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A Level H1 Mathematics Practice Paper 2

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TuitionGoWhere Practice Paper - Maths H1 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Mathematics (H1)
Level: A-Level (8865)
Paper: Practice Paper - Version 2 of 5
Topic Focus: Statistics and Probability
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 (GC) where appropriate.
Unsupported answers from a GC are allowed unless the question specifically states otherwise.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of 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 screens. The probability that a screen is defective is 0.04. A random sample of 25 screens is selected.
Let $X$ be the number of defective screens in the sample.

(a) State the distribution of $X$ , specifying the parameters.
[1]

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

2. In a certain population, 30% of adults prefer public transport over private cars. A random sample of 10 adults is chosen.

(a) Find the probability that more than 7 adults prefer public transport.
[2]

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

3. Events $A$ and $B$ are defined such that $P(A) = 0.6$ , $P(B) = 0.5$ , and $P(A \cap B) = 0.2$ .

(a) Find $P(A \cup B)$ .
[1]

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

4. The masses of apples sold at a supermarket are normally distributed with mean 150 g and standard deviation 12 g. An apple is selected at random.

(a) Find the probability that the mass of the apple is between 140 g and 165 g.
[2]

(b) Find the mass $m$ such that 10% of the apples have a mass greater than $m$ .
[2]

5. Two independent random variables $X$ and $Y$ are defined as follows:
$X \sim N(20, 3^2)$
$Y \sim N(15, 4^2)$

Let $W = 2X - Y$ .

(a) Find $E(W)$ .
[1]

(b) Find $Var(W)$ .
[2]

Section B: Sampling and Estimation (20 Marks)

6. A researcher wishes to estimate the mean height of students in a large college. He takes a random sample of 50 students. The heights, $h$ cm, are summarized as follows:
$\sum h = 8250$
$\sum h^2 = 1,361,500$

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

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

7. The daily sales of a bakery follow a normal distribution with unknown mean $\mu$ and known standard deviation $\sigma = 15$ units. A random sample of 36 days is taken, and the sample mean is found to be 120 units.

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

(b) State, with a reason, whether the value 115 is a plausible value for the population mean.
[1]

8. A machine fills bottles with juice. The volume of juice in a bottle is normally distributed with mean 500 ml and standard deviation 5 ml.

(a) A quality control officer takes a random sample of 16 bottles. Find the probability that the mean volume of these 16 bottles is less than 498 ml.
[3]

(b) Explain why the Central Limit Theorem is not required in part (a).
[1]

9. The weights of a certain breed of dog are normally distributed with mean 25 kg and variance 9 kg $^2$ .

(a) Find the probability that a randomly selected dog weighs more than 28 kg.
[2]

(b) Find the probability that the mean weight of a random sample of 9 dogs is more than 28 kg.
[3]

10. A surveyor wants to select a sample of 20 residents from a housing estate of 200 residents to interview about noise levels.

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

(b) Suggest one advantage of simple random sampling over convenience sampling.
[1]

Section C: Hypothesis Testing and Regression (20 Marks)

11. 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 a hypothesis test at the 5% significance level. State your conclusion in the context of the question.
[4]

12. The table below shows the age ( $x$ years) and the reaction time ( $y$ milliseconds) of 6 participants in a driving simulation test.

Age ( $x$ )	20	30	40	50	60	70
Reaction Time ( $y$ )	250	280	310	350	400	450

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

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

13. Refer to the data in Question 12.

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

(b) Estimate the reaction time for a participant aged 45 years.
[1]

(c) Explain why it might be unreliable to use this regression line to estimate the reaction time of a 90-year-old participant.
[1]

14. A two-tail hypothesis test is conducted at the 10% significance level. The test statistic $Z$ is calculated to be 1.85.

(a) Find the critical values for this test.
[2]

(b) State whether the null hypothesis should be rejected. Give a reason.
[2]

15. The time taken by students to complete a puzzle is normally distributed with mean $\mu$ minutes and standard deviation 3 minutes. A teacher believes that the mean time has increased from the historical value of 10 minutes. She takes a sample of 25 students.

(a) Find the critical region for the sample mean $\bar{X}$ at the 5% significance level.
[3]

(b) If the sample mean is 11.2 minutes, what is the conclusion of the test?
[1]

16. In a large population, 40% of voters support Party A. A pollster takes a random sample of 100 voters.

(a) State the approximate distribution of the sample proportion $\hat{P}$ of voters supporting Party A.
[2]

(b) Find the probability that the sample proportion is greater than 0.45.
[3]

17. The heights of men in a country are normally distributed with mean 175 cm and standard deviation 7 cm. The heights of women are normally distributed with mean 162 cm and standard deviation 6 cm.

(a) A man and a woman are selected at random. Find the probability that the man is taller than the woman.
[3]

(b) Two men are selected at random. Find the probability that their total height is greater than 360 cm.
[3]

18. A discrete random variable $X$ has the following probability distribution:

$x$	1	2	3	4
$P(X=x)$	0.1	0.3	0.4	0.2

(a) Find $E(X)$ .
[2]

(b) Find $Var(X)$ .
[3]

19. A factory produces two types of widgets, Type A and Type B. The probability that a Type A widget is defective is 0.02, and for Type B it is 0.05. 60% of the widgets produced are Type A, and 40% are Type B.

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

(b) Find the probability that a randomly selected widget is defective.
[2]

20. The weekly expenditure on groceries for households in a town is normally distributed with mean $\$ 150 $and standard deviation$ $30$.

(a) Find the probability that a randomly selected household spends more than $\$ 200$ on groceries in a week.
[2]

(b) Find the expenditure amount $k$ such that 25% of households spend less than $k$ .
[2]

End of Paper

Answers

TuitionGoWhere Practice Paper - Maths H1 A-Level (Answer Key)

Version 2 of 5

Section A: Probability and Distributions

1. (a) $X \sim B(25, 0.04)$
[1]

(b) $P(X=2) = \binom{25}{2} (0.04)^2 (0.96)^{23} \approx 0.187$
[2]
(1 mark for correct substitution, 1 mark for answer)

(c) $P(X \ge 1) = 1 - P(X=0) = 1 - (0.96)^{25} \approx 1 - 0.360 = 0.640$
[2]
(1 mark for method $1-P(X=0)$ , 1 mark for answer)

2. (a) Let $Y \sim B(10, 0.3)$ .
$P(Y > 7) = P(Y=8) + P(Y=9) + P(Y=10)$
$= \binom{10}{8}(0.3)^8(0.7)^2 + \binom{10}{9}(0.3)^9(0.7)^1 + \binom{10}{10}(0.3)^{10}$
$\approx 0.00145 + 0.00014 + 0.00001 = 0.00160$
[2]
(1 mark for correct sum setup, 1 mark for answer)

(b) $E(Y) = np = 10 \times 0.3 = 3$
[1]

3. (a) $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.6 + 0.5 - 0.2 = 0.9$
[1]

(b) Check if $P(A \cap B) = P(A)P(B)$ .
$P(A)P(B) = 0.6 \times 0.5 = 0.3$ .
Since $0.2 \neq 0.3$ , $A$ and $B$ are not independent.
[2]
(1 mark for calculation of product, 1 mark for conclusion)

(c) $P(A | B') = \frac{P(A \cap B')}{P(B')}$ .
$P(B') = 1 - 0.5 = 0.5$ .
$P(A \cap B') = P(A) - P(A \cap B) = 0.6 - 0.2 = 0.4$ .
$P(A | B') = \frac{0.4}{0.5} = 0.8$ .
[2]
(1 mark for numerator/denominator logic, 1 mark for answer)

4. Let $M \sim N(150, 12^2)$ .

(a) $P(140 < M < 165)$ .
Using GC: normalcdf(140, 165, 150, 12) $\approx 0.691$ .
[2]

(b) $P(M > m) = 0.10 \Rightarrow P(M < m) = 0.90$ .
Using GC: invNorm(0.90, 150, 12) $\approx 165.38$ .
$m \approx 165$ g (3 s.f.).
[2]

5. $X \sim N(20, 9)$ , $Y \sim N(15, 16)$ . Independent.

(a) $E(W) = E(2X - Y) = 2E(X) - E(Y) = 2(20) - 15 = 40 - 15 = 25$ .
[1]

(b) $Var(W) = Var(2X - Y) = 2^2 Var(X) + (-1)^2 Var(Y) = 4(9) + 1(16) = 36 + 16 = 52$ .
[2]
(1 mark for formula $4Var(X) + Var(Y)$ , 1 mark for answer)

Section B: Sampling and Estimation

6. $n=50, \sum h = 8250, \sum h^2 = 1,361,500$ .

(a) Unbiased estimate of mean $\bar{x} = \frac{8250}{50} = 165$ cm.
[1]

(b) Unbiased estimate of variance $s^2 = \frac{n}{n-1} \left( \frac{\sum h^2}{n} - \bar{x}^2 \right)$ .
$s^2 = \frac{50}{49} \left( \frac{1,361,500}{50} - 165^2 \right)$
$s^2 = \frac{50}{49} (27,230 - 27,225) = \frac{50}{49} (5) \approx 5.10$ .
[3]
(1 mark for formula, 1 mark for substitution, 1 mark for answer)

7. $\sigma = 15, n=36, \bar{x} = 120$ . 95% CI.

(a) Formula: $\bar{x} \pm z \frac{\sigma}{\sqrt{n}}$ .
$z_{0.025} = 1.96$ .
Margin of error $= 1.96 \times \frac{15}{\sqrt{36}} = 1.96 \times 2.5 = 4.9$ .
CI: $120 \pm 4.9 \Rightarrow (115.1, 124.9)$ .
[3]
(1 mark for standard error, 1 mark for z-value/margin, 1 mark for interval)

(b) Yes, 115 is not in the interval $(115.1, 124.9)$ , so it is not a plausible value at the 95% confidence level.
(Note: 115 is very close, but strictly outside. If student says "No" with correct reasoning, accept. If student calculates test stat, also accept.)
[1]

8. $V \sim N(500, 5^2)$ . Sample $n=16$ .

(a) Distribution of sample mean $\bar{V} \sim N(500, \frac{5^2}{16}) = N(500, 1.5625)$ .
$P(\bar{V} < 498)$ .
$Z = \frac{498 - 500}{5/\sqrt{16}} = \frac{-2}{1.25} = -1.6$ .
$P(Z < -1.6) \approx 0.0548$ .
[3]
(1 mark for dist of mean, 1 mark for standardizing, 1 mark for prob)

(b) The Central Limit Theorem is not required because the population distribution is already normal. The sample mean of a normal population is always normal, regardless of sample size.
[1]

9. $W \sim N(25, 9)$ . $\sigma = 3$ .

(a) $P(W > 28)$ .
$Z = \frac{28-25}{3} = 1$ .
$P(Z > 1) = 1 - 0.8413 = 0.1587$ .
[2]

(b) Sample $n=9$ . $\bar{W} \sim N(25, \frac{9}{9}) = N(25, 1)$ .
$P(\bar{W} > 28)$ .
$Z = \frac{28-25}{1} = 3$ .
$P(Z > 3) \approx 0.00135$ .
[3]
(1 mark for new variance, 1 mark for Z, 1 mark for prob)

10. (a) Assign each of the 200 residents a unique number from 1 to 200. Use a random number generator to produce 20 distinct integers between 1 and 200. Select the residents corresponding to these numbers.
[2]
(1 mark for numbering, 1 mark for random selection of distinct numbers)

(b) Simple random sampling ensures every resident has an equal chance of being selected, reducing selection bias. Convenience sampling may over-represent certain groups (e.g., those home during the day).
[1]

Section C: Hypothesis Testing and Regression

11. $\mu_0 = 1200, \sigma = 100, n=50, \bar{x} = 1180$ . $\alpha = 0.05$ .

(a) $H_0: \mu = 1200$
$H_1: \mu < 1200$
[2]

(b) Test statistic $Z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}} = \frac{1180 - 1200}{100/\sqrt{50}} = \frac{-20}{14.14} \approx -1.414$ .
Critical value for one-tail 5%: $-1.645$ .
Since $-1.414 > -1.645$ (or P-value $0.0786 > 0.05$ ), we do not reject $H_0$ .
Conclusion: There is insufficient evidence at the 5% level to suggest the mean lifetime is less than 1200 hours.
[4]
(1 mark for Z calc, 1 mark for critical value/p-value, 1 mark for comparison, 1 mark for context conclusion)

12. (a) Using GC: $r \approx 0.986$ .
[2]

(b) There is a strong, positive, linear correlation between age and reaction time. As age increases, reaction time tends to increase.
[1]

13. (a) Using GC: $y = 196.67 + 3.57x$ (values approx).
Exact: $b = \frac{S_{xy}}{S_{xx}}$ , $a = \bar{y} - b\bar{x}$ .
$\bar{x} = 45, \bar{y} = 340$ .
$S_{xx} = 1750, S_{xy} = 6250$ .
$b = 6250/1750 = 3.57$ .
$a = 340 - 3.57(45) = 179.35$ (Check: GC gives $a \approx 179.3, b \approx 3.57$ ).
Equation: $y = 179.3 + 3.57x$ .
[2]

(b) $y = 179.3 + 3.57(45) = 340$ ms.
[1]

(c) Age 90 is outside the range of the data (20-70). This is extrapolation, and the linear relationship may not hold for older ages.
[1]

14. Two-tail, $\alpha = 0.10$ .

(a) Critical values are $\pm z_{0.05} = \pm 1.645$ .
[2]

(b) Test statistic $Z = 1.85$ .
Since $1.85 > 1.645$ , the result falls in the critical region.
Reject $H_0$ .
[2]

15. $\mu_0 = 10, \sigma = 3, n=25$ . One-tail (increase), $\alpha = 0.05$ .

(a) Critical region for $\bar{X}$ .
Critical $Z = 1.645$ .
$\bar{X}_{crit} = \mu_0 + 1.645 \frac{\sigma}{\sqrt{n}} = 10 + 1.645 \frac{3}{5} = 10 + 0.987 = 10.987$ .
Critical region: $\bar{X} > 10.99$ (2 d.p.).
[3]
(1 mark for SE, 1 mark for Z, 1 mark for boundary)

(b) $11.2 > 10.99$ , so reject $H_0$ . There is evidence the mean time has increased.
[1]

16. $p = 0.4, n=100$ .

(a) $\hat{P} \sim N(p, \frac{p(1-p)}{n}) = N(0.4, \frac{0.4(0.6)}{100}) = N(0.4, 0.0024)$ .
[2]
(1 mark for mean, 1 mark for variance)

(b) $P(\hat{P} > 0.45)$ .
$Z = \frac{0.45 - 0.4}{\sqrt{0.0024}} = \frac{0.05}{0.04899} \approx 1.02$ .
$P(Z > 1.02) = 1 - 0.8461 = 0.1539$ .
[3]

17. $M \sim N(175, 7^2)$ , $W \sim N(162, 6^2)$ .

(a) Let $D = M - W$ .
$E(D) = 175 - 162 = 13$ .
$Var(D) = 7^2 + 6^2 = 49 + 36 = 85$ .
$D \sim N(13, 85)$ .
$P(D > 0)$ .
$Z = \frac{0 - 13}{\sqrt{85}} = \frac{-13}{9.22} \approx -1.41$ .
$P(Z > -1.41) = P(Z < 1.41) \approx 0.9207$ .
[3]

(b) Let $T = M_1 + M_2$ .
$E(T) = 175 + 175 = 350$ .
$Var(T) = 7^2 + 7^2 = 98$ .
$T \sim N(350, 98)$ .
$P(T > 360)$ .
$Z = \frac{360 - 350}{\sqrt{98}} = \frac{10}{9.90} \approx 1.01$ .
$P(Z > 1.01) = 1 - 0.8438 = 0.1562$ .
[3]

18. (a) $E(X) = \sum x P(x) = 1(0.1) + 2(0.3) + 3(0.4) + 4(0.2) = 0.1 + 0.6 + 1.2 + 0.8 = 2.7$ .
[2]

(b) $E(X^2) = 1^2(0.1) + 2^2(0.3) + 3^2(0.4) + 4^2(0.2) = 0.1 + 1.2 + 3.6 + 3.2 = 8.1$ .
$Var(X) = E(X^2) - [E(X)]^2 = 8.1 - (2.7)^2 = 8.1 - 7.29 = 0.81$ .
[3]

19. (a) Tree Diagram:
First branch: Type A (0.6), Type B (0.4).
Second branch from A: Defective (0.02), Not Def (0.98).
Second branch from B: Defective (0.05), Not Def (0.95).
[2]

(b) $P(D) = P(A \cap D) + P(B \cap D) = (0.6)(0.02) + (0.4)(0.05) = 0.012 + 0.020 = 0.032$ .
[2]

20. $E \sim N(150, 30^2)$ .

(a) $P(E > 200)$ .
$Z = \frac{200 - 150}{30} = \frac{50}{30} = 1.67$ .
$P(Z > 1.67) = 1 - 0.9525 = 0.0475$ .
[2]

(b) $P(E < k) = 0.25$ .
Using GC: invNorm(0.25, 150, 30) $\approx 129.77$ .
$k \approx \$ 129.77$.
[2]

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