Secondary 3 Elementary Mathematics Statistics Probability Quiz

Secondary 3 Elementary Mathematics Quiz - Statistics Probability

Answer Key and Marking Scheme

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Section A: Data Handling and Representation (Questions 1–5)

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1. Stem-and-leaf diagram — heights of 15 students.

(a) Median height

There are 15 data values. The median is the 8th value.

Ordered data: 152, 154, 156, 158, 160, 161, 163, 165, 167, 169, 172, 174, 176, 178, 181

Median = 165 cm [1]

(b) Interquartile range

Lower quartile (Q1) = 4th value = 158 cm
Upper quartile (Q3) = 12th value = 174 cm

IQR = Q3 − Q1 = 174 − 158 = 16 cm [2]

• 1 mark for correct Q1 or Q3
• 1 mark for correct IQR

(c) Advantage of stem-and-leaf diagram

Any one of the following:

• The original data values can be recovered (raw data is preserved).
• The shape of the distribution can be seen.
• It is quick to construct for small data sets. [1]

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2. Grouped frequency distribution — test scores.

(a) Modal class

The class with the highest frequency is 31–40. [1]

(b) Estimate of the mean

Score (x)	Midpoint (m)	Frequency (f)	f × m
10–20	15	4	60
21–30	25.5	8	204
31–40	35.5	14	497
41–50	45.5	10	455
51–60	55.5	4	222
Total		40	1438

Mean = Σ(f × m) / Σf = 1438 / 40 = 35.95 [3]

• 1 mark for correct midpoints
• 1 mark for correct Σ(f × m)
• 1 mark for correct mean

(c) Median class

There are 40 students. The median lies between the 20th and 21st values.

Cumulative frequencies: 4, 12, 26, 36, 40

Both the 20th and 21st values fall in the class 31–40. [1]

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3. Box-and-whisker plot — hours of homework.

(a) Interquartile range

IQR = Q3 − Q1 = 12 − 5 = 7 hours [1]

(b) Box-and-whisker plot

0    2    4    6    8    10   12   14   16   18   20
|----+----+----+----+----+----+----+----+----+----|
|    |    |    |    |    |    |    |    |    |    |
|----+----+----+----+----+----+----+----+----+----|
         |---------|    |    |--------------|
         5    |    8   12        18
              |
              2

• Minimum = 2, Q1 = 5, Median = 8, Q3 = 12, Maximum = 18 [3]

• 1 mark for correct median line

• 1 mark for correct box (Q1 to Q3)

• 1 mark for correct whiskers

(c) Skewness

The median (8) is closer to Q1 (5) than to Q3 (12), and the right whisker (18 − 12 = 6) is longer than the left whisker (5 − 2 = 3). The distribution is positively skewed (skewed to the right). [1]

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4. Cumulative frequency curve — masses of 60 adults.

(a) Median mass

Median corresponds to cumulative frequency = 30.

From the curve, at cumulative frequency 30, the mass is approximately 68 kg. [1]

(b) 70th percentile

70th percentile corresponds to cumulative frequency = 0.70 × 60 = 42.

From the curve, at cumulative frequency 42, the mass is approximately 73 kg. [1]

(c) Number of adults with mass less than 65 kg

From the curve, at mass = 65 kg, the cumulative frequency is approximately 22. [1]

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5. Histogram — ages of sports club members.

(a) Frequency for class 10–20

Frequency density = Frequency / Class width

4.0 = Frequency / 10

Frequency = 4.0 × 10 = 40 [1]

(b) Frequency density for class 20–30

Class width = 30 − 20 = 10

Frequency density = 30 / 10 = 3.0 [2]

• 1 mark for correct class width
• 1 mark for correct frequency density

(c) Frequency for class 40–60

Total members = 120

Known frequencies:

• 10–20: 40
• 20–30: 30
• 60–80: 12

From the histogram, the frequency density for 30–40 is 5.0, so frequency = 5.0 × 10 = 50.

Frequency for 40–60 = 120 − (40 + 30 + 50 + 12) = 120 − 132 = −12

Note: There is an inconsistency in the problem data. Assuming the total is correct and the class 30–40 has frequency density 4.0 (frequency = 40), then:

Frequency for 40–60 = 120 − (40 + 30 + 40 + 12) = 120 − 122 = −2

Revised interpretation: If the class 30–40 has frequency 30 (frequency density 3.0), then:

Frequency for 40–60 = 120 − (40 + 30 + 30 + 12) = 120 − 112 = 8 [2]

• 1 mark for correct approach
• 1 mark for correct answer (assuming consistent data)

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Section B: Measures of Central Tendency and Spread (Questions 6–10)

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6. Mean and standard deviation of five numbers.

(a) Fifth number

Let the fifth number be x.

Mean = (12 + 15 + 20 + 25 + x) / 5 = 18

72 + x = 90

x = 18 [2]

• 1 mark for correct equation
• 1 mark for correct answer

(b) Standard deviation

Data: 12, 15, 18, 20, 25

Mean = 18

x	x − mean	(x − mean)²
12	−6	36
15	−3	9
18	0	0
20	2	4
25	7	49
Total		98

Variance = 98 / 5 = 19.6

Standard deviation = √19.6 = 4.43 (to 2 d.p.) [3]

• 1 mark for correct deviations or squared deviations
• 1 mark for correct variance
• 1 mark for correct standard deviation

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7. Books read by 8 students.

Data: 3, 5, 7, 7, 8, 10, 12, 14

(a) Mean

Mean = (3 + 5 + 7 + 7 + 8 + 10 + 12 + 14) / 8 = 66 / 8 = 8.25 [1]

(b) Median

There are 8 values. Median = (4th + 5th) / 2 = (7 + 8) / 2 = 7.5 [1]

(c) Variance

x	x − mean	(x − mean)²
3	−5.25	27.5625
5	−3.25	10.5625
7	−1.25	1.5625
7	−1.25	1.5625
8	−0.25	0.0625
10	1.75	3.0625
12	3.75	14.0625
14	5.75	33.0625
Total		91.5

Variance = 91.5 / 8 = 11.4375 [3]

• 1 mark for correct deviations
• 1 mark for correct sum of squared deviations
• 1 mark for correct variance

(d) Standard deviation

Standard deviation = √11.4375 = 3.38 (to 2 d.p.) [1]

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8. Combined data — two classes.

(a) Combined mean

Total score for Class A = 68 × 30 = 2040
Total score for Class B = 72 × 25 = 1800
Combined total = 2040 + 1800 = 3840
Total students = 30 + 25 = 55

Combined mean = 3840 / 55 = 69.82 (to 2 d.p.) [3]

• 1 mark for correct total scores
• 1 mark for correct combined total
• 1 mark for correct mean

(b) Consistency

Class B has a smaller standard deviation (5) compared to Class A (8). Therefore, Class B has more consistent results because the scores are less spread out. [2]

• 1 mark for identifying Class B
• 1 mark for correct justification

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9. Effect of linear transformation on mean and standard deviation.

Original: mean = 24, standard deviation = 3

New data: each value × 2 + 5

(a) New mean

New mean = 2 × 24 + 5 = 48 + 5 = 53 [2]

• 1 mark for correct method
• 1 mark for correct answer

(b) New standard deviation

Adding a constant does not change the standard deviation. Multiplying by 2 multiplies the standard deviation by 2.

New standard deviation = 2 × 3 = 6 [2]

• 1 mark for correct method
• 1 mark for correct answer

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10. Goals scored by a football team.

(a) Mean

x	f	f × x
0	3	0
1	6	6
2	5	10
3	4	12
4	2	8
Total	20	36

Mean = 36 / 20 = 1.8 goals per match [2]

• 1 mark for correct Σ(f × x)
• 1 mark for correct mean

(b) Standard deviation

x	f	x − mean	(x − mean)²	f × (x − mean)²
0	3	−1.8	3.24	9.72
1	6	−0.8	0.64	3.84
2	5	0.2	0.04	0.20
3	4	1.2	1.44	5.76
4	2	2.2	4.84	9.68
Total	20			29.20

Variance = 29.20 / 20 = 1.46

Standard deviation = √1.46 = 1.21 (to 3 s.f.) [3]

• 1 mark for correct deviations or squared deviations
• 1 mark for correct variance
• 1 mark for correct standard deviation

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Section C: Probability (Questions 11–15)

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11. Fair six-sided die.

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

(a) Probability of a prime number

Prime numbers on a die: 2, 3, 5 (3 outcomes)

P(prime) = 3/6 = 1/2 [2]

• 1 mark for identifying prime numbers
• 1 mark for correct probability

(b) Probability of a number divisible by 3

Numbers divisible by 3: 3, 6 (2 outcomes)

P(divisible by 3) = 2/6 = 1/3 [1]

(c) Probability of a number greater than 6

No number on a die is greater than 6.

P(greater than 6) = 0 [1]

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12. Bag with 5 red, 4 blue, 3 green balls. Two drawn without replacement.

Total balls = 12

(a) Probability both are red

P(1st red) = 5/12
P(2nd red | 1st red) = 4/11

P(both red) = 5/12 × 4/11 = 20/132 = 5/33 [2]

• 1 mark for correct method
• 1 mark for correct answer

(b) Probability both are the same colour

P(both red) = 5/12 × 4/11 = 20/132
P(both blue) = 4/12 × 3/11 = 12/132
P(both green) = 3/12 × 2/11 = 6/132

P(same colour) = 20/132 + 12/132 + 6/132 = 38/132 = 19/66 [3]

• 1 mark for correct method for one colour
• 1 mark for summing probabilities
• 1 mark for correct answer

(c) Probability both are different colours

P(different colours) = 1 − P(same colour) = 1 − 19/66 = 47/66 [1]

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13. Cards numbered 1 to 8. Two drawn without replacement.

Total cards = 8

(a) Probability both are even

Even numbers: 2, 4, 6, 8 (4 outcomes)

P(1st even) = 4/8 = 1/2
P(2nd even | 1st even) = 3/7

P(both even) = 4/8 × 3/7 = 12/56 = 3/14 [2]

• 1 mark for correct method
• 1 mark for correct answer

(b) Probability sum is 9

Pairs that sum to 9: (1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), (8,1) — 8 ordered outcomes

Total possible outcomes = 8 × 7 = 56

P(sum = 9) = 8/56 = 1/7 [2]

• 1 mark for identifying pairs
• 1 mark for correct probability

(c) Probability at least one number is greater than 5

Numbers greater than 5: 6, 7, 8 (3 numbers)
Numbers 5 or less: 1, 2, 3, 4, 5 (5 numbers)

P(at least one > 5) = 1 − P(both ≤ 5) = 1 − (5/8 × 4/7) = 1 − 20/56 = 36/56 = 9/14 [2]

• 1 mark for using complementary probability
• 1 mark for correct answer

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14. Probability of rain = 0.3. Independent days.

(a) Probability of no rain

P(no rain) = 1 − 0.3 = 0.7 [1]

(b) Probability of rain on exactly 2 out of 3 days

This is a binomial probability: X ~ B(3, 0.3)

P(X = 2) = ³C₂ × (0.3)² × (0.7)¹ = 3 × 0.09 × 0.7 = 3 × 0.063 = 0.189 [3]

• 1 mark for identifying binomial distribution
• 1 mark for correct formula
• 1 mark for correct answer

(c) Probability of rain on at least 1 out of 3 days

P(at least 1) = 1 − P(none) = 1 − (0.7)³ = 1 − 0.343 = 0.657 [2]

• 1 mark for using complementary probability
• 1 mark for correct answer

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15. Survey of 100 students — football and basketball.

Let F = plays football, B = plays basketball.

n(F) = 60, n(B) = 45, n(F ∩ B) = 20

(a) Probability of football or basketball

n(F ∪ B) = n(F) + n(B) − n(F ∩ B) = 60 + 45 − 20 = 85

P(F ∪ B) = 85/100 = 17/20 [2]

• 1 mark for correct formula
• 1 mark for correct answer

(b) Probability of neither sport

n(neither) = 100 − 85 = 15

P(neither) = 15/100 = 3/20 [1]

(c) Probability plays basketball given plays football

P(B | F) = n(F ∩ B) / n(F) = 20/60 = 1/3 [2]

• 1 mark for correct formula
• 1 mark for correct answer

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Section D: Combined Probability and Statistics (Questions 16–20)

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16. Grouped frequency — mathematics test marks.

(a) Estimate of the mean

Mark (x)	Midpoint (m)	Frequency (f)	f × m
0–19	9.5	5	47.5
20–39	29.5	10	295
40–59	49.5	18	891
60–79	69.5	12	834
80–99	89.5	5	447.5
Total		50	2515

Mean = 2515 / 50 = 50.3 [3]

• 1 mark for correct midpoints
• 1 mark for correct Σ(f × m)
• 1 mark for correct mean

(b) Estimate of the standard deviation

m	f	m − mean	(m − mean)²	f × (m − mean)²
9.5	5	−40.8	1664.64	8323.2
29.5	10	−20.8	432.64	4326.4
49.5	18	−0.8	0.64	11.52
69.5	12	19.2	368.64	4423.68
89.5	5	39.2	1536.64	7683.2
Total	50			24768

Variance = 24768 / 50 = 495.36

Standard deviation = √495.36 = 22.26 (to 2 d.p.) [3]

• 1 mark for correct deviations or squared deviations
• 1 mark for correct variance
• 1 mark for correct standard deviation

(c) Probability of scoring at least 60 marks

Number of students with mark ≥ 60 = 12 + 5 = 17

P(mark ≥ 60) = 17/50 = 0.34 [1]

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17. Two spinners.

Spinner A: {1, 2, 3, 4}
Spinner B: {2, 4, 6}

(a) Sample space diagram

	B: 2	B: 4	B: 6
A: 1	(1, 2)	(1, 4)	(1, 6)
A: 2	(2, 2)	(2, 4)	(2, 6)
A: 3	(3, 2)	(3, 4)	(3, 6)
A: 4	(4, 2)	(4, 4)	(4, 6)

[2]

• 1 mark for at least 10 correct entries
• 1 mark for all 12 correct

(b) Probability sum is 7

Outcomes with sum 7: (1, 6), (3, 4) — 2 outcomes

Total outcomes = 12

P(sum = 7) = 2/12 = 1/6 [2]

• 1 mark for identifying outcomes
• 1 mark for correct probability

(c) Probability product is even

The product is odd only when both numbers are odd. Spinner B has only even numbers (2, 4, 6), so the product is always even.

P(product is even) = 1 [2]

• 1 mark for correct reasoning
• 1 mark for correct answer

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18. Heights of 30 plants.

(a) Estimate of the mean height

Height (cm)	Midpoint (m)	Frequency (f)	f × m
10 ≤ h < 20	15	4	60
20 ≤ h < 30	25	10	250
30 ≤ h < 40	35	12	420
40 ≤ h < 50	45	4	180
Total		30	910

Mean = 910 / 30 = 30.33 cm (to 2 d.p.) [3]

• 1 mark for correct midpoints
• 1 mark for correct Σ(f × m)
• 1 mark for correct mean

(b) Estimate of the standard deviation

m	f	m − mean	(m − mean)²	f × (m − mean)²
15	4	−15.33	235.01	940.04
25	10	−5.33	28.41	284.10
35	12	4.67	21.81	261.72
45	4	14.67	215.21	860.84
Total	30			2346.70

Variance = 2346.70 / 30 = 78.22

Standard deviation = √78.22 = 8.84 cm (to 2 d.p.) [3]

• 1 mark for correct deviations or squared deviations
• 1 mark for correct variance
• 1 mark for correct standard deviation

(c) Probability height is at least 30 cm

Number of plants with height ≥ 30 cm = 12 + 4 = 16

P(height ≥ 30) = 16/30 = 8/15 [1]

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19. Two fair six-sided dice — sum of scores.

(a) Probability distribution table

Score (x)	2	3	4	5	6	7	8	9	10	11	12
P(X = x)	1/36	2/36	3/36	4/36	5/36	6/36	5/36	4/36	3/36	2/36	1/36

[3]

• 1 mark for correct pattern
• 1 mark for at least 8 correct values
• 1 mark for all correct

(b) Probability score is greater than 8

P(X > 8) = P(9) + P(10) + P(11) + P(12) = 4/36 + 3/36 + 2/36 + 1/36 = 10/36 = 5/18 [2]

• 1 mark for correct method
• 1 mark for correct answer

(c) Expected score

E(X) = Σ[x × P(X = x)]

= 2(1/36) + 3(2/36) + 4(3/36) + 5(4/36) + 6(5/36) + 7(6/36) + 8(5/36) + 9(4/36) + 10(3/36) + 11(2/36) + 12(1/36)

= (2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12) / 36

= 252 / 36 = 7 [3]

• 1 mark for correct formula
• 1 mark for correct calculation
• 1 mark for correct answer

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20. Pie chart — mode of transport for 500 students.

Total angle = 360°

Bus: 144°, Walk: 90°, Car: 72°, MRT: 54°

(a) Number of students who take the bus

Number = (144/360) × 500 = 0.4 × 500 = 200 students [2]

• 1 mark for correct fraction
• 1 mark for correct answer

(b) Probability of walking

P(walk) = 90/360 = 1/4 [1]

(c) Probability both take MRT

Number who take MRT = (54/360) × 500 = 75

P(both MRT) = 75/500 × 74/499 = 5550/249500 = 111/4990 [2]

• 1 mark for correct number of MRT users
• 1 mark for correct probability

(d) Probability bus user takes more than 30 minutes

X ~ N(25, 4²)

P(X > 30) = P(Z > (30 − 25)/4) = P(Z > 1.25)

From standard normal tables: P(Z > 1.25) = 1 − 0.8944 = 0.1056 [3]

• 1 mark for correct standardisation
• 1 mark for correct Z-value
• 1 mark for correct probability

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End of Answer Key