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Secondary 3 Elementary Mathematics Statistics Probability Quiz
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Questions
Secondary 3 Elementary Mathematics Quiz - Statistics Probability
Name: _______________________
Class: _______________________
Date: _______________________
Score: _______ / 50
Duration: 60 minutes
Total Marks: 50
Instructions
- Answer all questions in the spaces provided.
- Show all working clearly. Marks are awarded for correct method even if the final answer is incorrect.
- Use a calculator where appropriate. Unless otherwise stated, give answers correct to 3 significant figures.
- The number of marks for each question is shown in brackets [ ].
- This quiz covers the topic Statistics and Probability only.
Section A: Data Handling and Representation (Questions 1–5)
Answer all questions in this section.
1. The following stem-and-leaf diagram shows the heights (in cm) of 15 students in a class.
Stem | Leaf
15 | 2 4 6 8
16 | 0 1 3 5 7 9
17 | 2 4 6 8
18 | 1
Key: 15 | 2 represents 152 cm
(a) Write down the median height. [1]
(b) Find the interquartile range of the heights. [2]
(c) State one advantage of using a stem-and-leaf diagram over a bar chart. [1]
2. The table below shows the distribution of test scores for 40 students.
| Score (x) | 10–20 | 21–30 | 31–40 | 41–50 | 51–60 |
|---|---|---|---|---|---|
| Frequency | 4 | 8 | 14 | 10 | 4 |
(a) Write down the modal class. [1]
(b) Calculate an estimate of the mean score. [3]
(c) In which class interval does the median lie? [1]
3. A survey was conducted on the number of hours 20 students spent on homework in a week. The results are summarised below.
| Minimum | 2 hours |
| Lower quartile | 5 hours |
| Median | 8 hours |
| Upper quartile | 12 hours |
| Maximum | 18 hours |
(a) Calculate the interquartile range. [1]
(b) Draw a box-and-whisker plot to represent this data on the grid provided below. [3]
0 2 4 6 8 10 12 14 16 18 20
|----+----+----+----+----+----+----+----+----+----|
| | | | | | | | | | |
|----+----+----+----+----+----+----+----+----+----|
| | | | | | | | | | |
|----+----+----+----+----+----+----+----+----+----|
(c) Comment on the skewness of the distribution. [1]
4. The cumulative frequency curve below shows the distribution of the masses (in kg) of 60 adults.
Cumulative
Frequency
60 | ●
| ●
50 | ●
| ●
40 | ●
| ●
30 | ●
| ●
20 |●
|
10 |
|
0 |----+----+----+----+----+----+----+----→ Mass (kg)
40 50 60 70 80 90 100 110
(a) Use the curve to estimate the median mass. [1]
(b) Use the curve to estimate the 70th percentile. [1]
(c) Find the number of adults whose mass is less than 65 kg. [1]
5. The histogram below shows the distribution of the ages of members in a sports club.
Frequency
Density
6 | ┌───┐
| │ │
5 | │ │
| │ │
4 | ┌───┐ │ │
| │ │ │ │
3 | │ │ │ │
| │ │ │ │
2 | │ │ │ │ ┌───┐
| │ │ │ │ │ │
1 | │ │ │ │ │ │ ┌───┐
| │ │ │ │ │ │ │ │
0 +--+--+--+--+--+--+--+--+--+--+--→ Age (years)
10 20 30 40 50 60 70 80
The frequency density for the class 10–20 is 4.0.
(a) State the frequency for the class 10–20. [1]
(b) The frequency for the class 20–30 is 30. Calculate the frequency density for this class. [2]
(c) Given that the total number of members is 120, and the class 60–80 has a frequency of 12, find the frequency for the class 40–60. [2]
Section B: Measures of Central Tendency and Spread (Questions 6–10)
Answer all questions in this section.
6. The mean of five numbers is 18. Four of the numbers are 12, 15, 20, and 25.
(a) Find the fifth number. [2]
(b) Calculate the standard deviation of the five numbers. [3]
7. The following data set shows the number of books read by 8 students in a month:
3, 5, 7, 7, 8, 10, 12, 14
(a) Find the mean number of books read. [1]
(b) Find the median number of books read. [1]
(c) Calculate the variance of the data set. [3]
(d) Calculate the standard deviation, giving your answer correct to 2 decimal places. [1]
8. Two classes, A and B, sat for the same mathematics test. The results are summarised below.
| Class A | Class B | |
|---|---|---|
| Mean | 68 | 72 |
| Standard deviation | 8 | 5 |
| Number of students | 30 | 25 |
(a) Calculate the combined mean score of all 55 students. [3]
(b) Which class has more consistent results? Justify your answer. [2]
9. A set of data has a mean of 24 and a standard deviation of 3. Each value in the data set is multiplied by 2 and then increased by 5.
(a) Find the new mean. [2]
(b) Find the new standard deviation. [2]
10. The table below shows the distribution of the number of goals scored by a football team in 20 matches.
| Number of goals (x) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Frequency (f) | 3 | 6 | 5 | 4 | 2 |
(a) Calculate the mean number of goals per match. [2]
(b) Calculate the standard deviation, giving your answer correct to 3 significant figures. [3]
Section C: Probability (Questions 11–15)
Answer all questions in this section.
11. A fair six-sided die is rolled once.
(a) Find the probability of obtaining a prime number. [2]
(b) Find the probability of obtaining a number that is divisible by 3. [1]
(c) Find the probability of obtaining a number greater than 6. [1]
12. A bag contains 5 red balls, 4 blue balls, and 3 green balls. Two balls are drawn at random without replacement.
(a) Find the probability that both balls are red. [2]
(b) Find the probability that the two balls are of the same colour. [3]
(c) Find the probability that the two balls are of different colours. [1]
13. A box contains 8 cards numbered 1, 2, 3, 4, 5, 6, 7, and 8. Two cards are drawn at random without replacement.
(a) Find the probability that both numbers are even. [2]
(b) Find the probability that the sum of the two numbers is 9. [2]
(c) Find the probability that at least one of the numbers is greater than 5. [2]
14. The probability that it rains on any given day in June is 0.3. Assume that the weather on each day is independent.
(a) Find the probability that it does not rain on a given day. [1]
(b) Find the probability that it rains on exactly 2 days out of 3 consecutive days. [3]
(c) Find the probability that it rains on at least 1 day out of 3 consecutive days. [2]
15. A survey of 100 students found that 60 students play football, 45 play basketball, and 20 play both sports.
(a) Find the probability that a randomly chosen student plays football or basketball. [2]
(b) Find the probability that a randomly chosen student plays neither sport. [1]
(c) Given that a student plays football, find the probability that the student also plays basketball. [2]
Section D: Combined Probability and Statistics (Questions 16–20)
Answer all questions in this section.
16. The table below shows the marks obtained by 50 students in a mathematics test.
| Mark (x) | 0–19 | 20–39 | 40–59 | 60–79 | 80–99 |
|---|---|---|---|---|---|
| Frequency | 5 | 10 | 18 | 12 | 5 |
(a) Calculate an estimate of the mean mark. [3]
(b) Calculate an estimate of the standard deviation, giving your answer correct to 2 decimal places. [3]
(c) A student is chosen at random. Find the probability that the student scored at least 60 marks. [1]
17. Two fair spinners are spun. Spinner A has four equal sectors labelled 1, 2, 3, and 4. Spinner B has three equal sectors labelled 2, 4, and 6.
(a) Complete the sample space diagram below showing all possible outcomes.
| Spinner B: 2 | Spinner B: 4 | Spinner B: 6 | |
|---|---|---|---|
| Spinner A: 1 | (1, 2) | ||
| Spinner A: 2 | |||
| Spinner A: 3 | |||
| Spinner A: 4 |
[2]
(b) Find the probability that the sum of the two numbers is 7. [2]
(c) Find the probability that the product of the two numbers is even. [2]
18. The heights of 30 plants (in cm) are recorded and summarised in the grouped frequency table below.
| Height (cm) | 10 ≤ h < 20 | 20 ≤ h < 30 | 30 ≤ h < 40 | 40 ≤ h < 50 |
|---|---|---|---|---|
| Frequency | 4 | 10 | 12 | 4 |
(a) Calculate an estimate of the mean height. [3]
(b) Calculate an estimate of the standard deviation, giving your answer correct to 2 decimal places. [3]
(c) A plant is selected at random. Find the probability that its height is at least 30 cm. [1]
19. A game involves rolling two fair six-sided dice. The score is the sum of the two numbers shown.
(a) Complete the probability distribution table for the score.
| Score (x) | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| P(X = x) | 1/36 | 2/36 | 1/36 |
[3]
(b) Find the probability that the score is greater than 8. [2]
(c) Find the expected score. [3]
20. In a school of 500 students, a survey was conducted on the mode of transport used to get to school. The results are shown in the pie chart below.
Bus: 144°
Walk: 90°
Car: 72°
MRT: 54°
(a) Find the number of students who take the bus. [2]
(b) Find the probability that a randomly chosen student walks to school. [1]
(c) Two students are chosen at random. Find the probability that both students take the MRT. [2]
(d) The mean travel time for students who take the bus is 25 minutes with a standard deviation of 4 minutes. Assuming the travel times are normally distributed, find the probability that a randomly chosen bus user takes more than 30 minutes. [3]
End of Quiz
Answers
Secondary 3 Elementary Mathematics Quiz - Statistics Probability
Answer Key and Marking Scheme
Section A: Data Handling and Representation (Questions 1–5)
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]
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]
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]
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]
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)
Section B: Measures of Central Tendency and Spread (Questions 6–10)
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
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]
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
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
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
Section C: Probability (Questions 11–15)
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]
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]
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
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
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
Section D: Combined Probability and Statistics (Questions 16–20)
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]
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
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]
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
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
End of Answer Key