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O Level Elementary Mathematics Statistics Probability Quiz

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O Level Elementary Mathematics From Real Exams Generated by DeepSeek V4 Pro Updated 2026-06-03

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Questions

O-Level Elementary Mathematics Quiz - Statistics Probability

Name: ________________________ Class: ________________________ Date: ________________________ Score: ______ / 40

Duration: 45 minutes Total Marks: 40

Instructions:

Answer ALL questions.
Show all working clearly. Marks are awarded for method.
Give non-exact answers to 3 significant figures unless otherwise stated.
Calculators may be used.

Section A: Data Handling and Analysis (Questions 1–10)

20 marks

1. The stem-and-leaf diagram shows the scores of 20 students in a Mathematics test.

Stem | Leaf
  3  | 2 5 8
  4  | 1 4 6 7 9
  5  | 0 2 3 5 5 8
  6  | 1 4 7
  7  | 0 3
Key: 3 | 2 means 32

(a) State the mode of the scores. [1 mark]

(b) Find the median score. [1 mark]

(c) Calculate the interquartile range. [2 marks]

2. The table shows the number of books read by a group of 50 students in a month.

Number of books	0	1	2	3	4	5
Frequency	8	12	15	7	5	3

(a) Find the mean number of books read. [2 marks]

(b) A student is chosen at random. Find the probability that the student read more than 3 books. [1 mark]

3. The cumulative frequency curve shows the distribution of the masses, in kg, of 80 parcels.

(Assume a cumulative frequency curve is provided with the following key points: lower quartile = 2.4 kg, median = 3.8 kg, upper quartile = 5.6 kg)

(a) Use the diagram to estimate the median mass. [1 mark]

(b) Find the interquartile range. [1 mark]

(c) A parcel is classified as "heavy" if its mass exceeds 6.0 kg. Estimate the number of heavy parcels. [2 marks]

4. The box-and-whisker plot summarises the heights, in cm, of plants in two different greenhouses, A and B.

Greenhouse A:  |---[   |   ]--------|
                12   18   24   30    38

Greenhouse B:  |------[  |  ]-------|
                14    22  28  34    42

(a) Compare the median heights of the plants in the two greenhouses. [1 mark]

(b) Which greenhouse has a greater range of heights? Justify your answer. [1 mark]

(c) Explain which greenhouse has more consistent plant heights. [1 mark]

5. The pie chart represents how a student spends her monthly allowance of $400.

(Assume a pie chart with sectors: Food 144°, Transport 72°, Entertainment 90°, Savings 54°)

(a) Calculate the amount spent on Food. [1 mark]

(b) Express the amount spent on Entertainment as a percentage of her total allowance. [1 mark]

(c) The student decides to increase her Savings to 20% of her allowance. How much more must she save each month? [2 marks]

6. The histogram shows the distribution of times, in minutes, taken by 60 students to complete a puzzle.

Time (t minutes)	Frequency
0 < t ≤ 10	8
10 < t ≤ 20	18
20 < t ≤ 30	15
30 < t ≤ 40	12
40 < t ≤ 50	7

(a) State the modal class. [1 mark]

(b) Calculate an estimate of the mean time taken. [2 marks]

7. The dot diagram shows the number of siblings of each student in a class of 25.

Number of siblings: 0  1  2  3  4
                    •  •  •  •  •
                    •  •  •  •
                    •  •  •  •
                    •  •  •
                    •  •
                    •

(a) Find the median number of siblings. [1 mark]

(b) Calculate the mean number of siblings. [1 mark]

8. A set of 10 numbers has a mean of 15 and a standard deviation of 3.

(a) A new number, 27, is added to the set. Find the new mean. [2 marks]

(b) Without calculation, state whether the new standard deviation will be larger or smaller than 3. Explain your reasoning. [1 mark]

9. The table shows the marks obtained by 40 students in an English test.

Mark (x)	Frequency (f)
10 ≤ x < 20	5
20 ≤ x < 30	8
30 ≤ x < 40	12
40 ≤ x < 50	10
50 ≤ x < 60	5

Calculate an estimate of the standard deviation of the marks. [3 marks]

10. Two data sets, X and Y, are summarised as follows:

	Mean	Standard Deviation
Set X	45	8
Set Y	45	12

(a) Compare the means of the two data sets. [1 mark]

(b) Explain what the difference in standard deviations tells you about the two data sets. [1 mark]

Section B: Probability (Questions 11–20)

20 marks

11. A fair six-sided die is rolled once. Find the probability of obtaining:

(a) a prime number, [1 mark]

(b) a number greater than 4. [1 mark]

12. A bag contains 5 red balls, 3 blue balls, and 2 green balls. A ball is drawn at random from the bag.

(a) Find the probability that the ball is blue. [1 mark]

(b) Find the probability that the ball is not red. [1 mark]

13. A card is drawn at random from a standard pack of 52 playing cards. Find the probability that the card is:

(a) a King, [1 mark]

(b) a red card or a Queen. [2 marks]

14. Two fair coins are tossed. Using a possibility diagram or otherwise, find the probability of obtaining:

(a) exactly one head, [1 mark]

(b) at least one tail. [1 mark]

15. A box contains 6 white chocolates and 4 dark chocolates. Two chocolates are taken at random from the box, one after the other, without replacement.

(a) Draw a tree diagram to represent this situation. [2 marks]

(b) Find the probability that both chocolates are dark. [1 mark]

(c) Find the probability that exactly one of the chocolates is white. [2 marks]

16. The probability that it rains on any given day in a certain city is 0.3. The probability that a bus is late is 0.1 on a rainy day and 0.05 on a dry day.

(a) Draw a tree diagram to represent this information. [2 marks]

(b) Find the probability that on a randomly chosen day, the bus is late. [2 marks]

(c) Given that the bus is late, find the probability that it is a rainy day. [2 marks]

17. Events A and B are such that P(A) = 0.4, P(B) = 0.5, and P(A ∩ B) = 0.2.

(a) Find P(A ∪ B). [1 mark]

(b) Determine whether A and B are independent events. Justify your answer. [2 marks]

18. A spinner has three sections coloured red, blue, and yellow. The probabilities of landing on red and blue are 0.35 and 0.45 respectively.

(a) Find the probability of landing on yellow. [1 mark]

(b) The spinner is spun twice. Find the probability that it lands on the same colour both times. [2 marks]

19. In a class of 30 students, 18 study Mathematics, 15 study Physics, and 8 study both subjects. A student is chosen at random.

(a) Draw a Venn diagram to illustrate this information. [2 marks]

(b) Find the probability that the student studies neither Mathematics nor Physics. [1 mark]

(c) Find the probability that the student studies Mathematics but not Physics. [1 mark]

20. A game involves drawing a marble from a bag containing 4 red marbles and 6 blue marbles. If a red marble is drawn, the player wins $5. If a blue marble is drawn, the player loses$ 2.

(a) Calculate the expected gain (or loss) per game. [2 marks]

(b) Is this a fair game? Explain your answer. [1 mark]

END OF QUIZ

Answers

O-Level Elementary Mathematics Quiz - Statistics Probability

Answer Key and Marking Scheme

Section A: Data Handling and Analysis

1. (a) Mode = 55 [1 mark]

Award 1 mark for correct answer.

(b) Median = 52.5 [1 mark]

20 values; median is mean of 10th and 11th values: (52 + 53) ÷ 2 = 52.5.
Award 1 mark for correct answer.

(c) IQR = 16 [2 marks]

Q1 = median of lower half (10 values) = mean of 5th and 6th = (44 + 46) ÷ 2 = 45
Q3 = median of upper half = mean of 15th and 16th = (61 + 61) ÷ 2 = 61
IQR = 61 – 45 = 16
Award 1 mark for correct Q1 and Q3, 1 mark for correct IQR.

2. (a) Mean = 2.08 books [2 marks]

Σfx = 0(8) + 1(12) + 2(15) + 3(7) + 4(5) + 5(3) = 0 + 12 + 30 + 21 + 20 + 15 = 98
Mean = 98 ÷ 50 = 1.96
Award 1 mark for correct Σfx, 1 mark for correct mean.

(b) P(more than 3) = 8/50 = 4/25 or 0.16 [1 mark]

Students reading more than 3 books: 5 + 3 = 8
Award 1 mark for correct probability.

3. (a) Median mass = 3.8 kg [1 mark]

Award 1 mark for correct reading from cumulative frequency curve.

(b) IQR = 5.6 – 2.4 = 3.2 kg [1 mark]

Award 1 mark for correct answer.

(c) Number of heavy parcels = 12 [2 marks]

From cumulative frequency curve, number of parcels ≤ 6.0 kg ≈ 68
Number of heavy parcels = 80 – 68 = 12
Award 1 mark for correct reading from graph, 1 mark for correct number.

4. (a) Greenhouse B has a higher median (28 cm) than Greenhouse A (24 cm). [1 mark]

Award 1 mark for correct comparison with values.

(b) Greenhouse A has a greater range. Range A = 38 – 12 = 26 cm; Range B = 42 – 14 = 28 cm. Greenhouse B has the greater range. [1 mark]

Award 1 mark for correct identification with justification.
Note: Correction – Greenhouse B has greater range (28 > 26).

(c) Greenhouse A has more consistent heights because it has a smaller interquartile range (30 – 18 = 12 cm) compared to Greenhouse B (34 – 22 = 12 cm). Both have same IQR. Greenhouse A has smaller overall range (26 cm vs 28 cm), so Greenhouse A is slightly more consistent. [1 mark]

Award 1 mark for valid reasoning referencing spread.

5. (a) Amount on Food = (144°/360°) × $400 =$ 160 [1 mark]

Award 1 mark for correct answer.

(b) Percentage on Entertainment = (90°/360°) × 100% = 25% [1 mark]

Award 1 mark for correct percentage.

(c) Additional savings needed = $26 [2 marks]

Current savings = (54°/360°) × $400 =$ 60
Target savings = 20% × $400 =$ 80
Additional = $80 –$ 60 = $20
Award 1 mark for current savings, 1 mark for correct additional amount.

6. (a) Modal class = 10 < t ≤ 20 [1 mark]

Award 1 mark for correct class.

(b) Estimated mean = 24.5 minutes [2 marks]

Midpoints: 5, 15, 25, 35, 45
Σfx = 5(8) + 15(18) + 25(15) + 35(12) + 45(7) = 40 + 270 + 375 + 420 + 315 = 1420
Mean = 1420 ÷ 60 = 23.666... ≈ 23.7 minutes
Award 1 mark for correct Σfx, 1 mark for correct mean.

7. (a) Median = 2 siblings [1 mark]

25 students; 13th value is median. Counting from 0 siblings: 5 have 0, 7 have 1 (cumulative 12), next 6 have 2 (cumulative 18). 13th value is 2.
Award 1 mark for correct median.

(b) Mean = 1.8 siblings [1 mark]

Frequencies: 0:5, 1:7, 2:6, 3:4, 4:3
Σfx = 0(5) + 1(7) + 2(6) + 3(4) + 4(3) = 0 + 7 + 12 + 12 + 12 = 43
Mean = 43 ÷ 25 = 1.72
Award 1 mark for correct mean.

8. (a) New mean = 16.0909... ≈ 16.1 [2 marks]

Sum of original 10 numbers = 10 × 15 = 150
New sum = 150 + 27 = 177
New mean = 177 ÷ 11 = 16.0909... ≈ 16.1
Award 1 mark for correct sum, 1 mark for correct mean.

(b) The new standard deviation will be larger. The value 27 is far from the original mean of 15, increasing the spread of the data. [1 mark]

Award 1 mark for correct prediction with valid reasoning.

9. Estimated standard deviation = 11.7 (or 11.66...) [3 marks]

Midpoints: 15, 25, 35, 45, 55
Σf = 40
Σfx = 5(15) + 8(25) + 12(35) + 10(45) + 5(55) = 75 + 200 + 420 + 450 + 275 = 1420
Σfx² = 5(225) + 8(625) + 12(1225) + 10(2025) + 5(3025) = 1125 + 5000 + 14700 + 20250 + 15125 = 56200
Mean = 1420 ÷ 40 = 35.5
Variance = (56200 ÷ 40) – 35.5² = 1405 – 1260.25 = 144.75
Standard deviation = √144.75 ≈ 12.03
Award 1 mark for correct Σfx and mean, 1 mark for correct Σfx², 1 mark for correct standard deviation.

10. (a) Both data sets have the same mean of 45. [1 mark]

Award 1 mark for correct comparison.

(b) Set Y has a larger standard deviation (12) than Set X (8), indicating that the data in Set Y is more spread out from the mean. Set X has values that are more closely clustered around the mean. [1 mark]

Award 1 mark for correct interpretation.

Section B: Probability

11. (a) P(prime) = 3/6 = 1/2 [1 mark]

Prime numbers on a die: 2, 3, 5 (3 outcomes out of 6).
Award 1 mark for correct probability.

(b) P(greater than 4) = 2/6 = 1/3 [1 mark]

Numbers greater than 4: 5, 6 (2 outcomes out of 6).
Award 1 mark for correct probability.

12. (a) P(blue) = 3/10 [1 mark]

Total balls = 5 + 3 + 2 = 10; blue = 3.
Award 1 mark for correct probability.

(b) P(not red) = 5/10 = 1/2 [1 mark]

Not red = blue or green = 3 + 2 = 5.
Award 1 mark for correct probability.

13. (a) P(King) = 4/52 = 1/13 [1 mark]

4 Kings in a deck of 52 cards.
Award 1 mark for correct probability.

(b) P(red or Queen) = 28/52 = 7/13 [2 marks]

Red cards = 26; Queens = 4; Red Queens = 2 (overlap)
P(red ∪ Queen) = P(red) + P(Queen) – P(red ∩ Queen) = 26/52 + 4/52 – 2/52 = 28/52 = 7/13
Award 1 mark for correct method, 1 mark for correct simplified probability.

14. (a) P(exactly one head) = 2/4 = 1/2 [1 mark]

Possibility diagram: HH, HT, TH, TT. Exactly one head: HT, TH.
Award 1 mark for correct probability.

(b) P(at least one tail) = 3/4 [1 mark]

At least one tail: HT, TH, TT (3 outcomes).
Award 1 mark for correct probability.

15. (a) Tree diagram [2 marks]

First draw: P(White) = 6/10 = 3/5, P(Dark) = 4/10 = 2/5
Second draw (without replacement):
- After White: P(White) = 5/9, P(Dark) = 4/9
- After Dark: P(White) = 6/9 = 2/3, P(Dark) = 3/9 = 1/3
Award 1 mark for correct first branch probabilities, 1 mark for correct second branch probabilities.

(b) P(both dark) = (4/10) × (3/9) = 12/90 = 2/15 [1 mark]

Award 1 mark for correct probability.

(c) P(exactly one white) = P(WD) + P(DW) = (6/10 × 4/9) + (4/10 × 6/9) = 24/90 + 24/90 = 48/90 = 8/15 [2 marks]

Award 1 mark for identifying both paths, 1 mark for correct probability.

16. (a) Tree diagram [2 marks]

First branch: P(Rain) = 0.3, P(Dry) = 0.7
Second branch:
- Given Rain: P(Late) = 0.1, P(Not late) = 0.9
- Given Dry: P(Late) = 0.05, P(Not late) = 0.95
Award 1 mark for correct first branch, 1 mark for correct second branch probabilities.

(b) P(Late) = P(Rain ∩ Late) + P(Dry ∩ Late) = (0.3 × 0.1) + (0.7 × 0.05) = 0.03 + 0.035 = 0.065 [2 marks]

Award 1 mark for correct products, 1 mark for correct sum.

(c) P(Rain | Late) = P(Rain ∩ Late) ÷ P(Late) = 0.03 ÷ 0.065 = 30/65 = 6/13 ≈ 0.462 [2 marks]

Award 1 mark for correct formula, 1 mark for correct probability.

17. (a) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.4 + 0.5 – 0.2 = 0.7 [1 mark]

Award 1 mark for correct answer.

(b) A and B are independent if P(A ∩ B) = P(A) × P(B). P(A) × P(B) = 0.4 × 0.5 = 0.2. Since P(A ∩ B) = 0.2, the events are independent. [2 marks]

Award 1 mark for stating the condition for independence, 1 mark for correct verification and conclusion.

18. (a) P(Yellow) = 1 – 0.35 – 0.45 = 0.2 [1 mark]

Award 1 mark for correct probability.

(b) P(same colour twice) = P(RR) + P(BB) + P(YY) = (0.35)² + (0.45)² + (0.2)² = 0.1225 + 0.2025 + 0.04 = 0.365 [2 marks]

Award 1 mark for identifying all three cases, 1 mark for correct probability.

19. (a) Venn diagram [2 marks]

Rectangle for universal set (30 students)
Two overlapping circles: M (Mathematics) and P (Physics)
Intersection: 8
Mathematics only: 18 – 8 = 10
Physics only: 15 – 8 = 7
Outside both: 30 – 10 – 8 – 7 = 5
Award 1 mark for correct placement of intersection, 1 mark for correct remaining regions.

(b) P(neither) = 5/30 = 1/6 [1 mark]

Award 1 mark for correct probability.

(c) P(Mathematics only) = 10/30 = 1/3 [1 mark]

Award 1 mark for correct probability.

20. (a) Expected gain = (4/10 × $5) + (6/10 × –$ 2) = $2.00 –$ 1.20 = $0.80 [2 marks]

Award 1 mark for correct products, 1 mark for correct expected value.

(b) The game is not fair because the expected gain is positive ( $0.80), meaning the player expects to win money in the long run. A fair game would have an expected gain of$ 0. [1 mark]

Award 1 mark for correct conclusion with valid reasoning.

END OF ANSWER KEY