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A Level H1 Mathematics Practice Paper 3

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TuitionGoWhere Practice Paper - Maths H1 A-Level

TuitionGoWhere Exam Practice (AI)

Subject: Mathematics (H1)
Level: A-Level
Paper: Practice Paper (Version 3 of 5)
Duration: 1 hour 30 minutes
Total Marks: 60

Name: ________________________
Class: ________________________
Date: ________________________

Instructions to Candidates

Write your name and class on the spaces provided.
Answer all questions.
You are expected to use an approved graphing calculator.
Unless otherwise stated, give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees.
The total mark for this paper is 60.
This paper focuses on Section B: Probability and Statistics.

Section A: Probability and Counting Principles (15 Marks)

1. A committee of 4 people is to be selected from a group of 6 men and 5 women. (a) Find the number of different committees that can be formed if there are no restrictions. [1]

(b) Find the number of different committees that can be formed if the committee must contain at least 2 women. [2]

2. In a certain factory, 60% of the workers are skilled, and 40% are unskilled. Among the skilled workers, 10% are left-handed. Among the unskilled workers, 25% are left-handed. A worker is selected at random. (a) Draw a fully labelled probability tree diagram to represent this information. [2]

Drawing space

(b) Find the probability that the selected worker is left-handed. [1]

  




(c) Given that the selected worker is left-handed, find the probability that the worker is skilled. [2]

3. Events $A$ and $B$ are such that $P(A) = 0.4$ , $P(B) = 0.5$ , and $P(A \cup B) = 0.7$ . (a) Find $P(A \cap B)$ . [1]

(b) Determine, with a reason, whether events $A$ and $B$ are independent. [2]

  





(c) Find $P(A' | B)$. [2]

4. A bag contains 5 red balls and 3 blue balls. Two balls are drawn at random from the bag without replacement. (a) Find the probability that both balls are red. [1]

(b) Find the probability that the two balls are of different colours. [1]

Section B: Discrete Distributions (15 Marks)

5. The random variable $X$ follows a binomial distribution $B(12, 0.3)$ . (a) Find $P(X = 4)$ . [1]

(b) Find $P(X \le 2)$. [1]

  



(c) Find $P(X > 5)$. [1]

  



(d) Find the mean and variance of $X$. [2]

6. A manufacturer of light bulbs knows that 5% of the bulbs produced are defective. A random sample of 20 bulbs is selected. (a) State two conditions that must be satisfied for the number of defective bulbs in the sample to be modelled by a binomial distribution. [2]

(b) Find the probability that exactly 2 bulbs in the sample are defective. [1]

  



(c) Find the probability that at least 1 bulb in the sample is defective. [1]

  



(d) The manufacturer takes a second independent random sample of 20 bulbs. Find the probability that the total number of defective bulbs in the two samples combined is exactly 3. [2]

7. In a large population, 20% of individuals have a specific genetic marker. A researcher selects individuals at random until they find one with the marker. (a) Explain why this scenario is not modelled by a binomial distribution. [1]

(b) Find the probability that the researcher finds the first individual with the marker on the 4th selection. [1]

  



(c) Find the expected number of selections needed to find the first individual with the marker. [1]

Section C: Normal Distribution (15 Marks)

8. The heights of adult males in a certain country are normally distributed with a mean of 175 cm and a standard deviation of 8 cm. Let $H$ be the height of a randomly selected adult male. (a) Find $P(H < 180)$ . [1]

(b) Find the value of $h$ such that $P(H > h) = 0.15$. [2]

  




(c) Find the interquartile range of the heights. [2]

9. The weekly expenditure on groceries for households in a town is normally distributed with mean $\mu$ and standard deviation $\sigma$ . It is known that 10% of households spend less than $80 per week, and 25% of households spend more than $120 per week. (a) Write down two equations involving $\mu$ and $\sigma$ based on the given information. [2]

(b) Find the values of $\mu$ and $\sigma$. [3]

10. The masses of bags of rice produced by a machine are normally distributed with mean 5.0 kg and standard deviation 0.1 kg. (a) Find the probability that a randomly selected bag has a mass between 4.9 kg and 5.1 kg. [1]

(b) The machine is adjusted so that the mean mass becomes $\mu$ kg, but the standard deviation remains 0.1 kg. Find the new value of $\mu$ such that only 2% of the bags have a mass less than 4.8 kg. [2]

  





(c) Three bags are selected at random. Find the probability that exactly one of them has a mass less than 4.8 kg, assuming the original distribution ($\mu=5.0, \sigma=0.1$). [2]

Section D: Sampling and Hypothesis Testing (15 Marks)

11. A random sample of 50 students was taken to estimate the mean time spent on homework per week. The summary statistics are $\sum t = 650$ and $\sum t^2 = 9500$ , where $t$ is the time in hours. (a) Calculate the unbiased estimate of the population mean. [1]

(b) Calculate the unbiased estimate of the population variance. [2]

  




(c) State the distribution of the sample mean $\bar{T}$ if the population is normally distributed. [1]

12. A company claims that the mean lifetime of its batteries is 100 hours. A consumer group suspects that the mean lifetime is less than 100 hours. They take a random sample of 40 batteries and find the sample mean to be 96 hours. The population standard deviation is known to be 12 hours. (a) State the null and alternative hypotheses. [2]

(b) Perform a hypothesis test at the 5% significance level. State your conclusion in context. [4]

  









(c) Explain what is meant by a "Type I error" in the context of this question. [2]

13. The Central Limit Theorem states that for a large sample size $n$ , the sampling distribution of the mean is approximately normal, regardless of the population distribution. (a) State the minimum sample size generally accepted for the Central Limit Theorem to apply. [1]

(b) A population has a mean of 50 and a variance of 25. A sample of size 64 is taken. Find the probability that the sample mean is greater than 51. [3]

End of Paper

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Answers

TuitionGoWhere Practice Paper - Maths H1 A-Level (Answers)

Version 3 of 5 - Answer Key

Section A: Probability and Counting Principles

1. (a) Total people = $6 + 5 = 11$ . Select 4. $\binom{11}{4} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330$ Answer: 330 [1]

(b) "At least 2 women" means 2 women, 3 women, or 4 women.

2 Women, 2 Men: $\binom{5}{2}\binom{6}{2} = 10 \times 15 = 150$
3 Women, 1 Man: $\binom{5}{3}\binom{6}{1} = 10 \times 6 = 60$
4 Women, 0 Men: $\binom{5}{4}\binom{6}{0} = 5 \times 1 = 5$ Total = $150 + 60 + 5 = 215$ Answer: 215 [2]

2. (a) Tree Diagram:

First branch: Skilled (0.6), Unskilled (0.4)
From Skilled: Left-handed (0.1), Right-handed (0.9)
From Unskilled: Left-handed (0.25), Right-handed (0.75) [2] (1 for structure, 1 for correct probabilities)

(b) $P(\text{Left}) = P(\text{Skilled} \cap \text{Left}) + P(\text{Unskilled} \cap \text{Left})$ $= (0.6 \times 0.1) + (0.4 \times 0.25) = 0.06 + 0.10 = 0.16$ Answer: 0.16 [1]

(c) $P(\text{Skilled} | \text{Left}) = \frac{P(\text{Skilled} \cap \text{Left})}{P(\text{Left})}$ $= \frac{0.06}{0.16} = \frac{6}{16} = 0.375$ Answer: 0.375 [2]

3. (a) $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ $0.7 = 0.4 + 0.5 - P(A \cap B)$ $P(A \cap B) = 0.9 - 0.7 = 0.2$ Answer: 0.2 [1]

(b) Check independence: Is $P(A \cap B) = P(A)P(B)$ ? $P(A)P(B) = 0.4 \times 0.5 = 0.2$ Since $P(A \cap B) = 0.2$ , they are equal. Answer: Yes, they are independent because $P(A \cap B) = P(A)P(B)$ . [2]

(c) $P(A' | B) = \frac{P(A' \cap B)}{P(B)}$ $P(A' \cap B) = P(B) - P(A \cap B) = 0.5 - 0.2 = 0.3$ $P(A' | B) = \frac{0.3}{0.5} = 0.6$ Answer: 0.6 [2]

4. (a) $P(\text{Red, Red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$ Answer: $\frac{5}{14}$ or 0.357 [1]

(b) $P(\text{Different}) = P(\text{Red, Blue}) + P(\text{Blue, Red})$ $= \left(\frac{5}{8} \times \frac{3}{7}\right) + \left(\frac{3}{8} \times \frac{5}{7}\right) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}$ Answer: $\frac{15}{28}$ or 0.536 [1]

Section B: Discrete Distributions

5. $X \sim B(12, 0.3)$ (a) $P(X=4) = \binom{12}{4}(0.3)^4(0.7)^8 \approx 0.231$ Answer: 0.231 [1]

(b) $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$ Using calculator: binomcdf(12, 0.3, 2) $\approx 0.253$ Answer: 0.253 [1]

(c) $P(X > 5) = 1 - P(X \le 5)$ Using calculator: 1 - binomcdf(12, 0.3, 5) $\approx 1 - 0.882 = 0.118$ Answer: 0.118 [1]

(d) Mean $\mu = np = 12 \times 0.3 = 3.6$ Variance $\sigma^2 = np(1-p) = 12 \times 0.3 \times 0.7 = 2.52$ Answer: Mean = 3.6, Variance = 2.52 [2]

6. $X \sim B(20, 0.05)$ (a) Conditions:

Fixed number of trials ( $n=20$ ).
Constant probability of success ( $p=0.05$ ).
Trials are independent.
Two outcomes (defective/not defective). (Any two) [2]

(b) $P(X=2) = \binom{20}{2}(0.05)^2(0.95)^{18} \approx 0.189$ Answer: 0.189 [1]

(d) Let $X_1$ and $X_2$ be defective counts in sample 1 and 2. Total defective $Y = X_1 + X_2$ . Since samples are independent and $p$ is same, $Y \sim B(40, 0.05)$ . $P(Y=3) = \binom{40}{3}(0.05)^3(0.95)^{37} \approx 0.185$ Answer: 0.185 [2]

7. (a) The number of trials is not fixed; we are counting trials until the first success. This is a Geometric distribution, not Binomial. [1]

(b) Let $Y$ be the trial of first success. $P(Y=4) = (1-0.2)^3 \times 0.2 = (0.8)^3 \times 0.2 = 0.512 \times 0.2 = 0.1024$ Answer: 0.102 [1]

Section C: Normal Distribution

8. $H \sim N(175, 8^2)$ (a) $P(H < 180) = \text{normalcdf}(-\infty, 180, 175, 8) \approx 0.734$ Answer: 0.734 [1]

(b) $P(H > h) = 0.15 \Rightarrow P(H < h) = 0.85$ $h = \text{invNorm}(0.85, 175, 8) \approx 183.28$ Answer: 183 cm (or 183.3) [2]

(c) Interquartile Range (IQR) = $Q_3 - Q_1$ . $Q_1 = \text{invNorm}(0.25, 175, 8) \approx 169.61$ $Q_3 = \text{invNorm}(0.75, 175, 8) \approx 180.39$ $IQR = 180.39 - 169.61 = 10.78$ Answer: 10.8 cm [2]

9. $X \sim N(\mu, \sigma^2)$ (a)

$P(X < 80) = 0.10 \Rightarrow \frac{80 - \mu}{\sigma} = z_{0.10} \approx -1.2816$ $\Rightarrow 80 - \mu = -1.2816\sigma$
$P(X > 120) = 0.25 \Rightarrow P(X < 120) = 0.75 \Rightarrow \frac{120 - \mu}{\sigma} = z_{0.75} \approx 0.6745$ $\Rightarrow 120 - \mu = 0.6745\sigma$ [2]

(b) Subtract eq 1 from eq 2: $(120 - \mu) - (80 - \mu) = 0.6745\sigma - (-1.2816\sigma)$ $40 = 1.9561\sigma$ $\sigma = \frac{40}{1.9561} \approx 20.45$ Substitute $\sigma$ into eq 2: $120 - \mu = 0.6745(20.45)$ $120 - \mu = 13.79$ $\mu = 106.21$ Answer: $\mu \approx 106$ , $\sigma \approx 20.5$ [3]

10. $M \sim N(5.0, 0.1^2)$ (a) $P(4.9 < M < 5.1) = \text{normalcdf}(4.9, 5.1, 5.0, 0.1) \approx 0.683$ Answer: 0.683 [1]

(b) New $\mu$ , $\sigma=0.1$ . $P(M < 4.8) = 0.02$ . $\frac{4.8 - \mu}{0.1} = \text{invNorm}(0.02) \approx -2.0537$ $4.8 - \mu = -0.20537$ $\mu = 4.8 + 0.20537 = 5.00537$ Answer: 5.01 kg [2]

(c) Let $p = P(M < 4.8)$ with original $\mu=5.0$ . $p = \text{normalcdf}(-\infty, 4.8, 5.0, 0.1) \approx 0.0228$ Let $Y$ be number of bags < 4.8kg in 3 bags. $Y \sim B(3, 0.0228)$ . $P(Y=1) = \binom{3}{1}(0.0228)^1(1-0.0228)^2$ $= 3 \times 0.0228 \times (0.9772)^2 \approx 0.0656$ Answer: 0.0656 [2]

Section D: Sampling and Hypothesis Testing

11. $n=50, \sum t = 650, \sum t^2 = 9500$ (a) Unbiased estimate of mean $\bar{t} = \frac{650}{50} = 13$ Answer: 13 hours [1]

(b) Unbiased estimate of variance $s^2 = \frac{1}{n-1} \left( \sum t^2 - \frac{(\sum t)^2}{n} \right)$ $s^2 = \frac{1}{49} \left( 9500 - \frac{650^2}{50} \right) = \frac{1}{49} (9500 - 8450) = \frac{1050}{49} \approx 21.43$ Answer: 21.4 [2]

(c) $\bar{T} \sim N\left(\mu, \frac{\sigma^2}{n}\right)$ or approximately $N\left(13, \frac{21.43}{50}\right)$ Answer: Normal distribution [1]

12. (a) $H_0: \mu = 100$ $H_1: \mu < 100$ [2]

(b) Test statistic $Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$ $Z = \frac{96 - 100}{12/\sqrt{40}} = \frac{-4}{1.897} \approx -2.108$ Critical value for 1-tail 5% test: $z_{0.05} = -1.645$ Since $-2.108 < -1.645$ , the result is in the critical region. Conclusion: Reject $H_0$ . There is sufficient evidence at the 5% level to suggest the mean lifetime is less than 100 hours. [4]

(c) A Type I error occurs if we reject $H_0$ when it is actually true. Context: Concluding that the mean battery life is less than 100 hours when it is actually 100 hours. [2]

13. (a) $n > 30$ [1]

(b) Population: $\mu=50, \sigma^2=25 \Rightarrow \sigma=5$ . Sample: $n=64$ . Sampling distribution of mean: $\bar{X} \sim N\left(50, \frac{25}{64}\right)$ . Standard error $SE = \frac{5}{\sqrt{64}} = \frac{5}{8} = 0.625$ . $P(\bar{X} > 51) = P\left(Z > \frac{51 - 50}{0.625}\right) = P(Z > 1.6)$ $P(Z > 1.6) = 1 - 0.9452 = 0.0548$ Answer: 0.0548 [3]