A Level H1 Mathematics Practice Paper 1

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TuitionGoWhere Practice Paper - Maths H1 A-Level

TuitionGoWhere Practice Paper (AI)
Version: 1 of 5
Subject: Mathematics H1 (8865)
Level: A-Level
Paper: Practice Paper - Statistics & Probability Focus
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. Write your answers in the spaces provided in this booklet.
4. You are expected to use an approved graphing calculator (GC).
5. Unless otherwise specified, give non-exact numerical answers correct to 3 significant figures.
6. The total mark for this paper is 60.

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Section A: Probability and Counting Principles (15 Marks)

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

(ii) Find the number of different ways the committee can be chosen if it must contain at least 2 women.
[2]

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

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

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

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3. A bag contains 4 red balls, 3 blue balls, and 2 green balls. Two balls are drawn at random from the bag without replacement.
(i) Draw a tree diagram to represent the possible outcomes and their probabilities for the first two draws.
[2]

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

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Section B: Discrete and Continuous Distributions (25 Marks)

4. The random variable $X$ follows a binomial distribution $B(15, 0.3)$.
(i) Find $P(X = 4)$.
[1]

(ii) Find $P(X \ge 2)$.
[2]

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

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5. The heights of adult males in a certain population are normally distributed with mean $175$ cm and standard deviation $8$ cm. A man is chosen at random from this population.
(i) Find the probability that his height is between $170$ cm and $185$ cm.
[2]

(ii) Find the height $h$ such that $10\%$ of the population is taller than $h$.
[2]

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6. The random variable $Y$ is normally distributed with mean $\mu$ and variance $25$. It is given that $P(Y < 40) = 0.9332$.
(i) Find the value of $\mu$.
[3]

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7. Let $X_1$ and $X_2$ be two independent random variables such that $X_1 \sim N(10, 4)$ and $X_2 \sim N(15, 9)$.
(i) State the distribution of $W = X_1 + X_2$.
[2]

(ii) Find $P(W > 28)$.
[3]

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8. A manufacturer produces light bulbs. The lifetime of a bulb, $L$ hours, is normally distributed with mean $1200$ and standard deviation $100$.
(i) Find the probability that a randomly selected bulb lasts more than $1350$ hours.
[2]

(ii) The manufacturer wants to offer a warranty such that only $5\%$ of bulbs fail within the warranty period. Find the maximum warranty period (in hours) they should offer.
[3]

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Section C: Sampling, Estimation, and Hypothesis Testing (20 Marks)

9. A random sample of 50 students was taken to estimate the mean time spent on homework per week. The summary statistics are:
$\sum t = 650, \quad \sum t^2 = 9200$
(i) Calculate the unbiased estimate of the population mean.
[1]

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

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10. The masses of bags of rice produced by a machine are normally distributed with a known standard deviation of $0.5$ kg. The machine is set to produce bags with a mean mass of $5.0$ kg.
A random sample of 40 bags is taken, and the mean mass is found to be $4.85$ kg.
(i) State the null and alternative hypotheses to test if the mean mass has decreased.
[2]

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

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11. A company claims that the mean battery life of their new smartphone model is $20$ hours. A consumer group suspects the mean life is actually less than $20$ hours.
They take a large sample of $100$ phones and find the sample mean is $19.5$ hours with a sample standard deviation of $2.5$ hours.
(i) Explain why the Central Limit Theorem is applicable in this context.
[1]

(ii) Test the company's claim at the $1\%$ significance level.
[4]

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12. In a survey of 200 residents, 120 stated they support a new local park project.
(i) Find the sample proportion $\hat{p}$ of residents who support the project.
[1]

(ii) Construct a $95\%$ confidence interval for the true population proportion of supporters.
[3]
(Note: Use the formula $\hat{p} \pm z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ where $z=1.96$ for 95% confidence)

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End of Paper

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

Version: 1 of 5
Subject: Mathematics H1 (8865)

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Section A: Probability and Counting Principles

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

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

(ii) 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, independent because $P(A \cap B) = P(A)P(B)$. [2]

(iii) $P(A | B') = \frac{P(A \cap B')}{P(B')}$
$P(B') = 1 - 0.5 = 0.5$
$P(A \cap B') = P(A) - P(A \cap B) = 0.4 - 0.2 = 0.2$
$P(A | B') = \frac{0.2}{0.5} = 0.4$
Answer: 0.4 [2]

3.
(i) Tree Diagram:

• 1st Draw: R (4/9), B (3/9), G (2/9)
• 2nd Draw (if R first): R (3/8), B (3/8), G (2/8)
• 2nd Draw (if B first): R (4/8), B (2/8), G (2/8)
• 2nd Draw (if G first): R (4/8), B (3/8), G (1/8)
  (Diagram should show these branches clearly) [2]

(ii) P(Different Colours) = 1 - P(Same Colour)
P(RR) = $\frac{4}{9} \times \frac{3}{8} = \frac{12}{72}$
P(BB) = $\frac{3}{9} \times \frac{2}{8} = \frac{6}{72}$
P(GG) = $\frac{2}{9} \times \frac{1}{8} = \frac{2}{72}$
P(Same) = $\frac{12+6+2}{72} = \frac{20}{72}$
P(Different) = $1 - \frac{20}{72} = \frac{52}{72} = \frac{13}{18} \approx 0.722$
Answer: 13/18 or 0.722 [2]

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Section B: Discrete and Continuous Distributions

4. $X \sim B(15, 0.3)$
(i) $P(X=4) = \binom{15}{4}(0.3)^4(0.7)^{11} \approx 0.2186$
Answer: 0.219 [1]

(ii) $P(X \ge 2) = 1 - P(X \le 1) = 1 - [P(X=0) + P(X=1)]$
$P(X=0) = (0.7)^{15} \approx 0.0047$
$P(X=1) = 15(0.3)(0.7)^{14} \approx 0.0305$
$P(X \ge 2) = 1 - (0.0047 + 0.0305) = 1 - 0.0352 = 0.9648$
Answer: 0.965 [2]

(iii) Mean $= np = 15 \times 0.3 = 4.5$
Variance $= np(1-p) = 15 \times 0.3 \times 0.7 = 3.15$
Answer: Mean 4.5, Variance 3.15 [2]

5. $H \sim N(175, 8^2)$
(i) $P(170 < H < 185)$
Using GC: normalcdf(170, 185, 175, 8) $\approx 0.6284$
Answer: 0.628 [2]

(ii) $P(H > h) = 0.10 \Rightarrow P(H < h) = 0.90$
Using GC: invNorm(0.90, 175, 8) $\approx 185.24$
Answer: 185 cm [2]

6. $Y \sim N(\mu, 25) \Rightarrow \sigma = 5$.
$P(Y < 40) = 0.9332$.
Standardizing: $P(Z < \frac{40-\mu}{5}) = 0.9332$.
From tables/GC, $z$-score for 0.9332 is approx $1.5$.
$\frac{40-\mu}{5} = 1.5 \Rightarrow 40 - \mu = 7.5 \Rightarrow \mu = 32.5$.
Answer: 32.5 [3]

7. $X_1 \sim N(10, 4)$, $X_2 \sim N(15, 9)$. Independent.
(i) $W = X_1 + X_2$.
$E(W) = 10 + 15 = 25$.
$Var(W) = 4 + 9 = 13$.
$W \sim N(25, 13)$.
Answer: $N(25, 13)$ [2]

(ii) $P(W > 28)$.
Using GC: normalcdf(28, 1E99, 25, \sqrt{13})
$Z = \frac{28-25}{\sqrt{13}} \approx 0.832$.
$P(Z > 0.832) \approx 0.2026$.
Answer: 0.203 [3]

8. $L \sim N(1200, 100^2)$.
(i) $P(L > 1350)$.
Using GC: normalcdf(1350, 1E99, 1200, 100) $\approx 0.0668$.
Answer: 0.0668 [2]

(ii) Find $w$ such that $P(L < w) = 0.05$.
Using GC: invNorm(0.05, 1200, 100) $\approx 1035.5$.
Answer: 1036 hours (or 1035.5) [3]

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Section C: Sampling, Estimation, and Hypothesis Testing

9. $n=50, \sum t = 650, \sum t^2 = 9200$.
(i) Unbiased estimate of mean $\bar{t} = \frac{650}{50} = 13$.
Answer: 13 [1]

(ii) Unbiased estimate of variance $s^2 = \frac{1}{n-1} \left( \sum t^2 - \frac{(\sum t)^2}{n} \right)$.
$s^2 = \frac{1}{49} \left( 9200 - \frac{650^2}{50} \right) = \frac{1}{49} (9200 - 8450) = \frac{750}{49} \approx 15.31$.
Answer: 15.3 [3]

10. $\sigma = 0.5$ (known), $n=40$, $\bar{x} = 4.85$.
(i) $H_0: \mu = 5.0$, $H_1: \mu < 5.0$. [2]

(ii) Test Statistic $Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} = \frac{4.85 - 5.0}{0.5/\sqrt{40}} = \frac{-0.15}{0.07906} \approx -1.897$.
P-value $= P(Z < -1.897) \approx 0.0289$.
Since $0.0289 < 0.05$, we reject $H_0$.
Conclusion: There is sufficient evidence at the 5% level to suggest the mean mass has decreased. [4]

11. $n=100$ (large), $\bar{x} = 19.5$, $s = 2.5$.
(i) CLT applies because the sample size $n=100$ is large ($>30$), so the sampling distribution of the mean is approximately normal regardless of the population distribution. [1]

(ii) $H_0: \mu = 20$, $H_1: \mu < 20$.
Test Statistic $Z = \frac{19.5 - 20}{2.5/\sqrt{100}} = \frac{-0.5}{0.25} = -2.0$.
P-value $= P(Z < -2.0) \approx 0.0228$.
Significance level $1\% = 0.01$.
Since $0.0228 > 0.01$, we do not reject $H_0$.
Conclusion: There is insufficient evidence at the 1% level to reject the company's claim. [4]

12. $n=200, x=120$.
(i) $\hat{p} = \frac{120}{200} = 0.6$. [1]

(ii) 95% CI: $\hat{p} \pm 1.96 \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$.
Standard Error $= \sqrt{\frac{0.6 \times 0.4}{200}} = \sqrt{0.0012} \approx 0.03464$.
Margin of Error $= 1.96 \times 0.03464 \approx 0.0679$.
CI: $0.6 \pm 0.0679 \Rightarrow (0.532, 0.668)$.
Answer: $(0.532, 0.668)$ [3]