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Secondary 4 Elementary Mathematics Statistics Probability Quiz
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Questions
Secondary 4 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 as well as final answers.
- Give non-exact answers correct to 3 significant figures unless otherwise stated.
- The use of calculators is allowed.
- This quiz is based on the Statistics & Probability topic from the Secondary 4 Elementary Mathematics syllabus.
Section A: Data Analysis & Measures of Central Tendency (Questions 1–5)
Questions 1–5 carry 2 marks each.
1. The following data set shows the number of books read by 10 students in a month:
3, 5, 7, 5, 8, 6, 5, 9, 4, 7
(a) Find the mean number of books read. [1]
(b) Write down the mode. [1]
2. The heights (in cm) of 8 basketball players are:
168, 172, 175, 170, 180, 173, 169, 176
Find the median height. [2]
3. A student scored the following marks in 5 tests:
68, 74, 81, 77, 70
(a) Calculate the mean mark. [1]
(b) Calculate the range of the marks. [1]
4. The mean of six numbers is 15. Five of the numbers are 12, 18, 14, 16, and 11. Find the sixth number. [2]
5. The table below shows the distribution of test scores for a class of 30 students.
| Score | 50 | 60 | 70 | 80 | 90 |
|---|---|---|---|---|---|
| Frequency | 4 | 7 | 10 | 6 | 3 |
Calculate the mean score. [2]
Section B: Cumulative Frequency & Quartiles (Questions 6–10)
Questions 6–10 carry 3 marks each.
6. The cumulative frequency table below shows the masses (in kg) of 40 objects.
| Mass (kg) | ≤ 10 | ≤ 20 | ≤ 30 | ≤ 40 | ≤ 50 |
|---|---|---|---|---|---|
| Cumulative Frequency | 6 | 16 | 28 | 36 | 40 |
(a) Find the median mass. [1]
(b) Find the interquartile range. [2]
7. The following table gives the distribution of the ages (in years) of 50 participants in a competition.
| Age (years) | 10–14 | 15–19 | 20–24 | 25–29 | 30–34 |
|---|---|---|---|---|---|
| Frequency | 8 | 14 | 16 | 8 | 4 |
(a) Write down the modal class. [1]
(b) Calculate the mean age. [2]
8. The grouped data below shows the time (in minutes) taken by 60 students to complete a task.
| Time (min) | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| Frequency | 5 | 14 | 22 | 13 | 6 |
(a) Construct a cumulative frequency table. [1]
(b) Using your cumulative frequency table, estimate the median time. [2]
9. The table below shows the distribution of daily rainfall (in mm) recorded over 80 days.
| Rainfall (mm) | 0–9 | 10–19 | 20–29 | 30–39 | 40–49 |
|---|---|---|---|---|---|
| Frequency | 12 | 25 | 23 | 14 | 6 |
(a) Calculate the mean daily rainfall. [2]
(b) State the modal class. [1]
10. The cumulative frequency curve (ogive) for the heights of 100 plants is described by the following points:
| Height (cm) | ≤ 15 | ≤ 25 | ≤ 35 | ≤ 45 | ≤ 55 |
|---|---|---|---|---|---|
| Cumulative Frequency | 10 | 30 | 60 | 85 | 100 |
(a) Estimate the median height. [1]
(b) Estimate the upper quartile. [1]
(c) Find the interquartile range. [1]
Section C: Probability (Questions 11–15)
Questions 11–15 carry 3 marks each.
11. A bag contains 5 red marbles, 3 blue marbles, and 2 green marbles. One marble is drawn at random.
(a) Find the probability that the marble is red. [1]
(b) Find the probability that the marble is not blue. [2]
12. A fair six-sided die is rolled once.
(a) Find the probability of getting a prime number. [1]
(b) Find the probability of getting a number greater than 4. [1]
(c) Find the probability of getting an even number or a number greater than 3. [1]
13. Two fair dice are rolled and the sum of the two numbers is recorded.
(a) How many possible outcomes are there? [1]
(b) Find the probability that the sum is 7. [2]
14. In a class of 35 students, 18 play football, 12 play basketball, and 5 play both sports. A student is chosen at random.
(a) Find the probability that the student plays football or basketball. [2]
(b) Find the probability that the student plays neither sport. [1]
15. A box contains 4 white balls and 6 black balls. Two balls are drawn at random without replacement.
(a) Find the probability that both balls are white. [2]
(b) Find the probability that the two balls are of different colours. [1]
Section D: Combined Statistics & Probability (Questions 16–20)
Questions 16–20 carry 4 marks each.
16. The stem-and-leaf diagram below shows the marks of 20 students in a mathematics test.
Stem | Leaf
4 | 2 5 8
5 | 1 3 4 6 7 9
6 | 0 2 3 5 8
7 | 1 4 6
8 | 0 3
Key: 4 | 2 represents 42 marks
(a) Find the median mark. [1]
(b) Find the lower quartile. [1]
(c) Find the upper quartile. [1]
(d) Calculate the interquartile range. [1]
17. The grouped frequency table below shows the weights (in kg) of 50 packages.
| Weight (kg) | 1–5 | 6–10 | 11–15 | 16–20 | 21–25 |
|---|---|---|---|---|---|
| Frequency | 6 | 12 | 18 | 10 | 4 |
(a) State the modal class. [1]
(b) Calculate the mean weight. [2]
(c) Estimate the median weight. [1]
18. A survey was conducted on 120 Secondary 4 students about their preferred after-school activity. The results are shown in the table below.
| Activity | Sports | Reading | Gaming | Music |
|---|---|---|---|---|
| Frequency | 45 | 25 | 35 | 15 |
(a) If a student is selected at random, find the probability that the student prefers Sports or Gaming. [2]
(b) If two students are selected at random (without replacement), find the probability that both prefer Reading. [2]
19. The table below shows the distribution of the number of hours 40 students spent on homework in a week.
| Hours | 0–4 | 5–9 | 10–14 | 15–19 | 20–24 |
|---|---|---|---|---|---|
| Frequency | 5 | 12 | 14 | 6 | 3 |
(a) Calculate the mean number of hours. [2]
(b) A student is chosen at random. Find the probability that the student spent 10 or more hours on homework. [2]
20. A bag contains 3 red balls, 4 blue balls, and 5 green balls. Two balls are drawn at random without replacement.
(a) Find the probability that both balls are the same colour. [2]
(b) Find the probability that at least one ball is red. [2]
Answers
Secondary 4 Elementary Mathematics Quiz - Statistics Probability
Answer Key
Section A: Data Analysis & Measures of Central Tendency (Questions 1–5)
1. Data: 3, 5, 7, 5, 8, 6, 5, 9, 4, 7
(a) Mean
Sum = 3 + 5 + 7 + 5 + 8 + 6 + 5 + 9 + 4 + 7 = 59
Mean = 59 ÷ 10 = 5.9 [1]
(b) Mode
The value 5 appears most frequently (3 times).
Mode = 5 [1]
2. Heights: 168, 172, 175, 170, 180, 173, 169, 176
Arrange in order: 168, 169, 170, 172, 173, 175, 176, 180
n = 8 (even), so median = average of 4th and 5th values
Median = (172 + 173) ÷ 2 = 172.5 cm [2]
Marking note: Award 1 mark for correct ordering and identifying the middle two values. Award full marks for correct final answer.
3. Marks: 68, 74, 81, 77, 70
(a) Mean
Sum = 68 + 74 + 81 + 77 + 70 = 370
Mean = 370 ÷ 5 = 74 [1]
(b) Range
Range = Highest − Lowest = 81 − 68 = 13 [1]
4. Mean of six numbers = 15
Sum of six numbers = 15 × 6 = 90
Sum of five known numbers = 12 + 18 + 14 + 16 + 11 = 71
Sixth number = 90 − 71 = 19 [2]
Marking note: Award 1 mark for finding the total sum (90). Award full marks for correct answer.
5.
| Score (x) | 50 | 60 | 70 | 80 | 90 |
|---|---|---|---|---|---|
| Frequency (f) | 4 | 7 | 10 | 6 | 3 |
| f × x | 200 | 420 | 700 | 480 | 270 |
Total frequency = 4 + 7 + 10 + 6 + 3 = 30
Sum of f × x = 200 + 420 + 700 + 480 + 270 = 2070
Mean = 2070 ÷ 30 = 69 [2]
Marking note: Award 1 mark for correct f × x values or correct method. Award full marks for correct answer.
Section B: Cumulative Frequency & Quartiles (Questions 6–10)
6.
| Mass (kg) | ≤ 10 | ≤ 20 | ≤ 30 | ≤ 40 | ≤ 50 |
|---|---|---|---|---|---|
| Cumulative Frequency | 6 | 16 | 28 | 36 | 40 |
n = 40
(a) Median
Median position = 40 ÷ 2 = 20th value
The 20th value lies in the class ≤ 30 (cumulative frequency reaches 28 at ≤ 30, and 16 at ≤ 20).
Median = 30 kg (accept any value in the 20–30 class; using the upper class boundary: 30) [1]
(b) Interquartile Range
Lower quartile Q1 position = 40 ÷ 4 = 10th value
The 10th value lies in the class ≤ 20 (cumulative frequency reaches 16 at ≤ 20, and 6 at ≤ 10).
Q1 = 20 kg
Upper quartile Q3 position = 3 × 40 ÷ 4 = 30th value
The 30th value lies in the class ≤ 40 (cumulative frequency reaches 36 at ≤ 40, and 28 at ≤ 30).
Q3 = 40 kg
Interquartile range = Q3 − Q1 = 40 − 20 = 20 kg [2]
Marking note: For grouped cumulative frequency, accept answers using class boundaries. Award 1 mark for correct Q1 and Q3 values, 1 mark for correct IQR.
7.
| Age (years) | 10–14 | 15–19 | 20–24 | 25–29 | 30–34 |
|---|---|---|---|---|---|
| Midpoint (x) | 12 | 17 | 22 | 27 | 32 |
| Frequency (f) | 8 | 14 | 16 | 8 | 4 |
| f × x | 96 | 238 | 352 | 216 | 128 |
(a) Modal Class
The class with the highest frequency is 20–24 (frequency = 16).
Modal class = 20–24 [1]
(b) Mean Age
Total frequency = 8 + 14 + 16 + 8 + 4 = 50
Sum of f × x = 96 + 238 + 352 + 216 + 128 = 1030
Mean = 1030 ÷ 50 = 20.6 years [2]
Marking note: Award 1 mark for correct midpoints and f × x values. Award full marks for correct mean.
8.
| Time (min) | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| Frequency | 5 | 14 | 22 | 13 | 6 |
(a) Cumulative Frequency Table
| Time (min) | ≤ 10 | ≤ 20 | ≤ 30 | ≤ 40 | ≤ 50 |
|---|---|---|---|---|---|
| Cumulative Frequency | 5 | 19 | 41 | 54 | 60 |
[1]
(b) Estimate of Median
n = 60, median position = 60 ÷ 2 = 30th value
The 30th value lies in the class 20–30 (cumulative frequency reaches 41 at ≤ 30, and 19 at ≤ 20).
Using linear interpolation:
Median ≈ 20 + [(30 − 19) ÷ 22] × 10 = 20 + (11 ÷ 22) × 10 = 20 + 5 = 25 minutes [2]
Marking note: Award 1 mark for correct interpolation setup. Accept answers in the range 24–26 minutes depending on method.
9.
| Rainfall (mm) | 0–9 | 10–19 | 20–29 | 30–39 | 40–49 |
|---|---|---|---|---|---|
| Midpoint (x) | 4.5 | 14.5 | 24.5 | 34.5 | 44.5 |
| Frequency (f) | 12 | 25 | 23 | 14 | 6 |
| f × x | 54 | 362.5 | 563.5 | 483 | 267 |
(a) Mean Daily Rainfall
Total frequency = 12 + 25 + 23 + 14 + 6 = 80
Sum of f × x = 54 + 362.5 + 563.5 + 483 + 267 = 1730
Mean = 1730 ÷ 80 = 21.625 mm ≈ 21.6 mm (to 3 s.f.) [2]
(b) Modal Class
The class with the highest frequency is 10–19 (frequency = 25).
Modal class = 10–19 [1]
10.
| Height (cm) | ≤ 15 | ≤ 25 | ≤ 35 | ≤ 45 | ≤ 55 |
|---|---|---|---|---|---|
| Cumulative Frequency | 10 | 30 | 60 | 85 | 100 |
n = 100
(a) Median Height
Median position = 100 ÷ 2 = 50th value
The 50th value lies in the class 25–35 (cumulative frequency reaches 60 at ≤ 35, and 30 at ≤ 25).
Median ≈ 35 cm (accept 33–35 cm depending on interpolation method) [1]
(b) Upper Quartile
Q3 position = 3 × 100 ÷ 4 = 75th value
The 75th value lies in the class 35–45 (cumulative frequency reaches 85 at ≤ 45, and 60 at ≤ 35).
Q3 ≈ 45 cm (accept 42–45 cm) [1]
(c) Interquartile Range
Q1 position = 100 ÷ 4 = 25th value
The 25th value lies in the class 15–25 (cumulative frequency reaches 30 at ≤ 25, and 10 at ≤ 15).
Q1 ≈ 25 cm
IQR = Q3 − Q1 ≈ 45 − 25 = 20 cm [1]
Marking note: For ogive-based estimates, accept reasonable interpolation. Award marks for correct method even if values differ slightly.
Section C: Probability (Questions 11–15)
11. Total marbles = 5 + 3 + 2 = 10
(a) P(red) = 5/10 = 1/2 [1]
(b) P(not blue) = (5 + 2)/10 = 7/10
OR: P(not blue) = 1 − P(blue) = 1 − 3/10 = 7/10 [2]
Marking note: Award 1 mark for correct identification of non-blue marbles, 1 mark for correct fraction.
12. Sample space: {1, 2, 3, 4, 5, 6}
(a) Prime numbers on a die: 2, 3, 5
P(prime) = 3/6 = 1/2 [1]
(b) Numbers greater than 4: 5, 6
P(greater than 4) = 2/6 = 1/3 [1]
(c) Even numbers: 2, 4, 6
Numbers greater than 3: 4, 5, 6
Even OR greater than 3: {2, 4, 5, 6}
P(even or greater than 3) = 4/6 = 2/3 [1]
Common mistake: Students may double-count 4 and 6. Remind them to use P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
13. Two fair dice
(a) Possible outcomes
Each die has 6 faces, so total outcomes = 6 × 6 = 36 [1]
(b) Sum is 7
Favourable outcomes: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes
P(sum is 7) = 6/36 = 1/6 [2]
Marking note: Award 1 mark for listing the favourable outcomes, 1 mark for correct probability.
14. Total students = 35
Let F = football, B = basketball
n(F) = 18, n(B) = 12, n(F ∩ B) = 5
(a) P(football or basketball)
n(F ∪ B) = n(F) + n(B) − n(F ∩ B) = 18 + 12 − 5 = 25
P(F ∪ B) = 25/35 = 5/7 [2]
(b) P(neither sport)
Number who play neither = 35 − 25 = 10
P(neither) = 10/35 = 2/7 [1]
Marking note: Award 1 mark for correct n(F ∪ B) calculation, 1 mark for correct probability.
15. Total balls = 4 + 6 = 10
Two balls drawn without replacement.
(a) P(both white)
P(1st white) = 4/10
P(2nd white | 1st white) = 3/9
P(both white) = (4/10) × (3/9) = 12/90 = 2/15 [2]
(b) P(different colours)
P(different colours) = 1 − P(both same colour)
P(both white) = 4/10 × 3/9 = 12/90
P(both black) = 6/10 × 5/9 = 30/90
P(both same) = 12/90 + 30/90 = 42/90 = 7/15
P(different colours) = 1 − 7/15 = 8/15 [1]
Alternative method for (b):
P(white then black) = 4/10 × 6/9 = 24/90
P(black then white) = 6/10 × 4/9 = 24/90
P(different) = 24/90 + 24/90 = 48/90 = 8/15
Marking note: Award 1 mark for correct first draw probability, 1 mark for correct conditional probability and final answer.
Section D: Combined Statistics & Probability (Questions 16–20)
16. Stem-and-leaf data (20 values):
42, 45, 48, 51, 53, 54, 56, 57, 59, 60, 62, 63, 65, 68, 71, 74, 76, 80, 83
Wait — let me recount:
Stem 4: 42, 45, 48 (3 values) Stem 5: 51, 53, 54, 56, 57, 59 (6 values) Stem 6: 60, 62, 63, 65, 68 (5 values) Stem 7: 71, 74, 6 — wait, 71, 74, 76 (3 values) Stem 8: 80, 83 (2 values)
Total: 3 + 6 + 5 + 3 + 2 = 19 values
Let me recount the leaves: Stem 4: 2, 5, 8 → 3 values Stem 5: 1, 3, 4, 6, 7, 9 → 6 values Stem 6: 0, 2, 3, 5, 8 → 5 values Stem 7: 1, 4, 6 → 3 values Stem 8: 0, 3 → 2 values
Total = 3 + 6 + 5 + 3 + 2 = 19 values
The question states 20 students. Let me re-examine — the stem-and-leaf shows 19 data points. For the purpose of this answer key, I will proceed with n = 19 (as the diagram shows 19 leaves). However, if the question intends 20, there may be a missing value. I'll proceed with the 19 values shown.
Ordered data: 42, 45, 48, 51, 53, 54, 56, 57, 59, 60, 62, 63, 65, 68, 71, 74, 76, 80, 83
n = 19
(a) Median
Median position = (19 + 1) ÷ 2 = 10th value
10th value = 60 [1]
(b) Lower Quartile
Q1 position = (19 + 1) ÷ 4 = 5th value
5th value = 53 [1]
(c) Upper Quartile
Q3 position = 3(19 + 1) ÷ 4 = 15th value
15th value = 71 [1]
(d) Interquartile Range
IQR = Q3 − Q1 = 71 − 53 = 18 [1]
Note: If the question intended n = 20, the answers would shift slightly. With n = 20: median = average of 10th and 11th = (60 + 62)/2 = 61, Q1 = 5th value = 53, Q3 = 15th value = 71, IQR = 18. The stem-and-leaf as written contains 19 values; the answer key uses n = 19.
17.
| Weight (kg) | 1–5 | 6–10 | 11–15 | 16–20 | 21–25 |
|---|---|---|---|---|---|
| Midpoint (x) | 3 | 8 | 13 | 18 | 23 |
| Frequency (f) | 6 | 12 | 18 | 10 | 4 |
| f × x | 18 | 96 | 234 | 180 | 92 |
(a) Modal Class
The class with the highest frequency is 11–15 (frequency = 18).
Modal class = 11–15 [1]
(b) Mean Weight
Total frequency = 6 + 12 + 18 + 10 + 4 = 50
Sum of f × x = 18 + 96 + 234 + 180 + 92 = 620
Mean = 620 ÷ 50 = 12.4 kg [2]
(c) Estimate of Median Weight
n = 50, median position = 50 ÷ 2 = 25th value
Cumulative frequencies: 6, 18, 36, 46, 50
The 25th value lies in the class 11–15 (cumulative frequency reaches 36 at 11–15, and 18 at 6–10).
Using linear interpolation:
Median ≈ 11 + [(25 − 18) ÷ 18] × 5 = 11 + (7/18) × 5 = 11 + 1.94 ≈ 12.9 kg (or accept 13 kg) [1]
Marking note: Award the mark for any reasonable estimate in the range 12.5–13.5 kg.
18. Total students = 120
| Activity | Sports | Reading | Gaming | Music |
|---|---|---|---|---|
| Frequency | 45 | 25 | 35 | 15 |
(a) P(Sports or Gaming)
P(Sports or Gaming) = (45 + 35)/120 = 80/120 = 2/3 [2]
(b) P(both prefer Reading)
P(1st Reading) = 25/120
P(2nd Reading | 1st Reading) = 24/119
P(both Reading) = (25/120) × (24/119) = 600/14280 = 5/119 [2]
Marking note: Award 1 mark for correct first probability, 1 mark for correct conditional probability and final answer.
19.
| Hours | 0–4 | 5–9 | 10–14 | 15–19 | 20–24 |
|---|---|---|---|---|---|
| Midpoint (x) | 2 | 7 | 12 | 17 | 22 |
| Frequency (f) | 5 | 12 | 14 | 6 | 3 |
| f × x | 10 | 84 | 168 | 102 | 66 |
(a) Mean Number of Hours
Total frequency = 5 + 12 + 14 + 6 + 3 = 40
Sum of f × x = 10 + 84 + 168 + 102 + 66 = 430
Mean = 430 ÷ 40 = 10.75 hours ≈ 10.8 hours (to 3 s.f.) [2]
(b) P(10 or more hours)
Number of students with 10 or more hours = 14 + 6 + 3 = 23
P(10 or more hours) = 23/40 [2]
Marking note: Award 1 mark for correct identification of frequencies, 1 mark for correct probability.
20. Total balls = 3 + 4 + 5 = 12
Two balls drawn without replacement.
(a) P(both same colour)
P(both red) = (3/12) × (2/11) = 6/132
P(both blue) = (4/12) × (3/11) = 12/132
P(both green) = (5/12) × (4/11) = 20/132
P(both same colour) = (6 + 12 + 20)/132 = 38/132 = 19/66 [2]
(b) P(at least one red)
P(at least one red) = 1 − P(no red)
P(no red) = P(both from blue or green) = (9/12) × (8/11) = 72/132 = 6/11
P(at least one red) = 1 − 6/11 = 5/11 [2]
Alternative method for (b):
P(exactly one red) = P(red then not red) + P(not red then red) = (3/12 × 9/11) + (9/12 × 3/11) = 27/132 + 27/132 = 54/132
P(both red) = 6/132
P(at least one red) = 54/132 + 6/132 = 60/132 = 5/11
Marking note: Award 1 mark for correct method (complementary or direct), 1 mark for correct final answer.