A Level H1 Mathematics Practice Paper 1

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TuitionGoWhere Practice Paper - Maths H1 A-Level

TuitionGoWhere Practice Paper (AI) — Version 1 of 5

Subject: Maths H1
Level: A-Level
Paper: Practice Paper (Statistics & Probability Topic Quiz)
Duration: 60 minutes
Total Marks: 40
Name: ________________________
Class: ________________________
Date: ________________________

Instructions

• This practice paper contains 20 questions on Statistics & Probability (Topic 3 of H1 Maths 8865).
• Answer all questions in the spaces provided.
• Show all working clearly. Calculator (non-CAS GC) is allowed.
• Total marks: 40. Section marks are shown at the end of each section.

Section A: Probability and Counting (Questions 1–6) [12 marks]

1. A committee of 3 students is to be selected from a class of 10 students. Find the number of possible committees. [1]

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2. A bag contains 5 red and 7 blue marbles. Two marbles are drawn at random without replacement. Draw a tree diagram to represent the possible outcomes and their probabilities. [2]

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3. Events A and B are such that $P(A) = 0.4$, $P(B) = 0.5$, and $P(A \cap B) = 0.2$. Find $P(A \cup B)$. [1]

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4. A fair die is rolled and a fair coin is tossed. Find the probability of obtaining an even number on the die and a head on the coin. [1]

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5. In a survey, 60% of residents use public transport. Two residents are selected at random. Find the probability that exactly one uses public transport. [2]

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6. Given $P(X) = 0.3$ and $P(Y \mid X) = 0.6$, find $P(X \cap Y)$. [1]

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Section B: Distributions (Questions 7–13) [14 marks]

7. The random variable $X \sim B(20, 0.3)$. Find $P(X > 13)$. [2]

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8. State the mean and variance of $X \sim B(15, 0.4)$. [2]

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9. The mass of a type of apple is normally distributed with mean 150 g and standard deviation 12 g. Find the probability that a randomly chosen apple has mass greater than 165 g. [2]

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10. For $X \sim N(50, 8^2)$, find the value of $x$ such that $P(X < x) = 0.95$. [2]

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11. Let $X \sim N(10, 4)$ and $Y \sim N(20, 9)$ be independent. Find $E(2X + 3Y)$ and $\text{Var}(2X + 3Y)$. [2]

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12. A random sample of 36 observations is taken from a normal population with mean $\mu$ and variance $\sigma^2$. State the distribution of the sample mean $\bar{X}$ and give its mean and variance. [2]

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13. Explain, with reference to the Central Limit Theorem, why the sample mean of a sample of size 50 from a non-normal population may be treated as approximately normal. [2]

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Section C: Data Analysis and Hypothesis Testing (Questions 14–20) [14 marks]

14. The following are the scores of 8 students in a test: 12, 15, 14, 10, 13, 16, 11, 13. Find the unbiased estimates of the population mean and variance. [3]

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15. The product moment correlation coefficient for the data below is $r = -0.78$. Comment on the strength and direction of the relationship.

Hours of exercise per week	2	5	1	8	3	6
Cholesterol level	220	200	230	180	210	190

[2]

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16. The scatter diagram for the data in Q15 is shown below.

Image pending generation: scatter for Q16.

Sketch the scatter diagram as shown on your calculator. [1]

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17. A researcher claims that the mean reaction time of drivers is 0.4 s. A random sample of 40 drivers gives a sample mean of 0.43 s and standard deviation 0.08 s. Test at the 5% significance level whether there is evidence that the mean reaction time differs from 0.4 s. [4]

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18. For the data in Q15, the equation of the least squares regression line of cholesterol level $y$ on exercise hours $x$ is $y = a + bx$. Given $\bar{x}=4.17$, $\bar{y}=205$, and $b = -6.2$, find the value of $a$. [1]

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19. State one reason why a high correlation between two variables does not imply causation. [1]

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20. A surveyor wishes to select a random sample of 25 households from a list of 200. Describe a method of simple random sampling he could use. [1]

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TuitionGoWhere Practice Paper - Maths H1 A-Level (Answers) — Version 1

Total Marks: 40

Section A: Probability and Counting (12 marks)

1. [1]
Number of committees = $\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
Teaching note: Order does not matter in a committee, so we use combinations not permutations.

2. [2]
Tree diagram:
First branch: R ($5/12$), B ($7/12$).
If R drawn: second R ($4/11$), B ($7/11$).
If B drawn: second R ($5/11$), B ($6/11$).
Marking: 1 mark for correct first-stage probabilities, 1 mark for correct second-stage conditional probabilities.

3. [1]
$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.5 - 0.2 = 0.7$.

4. [1]
$P(\text{even}) = 3/6 = 1/2$, $P(\text{head}) = 1/2$. Independent, so $P = 1/2 \times 1/2 = 1/4$.

5. [2]
Exactly one uses public transport: $0.6 \times 0.4 + 0.4 \times 0.6 = 0.24 + 0.24 = 0.48$.
Common mistake: Forgetting one of the two orders.

6. [1]
$P(X \cap Y) = P(X) \times P(Y \mid X) = 0.3 \times 0.6 = 0.18$.

Section B: Distributions (14 marks)

7. [2]
$X \sim B(20, 0.3)$. $P(X > 13) = 1 - P(X \leq 13)$. Using GC binomial CDF: $\approx 0.0015$.
Note: Do not use normal approximation (excluded in H1).

8. [2]
Mean = $np = 15 \times 0.4 = 6$. Variance = $np(1-p) = 15 \times 0.4 \times 0.6 = 3.6$.

9. [2]
$X \sim N(150, 12^2)$. $Z = (165-150)/12 = 1.25$. $P(X > 165) = P(Z > 1.25) = 1 - 0.8944 = 0.1056$.

10. [2]
$P(X < x) = 0.95$, $Z_{0.95} = 1.645$. $x = 50 + 1.645 \times 8 = 63.16$.

11. [2]
$E(2X+3Y) = 2E(X)+3E(Y) = 2(10)+3(20) = 80$.
$\text{Var}(2X+3Y) = 4\text{Var}(X)+9\text{Var}(Y) = 4(4)+9(9) = 16+81 = 97$.

12. [2]
$\bar{X} \sim N(\mu, \sigma^2/36)$. Mean = $\mu$, Variance = $\sigma^2/36$.

13. [2]
By CLT, for a sufficiently large sample ($n \ge 30$), the distribution of the sample mean is approximately normal regardless of population distribution. With $n=50$, we may treat $\bar{X}$ as approximately $N(\mu, \sigma^2/50)$.

Section C: Data Analysis and Hypothesis Testing (14 marks)

14. [3]
$\bar{x} = (12+15+14+10+13+16+11+13)/8 = 104/8 = 13$.
$s^2 = \frac{\sum x_i^2 - n\bar{x}^2}{n-1}$; $\sum x_i^2 = 144+225+196+100+169+256+121+169 = 1380$.
$s^2 = (1380 - 8 \times 169)/7 = (1380-1352)/7 = 28/7 = 4$.
Unbiased mean = 13, variance = 4.
Marking: 1 for mean, 2 for variance (using $n-1$).

15. [2]
$r = -0.78$ indicates a strong negative linear relationship: as exercise hours increase, cholesterol tends to decrease.
Marking: 1 for strength (strong), 1 for direction (negative).

16. [1]
Sketch must show 6 points with downward trend, labelled axes as in placeholder.
Marking: 1 for correct plotted points and labels.

17. [4]
$H_0: \mu = 0.4$, $H_1: \mu \neq 0.4$ (two-tailed).
Test statistic: $Z = \frac{0.43 - 0.4}{0.08/\sqrt{40}} = \frac{0.03}{0.01265} = 2.37$.
Critical value at 5% two-tailed: $\pm 1.96$.
Since $2.37 > 1.96$, reject $H_0$. There is evidence at 5% level that mean reaction time differs from 0.4 s.
Marking: 1 H₀/H₁, 1 calc, 1 crit, 1 conclusion.

18. [1]
$a = \bar{y} - b\bar{x} = 205 - (-6.2)(4.17) = 205 + 25.854 = 230.854 \approx 230.9$.

19. [1]
Example: confounding variable (e.g., diet affects both exercise and cholesterol).

20. [1]
Assign numbers 1–200 to households, use random number generator to pick 25 distinct numbers, select those households.