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A Level H1 Mathematics Practice Paper 2

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Questions

TuitionGoWhere Exam Practice (AI) - A-Level Maths H1

Subject: Mathematics (H1)
Level: A-Level
Paper: Practice Paper 2 of 5 (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.
Write your answers in the spaces provided in this booklet.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
You are expected to use an approved graphing calculator.
Unsupported answers from a graphing calculator are allowed unless the question specifically states otherwise.
Clear presentation in your working is essential.

Section A: Probability and Distributions (20 Marks)

1. A company manufactures smartphone cases. The probability that a case is defective is 0.04. A random sample of 25 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]

2. The weights of durians sold at a market are normally distributed with mean 1.8 kg and standard deviation 0.3 kg.

(a) Find the probability that a randomly chosen durian weighs more than 2.2 kg. [2]

(b) Find the weight $w$ such that 10% of durians weigh less than $w$ . [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 whether events $A$ and $B$ are independent, giving a reason for your answer. [2]

4. In a certain population, 60% of adults prefer tea over coffee. A random sample of 8 adults is chosen.

(a) Find the probability that more than 5 adults prefer tea. [2]

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

5. The time taken by students to complete a statistics quiz is normally distributed with mean 25 minutes and variance 16 minutes $^2$ .

(a) Find the probability that a student takes between 20 and 30 minutes. [2]

(b) If 100 students take the quiz, estimate how many students take more than 33 minutes. [1]

Section B: Sampling and Estimation (20 Marks)

6. A random sample of 10 students was asked how many hours they spent on social media last week. The results are summarized as follows: $\sum x = 120, \quad \sum x^2 = 1550$

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

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

7. The masses of packets of rice are normally distributed with mean $\mu$ kg and standard deviation 0.05 kg. A random sample of 16 packets has a mean mass of 4.98 kg.

(a) Find a 95% confidence interval for the population mean $\mu$ . [3]

(b) State whether the population mean is likely to be 5.00 kg, giving a reason. [1]

8. A surveyor wants to estimate the mean height of trees in a forest. He takes a random sample of 50 trees. The sample mean height is 12.5 m and the sample variance is 4.0 m $^2$ .

(a) Find the standard error of the mean. [2]

(b) Find the probability that the sample mean is within 0.5 m of the population mean. [3]

9. The daily revenue of a shop is normally distributed with mean $2000 and standard deviation $300.

(a) Find the probability that the mean daily revenue over a period of 9 days is less than $1900. [3]

(b) Explain why the Central Limit Theorem is not needed in this calculation. [1]

10. A manufacturer claims that the mean lifetime of their light bulbs is 1000 hours. A consumer group tests a random sample of 36 bulbs and finds a mean lifetime of 980 hours with a standard deviation of 60 hours.

(a) Calculate the test statistic for testing the manufacturer's claim. [2]

(b) State the distribution of the test statistic under the null hypothesis. [1]

Section C: Hypothesis Testing and Regression (20 Marks)

11. A teacher claims that the mean score of her class in a test is greater than 70. A random sample of 25 students has a mean score of 74 and a standard deviation of 10. Assume the scores are normally distributed.

Test the teacher's claim at the 5% significance level.

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

(b) Calculate the test statistic. [2]

12. The table below shows the advertising expenditure $x$ (in $1000s) and sales $y$ (in $10,000s) for 6 months.

Month	$x$	$y$
1	2	3
2	4	5
3	6	7
4	8	9
5	10	11
6	12	13

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

(b) Interpret the value of $r$ in the context of the data. [1]

13. Using the data from Question 12:

(a) Find the equation of the regression line of $y$ on $x$ in the form $y = a + bx$ . [2]

(b) Estimate the sales when the advertising expenditure is $15,000. [1]

14. A researcher investigates the relationship between study time ( $t$ hours) and exam marks ( $m$ ). The regression line of $m$ on $t$ is found to be $m = 40 + 5t$ .

(a) Interpret the gradient of the regression line. [1]

(b) Explain why it might not be appropriate to use this line to predict the mark of a student who studied for 0 hours. [1]

15. In a hypothesis test for the population mean, the null hypothesis is $H_0: \mu = 50$ and the alternative hypothesis is $H_1: \mu \neq 50$ . The test is carried out at the 10% significance level.

(a) Define what is meant by the "critical region" in this context. [1]

(b) If the test statistic falls in the critical region, what conclusion should be drawn? [1]

16. The number of customers entering a shop per hour follows a Poisson distribution with mean 8.

(a) Find the probability that exactly 6 customers enter in a given hour. [2]

(b) Find the probability that more than 10 customers enter in a given hour. [2]

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

(a) Draw a tree diagram to represent the outcomes. [2]

(b) Find the probability that both balls are red. [1]

18. The heights of male students in a school are normally distributed with mean 170 cm and standard deviation 10 cm.

(a) Find the probability that a randomly selected student is taller than 185 cm. [2]

(b) Find the height $h$ such that 90% of students are shorter than $h$ . [2]

19. A sample of 40 observations has $\sum x = 2000$ and $\sum x^2 = 105000$ .

(a) Calculate the sample mean. [1]

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

20. The lifespan of a battery is normally distributed with mean 20 hours and standard deviation 2 hours.

(a) Find the probability that a battery lasts between 18 and 22 hours. [2]

(b) If 5 batteries are chosen at random, find the probability that all 5 last between 18 and 22 hours. [2]

End of Paper

Answers

TuitionGoWhere Exam Practice (AI) - A-Level Maths H1

Answer Key and Marking Scheme - Practice Paper 2 of 5

Subject: Mathematics (H1)
Topic: Statistics and Probability

Section A: Probability and Distributions

1. (a) Let $X$ be the number of defective cases. $X \sim B(25, 0.04)$ . [1] (b) $P(X=2) = \binom{25}{2} (0.04)^2 (0.96)^{23} \approx 0.187$ . [2] (c) $P(X \ge 1) = 1 - P(X=0) = 1 - (0.96)^{25} \approx 1 - 0.360 = 0.640$ . [2]

2. Let $W$ be the weight of a durian. $W \sim N(1.8, 0.3^2)$ . (a) $P(W > 2.2) = P(Z > \frac{2.2-1.8}{0.3}) = P(Z > 1.33) \approx 0.0918$ . [2] (b) $P(W < w) = 0.1 \Rightarrow \frac{w-1.8}{0.3} = -1.282 \Rightarrow w = 1.8 - 0.3846 \approx 1.42$ kg. [2]

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

4. Let $Y$ be the number of adults preferring tea. $Y \sim B(8, 0.6)$ . (a) $P(Y > 5) = P(Y=6) + P(Y=7) + P(Y=8) \approx 0.209 + 0.0896 + 0.0168 = 0.315$ . [2] (b) $E(Y) = np = 8 \times 0.6 = 4.8$ . [1]

5. Let $T$ be the time taken. $T \sim N(25, 4^2)$ . (a) $P(20 < T < 30) = P(\frac{20-25}{4} < Z < \frac{30-25}{4}) = P(-1.25 < Z < 1.25) \approx 0.7887$ . [2] (b) $P(T > 33) = P(Z > \frac{33-25}{4}) = P(Z > 2) \approx 0.0228$ . Expected number $= 100 \times 0.0228 \approx 2$ or $3$ (depending on rounding, 2.28 rounds to 2). [1]

Section B: Sampling and Estimation

6. (a) Unbiased estimate of mean $\bar{x} = \frac{120}{10} = 12$ . [1] (b) Unbiased estimate of variance $s^2 = \frac{1}{n-1} (\sum x^2 - \frac{(\sum x)^2}{n}) = \frac{1}{9} (1550 - \frac{14400}{10}) = \frac{1}{9} (1550 - 1440) = \frac{110}{9} \approx 12.2$ . [2]

7. (a) 95% CI: $\bar{x} \pm z \frac{\sigma}{\sqrt{n}} = 4.98 \pm 1.96 \frac{0.05}{\sqrt{16}} = 4.98 \pm 1.96(0.0125) = 4.98 \pm 0.0245$ . Interval: $(4.9555, 5.0045)$ . [3] (b) Yes, 5.00 is within the confidence interval. [1]

8. (a) Standard error $SE = \frac{s}{\sqrt{n}} = \frac{\sqrt{4}}{\sqrt{50}} = \frac{2}{7.071} \approx 0.283$ . [2] (b) $P(|\bar{X} - \mu| < 0.5) = P(-0.5 < \bar{X} - \mu < 0.5) = P(\frac{-0.5}{0.283} < Z < \frac{0.5}{0.283}) = P(-1.77 < Z < 1.77) \approx 0.923$ . [3]

9. (a) Let $\bar{R}$ be the mean revenue over 9 days. $\bar{R} \sim N(2000, \frac{300^2}{9}) = N(2000, 100^2)$ . $P(\bar{R} < 1900) = P(Z < \frac{1900-2000}{100}) = P(Z < -1) \approx 0.1587$ . [3] (b) The population is already normally distributed, so the sample mean is exactly normal regardless of sample size. CLT is for large samples from non-normal populations. [1]

10. (a) Test statistic $z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} = \frac{980 - 1000}{60/\sqrt{36}} = \frac{-20}{10} = -2$ . [2] (b) Standard Normal Distribution, $N(0,1)$ . [1]

Section C: Hypothesis Testing and Regression

11. (a) $H_0: \mu = 70$ , $H_1: \mu > 70$ . [2] (b) $z = \frac{74 - 70}{10/\sqrt{25}} = \frac{4}{2} = 2$ . [2] (c) Critical value for 5% one-tail is 1.645. Since $2 > 1.645$ , reject $H_0$ . There is sufficient evidence to support the teacher's claim that the mean score is greater than 70. [3]

12. (a) Using GC, $r = 1$ . [2] (b) There is a perfect positive linear correlation between advertising expenditure and sales. [1]

13. (a) Using GC, $y = 1 + 1x$ (or $y = x + 1$ ). [2] (b) If $x=15$ , $y = 15 + 1 = 16$ . Sales = $160,000. [1] (c) This is extrapolation (outside the range of data 2-12), so it may not be reliable. [1]

14. (a) For every additional hour of study, the exam mark increases by 5 marks on average. [1] (b) Extrapolation to 0 hours may not be valid as the linear relationship might not hold at low study times, or a student might still get some marks by guessing. [1]

15. (a) The set of values for the test statistic for which the null hypothesis is rejected. [1] (b) Reject the null hypothesis. [1]

16. Let $C$ be the number of customers. $C \sim Po(8)$ . (a) $P(C=6) = \frac{e^{-8} 8^6}{6!} \approx 0.122$ . [2] (b) $P(C > 10) = 1 - P(C \le 10) \approx 1 - 0.8159 = 0.184$ . [2]

17. (a) Tree diagram: First branch Red (5/8), Blue (3/8). Second branch from Red: Red (4/7), Blue (3/7). From Blue: Red (5/7), Blue (2/7). [2] (b) $P(RR) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$ . [1]

18. Let $H$ be height. $H \sim N(170, 10^2)$ . (a) $P(H > 185) = P(Z > \frac{185-170}{10}) = P(Z > 1.5) \approx 0.0668$ . [2] (b) $P(H < h) = 0.9 \Rightarrow \frac{h-170}{10} = 1.282 \Rightarrow h = 170 + 12.82 = 182.8$ cm. [2]

19. (a) $\bar{x} = \frac{2000}{40} = 50$ . [1] (b) $s^2 = \frac{1}{39} (105000 - \frac{2000^2}{40}) = \frac{1}{39} (105000 - 100000) = \frac{5000}{39} \approx 128$ . [2]

20. Let $L$ be lifespan. $L \sim N(20, 2^2)$ . (a) $P(18 < L < 22) = P(-1 < Z < 1) \approx 0.6827$ . [2] (b) Let $p = 0.6827$ . Probability all 5 last between 18 and 22 hours is $p^5 = (0.6827)^5 \approx 0.148$ . [2]

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