A Level H1 Mathematics Practice Paper 5

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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 5 of 5)
Duration: 1 hour 30 minutes
Total Marks: 60

Name: __________________________
Class: __________________________
Date: __________________________

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Instructions to Candidates

1. Write your name, class, and date in the spaces above.
2. Answer all questions.
3. Use an approved graphing calculator (GC). Unsupported answers obtained from a GC are allowed unless the question specifically states otherwise.
4. 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.
5. The use of an approved graphing calculator is expected.
6. Marks are indicated in brackets [ ] at the end of each question or part question.

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Section A: Probability and Distributions [20 Marks]

1. A company manufactures smartphone batteries. The probability that a battery is defective is 0.05. A quality control inspector selects a random sample of 20 batteries.

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

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

(c) Find the probability that more than 1 battery in the sample is defective. [2]

2. In a certain university, 60% of the students are female. A random sample of 8 students is selected.

(a) Draw a fully labelled probability tree diagram to represent the gender of the first two students selected, assuming independence. [2]

(b) Find the probability that at least 5 of the 8 students are female. [3]

3. The time taken for a delivery driver to complete a route, $T$ minutes, is normally distributed with mean 45 and standard deviation 8.

(a) Find the probability that a randomly selected delivery takes less than 40 minutes. [2]

(b) Find the value of $k$ such that $P(T > k) = 0.15$. [3]

(c) The driver completes 5 independent routes in a day. Let $S$ be the total time taken for these 5 routes. Find $E(S)$ and $Var(S)$. [2]

4. Events $A$ and $B$ are such that $P(A) = 0.4$, $P(B) = 0.5$, and $P(A \cap B) = 0.2$.

(a) Determine whether events $A$ and $B$ are independent, showing your working. [2]

(b) Find $P(A \cup B')$. [3]

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Section B: Sampling and Estimation [20 Marks]

5. A researcher wishes to estimate the mean height of adult males in Singapore. He stands outside a shopping mall and measures the height of the first 50 men who walk past him.

(a) Explain why this sampling method may produce a biased estimate of the population mean. [1]

(b) Describe how the researcher could obtain a simple random sample of 50 men from a registry of 10,000 adult males. [2]

6. A random sample of 12 toddlers was taken to study the age at which they start walking. The ages (in months) are summarized as follows: $\sum x = 144, \quad \sum x^2 = 1750$

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

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

7. The masses of bags of rice produced by a factory are normally distributed with mean $\mu$ kg and standard deviation 0.5 kg. A random sample of 25 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]

(c) If the sample size was increased to 100, explain qualitatively how the probability in part (b) would change. [1]

8. A sample of 80 observations from a population with unknown distribution has a sample mean of 52.4 and a sample variance of 16.9.

(a) Explain why the Central Limit Theorem is applicable in this context. [1]

(b) Find a 95% confidence interval for the population mean. [4]

(Note: For 95% confidence, use $z = 1.96$)

9. The daily revenue of a café, $R$, is normally distributed with mean \800$and standard deviation$$120$.

(a) Find the probability that the mean daily revenue over a period of 9 days is less than \750$. [3]

(b) Find the probability that the total revenue over 9 days exceeds \7500$. [3]

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Section C: Hypothesis Testing and Regression [20 Marks]

10. 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 mean lifetime 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 context. [5]

11. The table below shows the advertising expenditure ($x$, in $000s) and monthly sales ($y$, in $000s) for a company over 6 months.

Month	$x$	$y$
1	2.0	15.0
2	3.5	22.0
3	4.0	24.5
4	5.0	28.0
5	6.5	35.0
6	8.0	40.0

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

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

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

(d) Estimate the monthly sales if the advertising expenditure is $7,000. Comment on the reliability of this estimate. [3]

(e) Explain why it might be inappropriate to use this regression line to estimate sales if the advertising expenditure is $20,000. [1]

12. A two-tail hypothesis test is conducted on the population mean $\mu$ of a normal distribution with known variance. The null hypothesis is $H_0: \mu = 50$. The critical region is defined as $\bar{X} < 48.5$ or $\bar{X} > 51.5$ for a sample size of $n=36$ and population standard deviation $\sigma = 6$.

(a) Find the level of significance of this test. [4]

(b) If the true population mean is actually 52, find the probability of making a Type II error. [Note: This topic is excluded in H1, so replace with standard interpretation] Correction for H1 Syllabus Compliance: (b) Explain what is meant by a "Type I error" in the context of this specific test. [2]

13. The weights of apples from Orchard A are normally distributed with mean 150g and variance 25g$^2$. The weights of apples from Orchard B are normally distributed with mean 145g and variance 36g$^2$.

One apple is selected at random from Orchard A and one from Orchard B. Let $W = W_A - W_B$.

(a) Find $E(W)$ and $Var(W)$. [2]

(b) Find the probability that the apple from Orchard A is heavier than the apple from Orchard B. [3]

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End of Paper

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TuitionGoWhere Practice Paper - Maths H1 A-Level

Answer Key and Marking Scheme Version 5 of 5

Section A: Probability and Distributions

1. (a) Let $X$ be the number of defective batteries in the sample. $X \sim B(20, 0.05)$ [1]

(b) $P(X=2) = \binom{20}{2} (0.05)^2 (0.95)^{18}$ $= 190 \times 0.0025 \times 0.3972...$ $= 0.1887...$ $\approx 0.189$ [2]

(c) $P(X > 1) = 1 - P(X \le 1)$ $= 1 - [P(X=0) + P(X=1)]$ $P(X=0) = (0.95)^{20} \approx 0.3585$ $P(X=1) = 20(0.05)(0.95)^{19} \approx 0.3774$ $P(X > 1) = 1 - (0.3585 + 0.3774) = 1 - 0.7359 = 0.2641$ $\approx 0.264$ [2]

2. (a) Tree Diagram:

• Stage 1: Branches F (0.6), M (0.4)
• Stage 2 (from F): Branches F (0.6), M (0.4)
• Stage 2 (from M): Branches F (0.6), M (0.4)
• Labels must include probabilities. [2]

(b) Let $Y$ be the number of females. $Y \sim B(8, 0.6)$. $P(Y \ge 5) = P(Y=5) + P(Y=6) + P(Y=7) + P(Y=8)$ Using GC or formula: $P(Y \ge 5) = 1 - P(Y \le 4)$ $P(Y \le 4) \approx 0.4059$ $P(Y \ge 5) = 1 - 0.4059 = 0.5941$ $\approx 0.594$ [3]

3. (a) $T \sim N(45, 8^2)$. $P(T < 40) = P(Z < \frac{40-45}{8}) = P(Z < -0.625)$ Using GC: $\approx 0.266$ [2]

(b) $P(T > k) = 0.15 \Rightarrow P(T < k) = 0.85$ Using inverse normal: $k = \text{invNorm}(0.85, 45, 8)$ $k \approx 53.29$ $k = 53.3$ (3 s.f.) [3]

(c) $S = T_1 + T_2 + T_3 + T_4 + T_5$ $E(S) = 5 \times E(T) = 5 \times 45 = 225$ [1] Since independent: $Var(S) = 5 \times Var(T) = 5 \times 8^2 = 5 \times 64 = 320$ [1]

4. (a) Check if $P(A \cap B) = P(A)P(B)$. $P(A)P(B) = 0.4 \times 0.5 = 0.2$ Given $P(A \cap B) = 0.2$. Since $0.2 = 0.2$, events A and B are independent. [2]

(b) $P(A \cup B') = P(A) + P(B') - P(A \cap B')$ Alternatively: $P(A \cup B') = 1 - P(A' \cap B)$ $P(A' \cap B) = P(B) - P(A \cap B) = 0.5 - 0.2 = 0.3$ $P(A \cup B') = 1 - 0.3 = 0.7$ OR $P(B') = 1 - 0.5 = 0.5$ Since independent, $P(A \cap B') = P(A)P(B') = 0.4 \times 0.5 = 0.2$ $P(A \cup B') = 0.4 + 0.5 - 0.2 = 0.7$ [3]

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Section B: Sampling and Estimation

5. (a) The sample is a convenience sample, not random. It only includes men visiting that specific mall at that specific time, which may not represent the general population (e.g., excludes those who work during the day, live elsewhere, or do not visit malls). [1]

(b) Assign each of the 10,000 men a unique number from 1 to 10,000. Use a random number generator to select 50 distinct numbers. Select the men corresponding to these numbers. [2]

6. (a) Unbiased estimate of mean $\bar{x} = \frac{\sum x}{n} = \frac{144}{12} = 12$ months. [1]

(b) Unbiased estimate of variance $s^2 = \frac{1}{n-1} \left( \sum x^2 - \frac{(\sum x)^2}{n} \right)$ $s^2 = \frac{1}{11} \left( 1750 - \frac{144^2}{12} \right)$ $s^2 = \frac{1}{11} (1750 - 1728)$ $s^2 = \frac{22}{11} = 2$ [3]

7. (a) Population $X \sim N(\mu, 0.5^2)$. Sample size $n=25$. $\bar{X} \sim N(\mu, \frac{0.5^2}{25}) = N(\mu, 0.01)$ [2]

(b) We want $P(\mu - 0.1 < \bar{X} < \mu + 0.1)$. Standardize: $Z = \frac{\bar{X} - \mu}{\sqrt{0.01}} = \frac{\bar{X} - \mu}{0.1}$ Limits: $\frac{-0.1}{0.1} = -1$ and $\frac{0.1}{0.1} = 1$. $P(-1 < Z < 1) = P(Z < 1) - P(Z < -1)$ Using GC: $\approx 0.8413 - 0.1587 = 0.6826$ $\approx 0.683$ [3]

(c) As $n$ increases, the variance of $\bar{X}$ decreases ($\frac{\sigma^2}{n}$). The distribution becomes narrower/tighter around $\mu$. Therefore, the probability that $\bar{X}$ is within a fixed distance of $\mu$ increases. [1]

8. (a) The sample size $n=80$ is large ($>30$). By the Central Limit Theorem, the sampling distribution of the sample mean is approximately normal, regardless of the population distribution. [1]

(b) $\bar{x} = 52.4$, $s^2 = 16.9 \Rightarrow s = \sqrt{16.9} \approx 4.11$. Standard Error $SE = \frac{s}{\sqrt{n}} = \frac{4.11}{\sqrt{80}} \approx 0.4595$. 95% CI: $\bar{x} \pm z \times SE$ $52.4 \pm 1.96(0.4595)$ $52.4 \pm 0.9006$ $(51.499, 53.301)$ $\approx (51.5, 53.3)$ [4]

9. (a) Let $\bar{R}$ be the mean revenue over 9 days. $\bar{R} \sim N(800, \frac{120^2}{9}) = N(800, 1600)$. SD = 40. $P(\bar{R} < 750) = P(Z < \frac{750-800}{40}) = P(Z < -1.25)$ Using GC: $\approx 0.1056$ $\approx 0.106$ [3]

(b) Let $S_9$ be total revenue. $S_9 = 9\bar{R}$. $E(S_9) = 9 \times 800 = 7200$. $Var(S_9) = 9^2 \times Var(\bar{R}) = 81 \times 1600 = 129600$. $SD(S_9) = \sqrt{129600} = 360$. Alternatively: $Var(S_9) = 9 \times Var(R) = 9 \times 120^2 = 129600$. $P(S_9 > 7500) = P(Z > \frac{7500-7200}{360}) = P(Z > \frac{300}{360}) = P(Z > 0.8333)$ Using GC: $\approx 0.2023$ $\approx 0.202$ [3]

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Section C: Hypothesis Testing and Regression

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

(b) Test statistic $Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$ $Z = \frac{1180 - 1200}{100/\sqrt{50}} = \frac{-20}{14.142} \approx -1.414$ Critical value for 1-tail 5% test: $z_{crit} = -1.645$. Since $-1.414 > -1.645$, the test statistic is not in the critical region. Alternatively, p-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% level to support the claim that the mean lifetime is less than 1200 hours. [5]

11. (a) Using GC: $r \approx 0.996$ (3 d.p.) [2]

(b) There is a very strong positive linear correlation between advertising expenditure and monthly sales. [1]

(c) Using GC for regression line $y$ on $x$: $m \approx 4.318$ $c \approx 5.818$ Equation: $y = 4.32x + 5.82$ (2 d.p.) [3]

(d) $x = 7$ (since x is in $000s). $y = 4.318(7) + 5.818 \approx 36.04$ Estimated sales: $36,040. Reliability: Reliable because $x=7$ is within the range of the data (interpolation) and $r$ is close to 1. [3]

(e) $x=20$ is well outside the range of the observed data (extrapolation). The linear relationship may not hold for such high expenditure. [1]

12. (a) Under $H_0$, $\bar{X} \sim N(50, \frac{6^2}{36}) = N(50, 1)$. Critical region: $\bar{X} < 48.5$ or $\bar{X} > 51.5$. $P(\text{Type I}) = P(\bar{X} < 48.5) + P(\bar{X} > 51.5)$ By symmetry: $2 \times P(\bar{X} > 51.5)$ $Z = \frac{51.5 - 50}{1} = 1.5$ $P(Z > 1.5) = 1 - P(Z < 1.5) \approx 1 - 0.9332 = 0.0668$ Level of significance $= 2 \times 0.0668 = 0.1336$ $\approx 13.4\%$ [4]

(b) A Type I error occurs if we reject the null hypothesis $H_0: \mu = 50$ when the true population mean is actually 50. In this context, it means concluding the mean is different from 50 when it is actually 50. [2]

13. (a) $W = W_A - W_B$ $E(W) = E(W_A) - E(W_B) = 150 - 145 = 5$ g [1] Since independent: $Var(W) = Var(W_A) + Var(W_B) = 25 + 36 = 61$ g$^2$ [1]

(b) We want $P(W_A > W_B) \Rightarrow P(W_A - W_B > 0) \Rightarrow P(W > 0)$. $W \sim N(5, 61)$. SD $= \sqrt{61} \approx 7.81$. $P(W > 0) = P(Z > \frac{0 - 5}{7.81}) = P(Z > -0.640)$ $P(Z > -0.640) = P(Z < 0.640)$ Using GC: $\approx 0.739$ [3]