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A Level H1 Mathematics Practice Paper 1

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TuitionGoWhere Exam Practice (AI)

A-Level H1 Mathematics (8865) - Practice Paper 1 (Version 1)

Topic Focus: Statistics and Probability

Subject: Mathematics
Level: A-Level H1
Paper: Practice Paper 1 (Statistics & Probability Focus)
Duration: 1 hour 30 minutes
Total Marks: 60

Name: __________________________
Class: __________________________
Date: __________________________

Instructions to Candidates:

Write your name, class, and date in the spaces provided.
Answer all questions.
You are expected to use an approved graphing calculator.
Unsupported answers from a calculator are generally allowed unless the question specifically states otherwise.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
The total mark for this paper is 60.

Section A: Probability and Distributions (20 Marks)

1. A company manufactures smartphone cases. The probability that a case is defective is 0.05. A random sample of 20 cases is selected.

(a) State the distribution of the number of defective cases in the sample, defining any variables used. [1]

(b) Find the probability that exactly 2 cases are defective. [2]

Answer space

2. In a certain population, 30% of adults prefer online shopping over in-store shopping. A random sample of 15 adults is chosen.

(a) Find the probability that fewer than 4 adults prefer online shopping. [2]

(b) Find the expected number of adults in the sample who prefer online shopping. [1]

Answer space

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 whether events $A$ and $B$ are independent, giving a reason for your answer. [2]

Answer space

4. A bag contains 5 red balls and 3 blue balls. Two balls are drawn at random without replacement.

(a) Draw a tree diagram to represent the possible outcomes and their probabilities. [2]

(b) Find the probability that the two balls are of different colours. [2]

Answer space

Section B: Sampling and Estimation (20 Marks)

5. A researcher wants to estimate the mean height of students in a large college. She takes a random sample of 50 students. The heights, $h$ cm, are summarised as follows: $\sum h = 8500, \quad \sum h^2 = 1,446,000$

(a) Calculate an unbiased estimate of the population mean height. [1]

(b) Calculate an unbiased estimate of the population variance. [3]

Answer space

6. The masses of bags of rice produced by a factory are normally distributed with mean $\mu$ kg and standard deviation 0.2 kg. A random sample of 16 bags is selected.

(a) State the distribution of the sample mean $\bar{X}$ . [2]

(b) Find the probability that the sample mean is within 0.1 kg of the population mean $\mu$ . [3]

Answer space

7. A surveyor wishes to select a sample of 20 residents from a list of 500 residents in a housing estate.

(a) Describe how the surveyor could use a random number generator to select a simple random sample. [2]

(b) Explain why stratified sampling might be more appropriate than simple random sampling if the estate consists of distinct blocks with different demographic profiles. [2]

Answer space

8. The daily income of freelance writers is normally distributed with mean $150 and standard deviation $40.

(a) Find the probability that a randomly selected writer earns more than $180 in a day. [2]

(b) A random sample of 25 writers is selected. Find the probability that their mean daily income is less than $140. [3]

Answer space

Section C: Hypothesis Testing and Regression (20 Marks)

9. A manufacturer claims that the mean lifetime of their light bulbs is 1200 hours. A consumer group suspects the mean lifetime is less than 1200 hours. They test a random sample of 50 bulbs and find a sample mean of 1180 hours. Assume the population standard deviation is known to be 100 hours.

(a) State the null and alternative hypotheses. [2]

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

Answer space

10. The table below shows the advertising expenditure $x$ ($000) and sales revenue $y$ ($000) for a company over 6 months.

Month	$x$	$y$
1	2.0	15.0
2	3.5	22.0
3	4.0	25.0
4	5.5	30.0
5	6.0	32.0
6	7.0	38.0

(a) Calculate the product moment correlation coefficient, $r$ . [2]

(b) Find the equation of the least squares regression line of $y$ on $x$ in the form $y = mx + c$ . [3]

(d) Estimate the sales revenue if the advertising expenditure is $4.5 thousand. Comment on the reliability of this estimate. [2]

Answer space

11. For a different dataset, the regression line of $y$ on $x$ is $y = 2.5x + 10$ and the regression line of $x$ on $y$ is $x = 0.3y + 2$ .

(a) Find the product moment correlation coefficient $r$ . [2]

(b) Explain why $r$ must be positive in this case. [1]

Answer space

End of Paper

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Answers

TuitionGoWhere Exam Practice (AI) - Answer Key

A-Level H1 Mathematics (8865) - Practice Paper 1 (Version 1)

Section A: Probability and Distributions

1. (a) Let $X$ be the number of defective cases. $X \sim B(20, 0.05)$ . [1] (b) $P(X=2) = \binom{20}{2}(0.05)^2(0.95)^{18} \approx 0.1887$ . [2] (c) $P(X > 1) = 1 - P(X \le 1) = 1 - [P(X=0) + P(X=1)]$ . $P(X=0) = (0.95)^{20} \approx 0.3585$ . $P(X=1) = 20(0.05)(0.95)^{19} \approx 0.3774$ . $P(X > 1) = 1 - (0.3585 + 0.3774) = 1 - 0.7359 = 0.2641$ . [2]

2. (a) Let $Y$ be the number of adults preferring online shopping. $Y \sim B(15, 0.3)$ . $P(Y < 4) = P(Y \le 3) = P(Y=0) + P(Y=1) + P(Y=2) + P(Y=3)$ . Using calculator: $P(Y \le 3) \approx 0.2969$ . [2] (b) $E(Y) = np = 15 \times 0.3 = 4.5$ . [1]

3. (a) $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ . $0.7 = 0.4 + 0.5 - P(A \cap B) \Rightarrow P(A \cap B) = 0.2$ . [1] (b) Check independence: $P(A)P(B) = 0.4 \times 0.5 = 0.2$ . Since $P(A \cap B) = 0.2$ , $P(A \cap B) = P(A)P(B)$ . Therefore, $A$ and $B$ are independent. [2] (c) $P(A | B') = \frac{P(A \cap B')}{P(B')}$ . $P(B') = 1 - 0.5 = 0.5$ . $P(A \cap B') = P(A) - P(A \cap B) = 0.4 - 0.2 = 0.2$ . $P(A | B') = \frac{0.2}{0.5} = 0.4$ . [2]

4. (a) Tree Diagram: First Draw: Red (5/8), Blue (3/8). If Red: Second Draw Red (4/7), Blue (3/7). If Blue: Second Draw Red (5/7), Blue (2/7). [2] (b) $P(\text{Different}) = P(RB) + P(BR)$ . $P(RB) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}$ . $P(BR) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}$ . $P(\text{Different}) = \frac{30}{56} = \frac{15}{28} \approx 0.536$ . [2]

Section B: Sampling and Estimation

5. (a) Unbiased estimate of mean $\bar{x} = \frac{8500}{50} = 170$ cm. [1] (b) Unbiased estimate of variance $s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right)$ . $s^2 = \frac{1}{49} \left( 1,446,000 - \frac{8500^2}{50} \right)$ . $s^2 = \frac{1}{49} (1,446,000 - 1,445,000) = \frac{1000}{49} \approx 20.41$ . [3]

6. (a) Population $X \sim N(\mu, 0.2^2)$ . Sample size $n=16$ . Sample mean $\bar{X} \sim N\left(\mu, \frac{0.2^2}{16}\right) = N(\mu, 0.0025)$ . Standard deviation of $\bar{X} = \sqrt{0.0025} = 0.05$ . [2] (b) We want $P(\mu - 0.1 < \bar{X} < \mu + 0.1)$ . Standardize: $Z = \frac{\bar{X} - \mu}{0.05}$ . Limits: $\frac{-0.1}{0.05} = -2$ and $\frac{0.1}{0.05} = 2$ . $P(-2 < Z < 2) = P(Z < 2) - P(Z < -2)$ . Using calculator/tables: $0.9772 - 0.0228 = 0.9544$ . [3]

7. (a) Assign each resident a unique number from 1 to 500. Use a random number generator to produce 20 distinct integers between 1 and 500. Select the residents corresponding to these numbers. [2] (b) Stratified sampling ensures that each distinct block (stratum) is represented in the sample in proportion to its size. This reduces sampling error if the demographic profiles (and thus the variable of interest) vary significantly between blocks, ensuring the sample is more representative of the whole population. [2]

8. (a) $X \sim N(150, 40^2)$ . $P(X > 180) = P\left(Z > \frac{180-150}{40}\right) = P(Z > 0.75)$ . $P(Z > 0.75) = 1 - 0.7734 = 0.2266$ . [2] (b) Sample mean $\bar{X} \sim N\left(150, \frac{40^2}{25}\right) = N(150, 64)$ . SD of $\bar{X} = \sqrt{64} = 8$ . $P(\bar{X} < 140) = P\left(Z < \frac{140-150}{8}\right) = P(Z < -1.25)$ . $P(Z < -1.25) = 0.1056$ . [3]

Section C: Hypothesis Testing and Regression

9. (a) $H_0: \mu = 1200$ . $H_1: \mu < 1200$ . [2] (b) Test statistic $Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} = \frac{1180 - 1200}{100/\sqrt{50}} = \frac{-20}{14.142} \approx -1.414$ . Critical value for 1-tail test at 5%: $z_{crit} = -1.645$ . Since $-1.414 > -1.645$ , the test statistic is not in the critical region. Alternatively, $p\text{-value} = P(Z < -1.414) \approx 0.0787$ . Since $0.0787 > 0.05$ , we do not reject $H_0$ . Conclusion: There is insufficient evidence at the 5% significance level to support the claim that the mean lifetime is less than 1200 hours. [4]

10. (a) Using GC: $r \approx 0.993$ . [2] (b) Using GC: $y = 4.64x + 5.95$ (values to 3 s.f.). [3] (c) For every additional $1000 spent on advertising, the sales revenue increases by approximately $4640 on average. [1] (d) Substitute $x=4.5$ : $y = 4.64(4.5) + 5.95 = 20.88 + 5.95 = 26.83$ . Estimated revenue is $26,830. Reliability: High, because $x=4.5$ is within the range of the observed data (interpolation) and $r$ is very close to 1, indicating a strong linear correlation. [2]

11. (a) The product of the gradients of the two regression lines is equal to $r^2$ . Gradient of $y$ on $x$ ( $b_{yx}$ ) = 2.5. Gradient of $x$ on $y$ ( $b_{xy}$ ) = 0.3. $r^2 = b_{yx} \times b_{xy} = 2.5 \times 0.3 = 0.75$ . $r = \sqrt{0.75} \approx 0.866$ . [2] (b) Both gradients (2.5 and 0.3) are positive, which indicates a positive correlation between $x$ and $y$ . Therefore, $r$ must be positive. [1]