A Level H1 Mathematics Practice Paper 4

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TuitionGoWhere Exam Practice (AI)

A-Level H1 Mathematics (8865) - Practice Paper

Topic: Statistics & Probability (Version 4 of 5)

Subject: Mathematics H1
Level: A-Level
Paper: Practice Paper (Topic 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. You are expected to use an approved graphing calculator (GC).
4. Unsupported answers from a GC are allowed unless the question specifically states otherwise.
5. 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.
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 5 members is to be formed from a group of 8 men and 6 women. (a) Find the number of different committees that can be formed if there are no restrictions. [1]

(b) Find the number of different committees that can be formed if the committee must contain exactly 3 men and 2 women. [2]

(c) Find the number of different committees that can be formed if the committee must contain at least 4 women. [2]

2. In a certain factory, 10% of the items produced are defective. A quality control inspector selects a random sample of 15 items. (a) State the distribution of the number of defective items in the sample, defining any variables used. [1]

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

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

3. Events $A$ and $B$ are defined such that $P(A) = 0.4$, $P(B) = 0.5$, and $P(A \cap B) = 0.2$. (a) Determine, with a reason, whether events $A$ and $B$ are independent. [2]

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

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

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Section B: Descriptive Statistics and Estimation (15 Marks)

4. A researcher records the time taken (in minutes) for 10 students to complete a puzzle. The data is summarized as follows: $\sum t = 120, \quad \sum t^2 = 1550$ (a) Calculate the mean time. [1]

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

5. The heights of a sample of 80 male students in a college are recorded. The data is coded using $h = \frac{x - 170}{5}$, where $x$ is the height in cm. The summarized coded data is: $\sum h = 40, \quad \sum h^2 = 120$ (a) Calculate the unbiased estimate of the population mean height. [3]

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

6. A simple random sample of size $n$ is taken from a large population with mean $\mu$ and variance $\sigma^2$. (a) State the expected value and variance of the sample mean $\bar{X}$. [2]

(b) Explain why the sample variance $S^2 = \frac{1}{n-1}\sum(X_i - \bar{X})^2$ is preferred over $\frac{1}{n}\sum(X_i - \bar{X})^2$ as an estimator for the population variance. [1]

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Section C: Normal Distribution and Sampling (15 Marks)

7. The masses of bags of rice produced by a machine are normally distributed with mean 5.0 kg and standard deviation 0.1 kg. (a) Find the probability that a randomly selected bag has a mass less than 4.9 kg. [2]

(b) Find the mass $m$ such that 95% of the bags have a mass greater than $m$. [3]

8. The weekly expenditure on groceries for families in a certain town is normally distributed with mean \150$and standard deviation$$40$. (a) A random sample of 16 families is selected. Find the probability that the mean expenditure of these 16 families is less than$$140$. [3]

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

(c) If the sample size was increased to 50 families, how would the standard error of the mean change? Justify your answer. [2]

9. Let $X$ and $Y$ be independent random variables such that $X \sim N(10, 4)$ and $Y \sim N(5, 9)$. (a) Find $E(2X - Y)$. [1]

(b) Find $Var(2X - Y)$. [2]

(c) State the distribution of $2X - Y$. [1]

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Section D: Hypothesis Testing and Regression (15 Marks)

10. A manufacturer claims that the mean lifetime of their light bulbs is 1000 hours. A consumer group suspects the mean lifetime is less than 1000 hours. They test a random sample of 50 bulbs and find a sample mean of 980 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]

11. The table below shows the age ($x$ years) and the selling price ($y$ hundred dollars) of 6 used cars.

Age ($x$)	2	3	5	7	8	10
Price ($y$)	15	14	11	9	8	5

(a) Draw a scatter diagram for this data. [2]

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

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

(d) Interpret the value of the gradient $b$ in the context of the question. [1]

(e) Explain why it might be inappropriate to use this regression line to estimate the price of a car that is 20 years old. [1]

[End of Paper]

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TuitionGoWhere Exam Practice (AI) - Answer Key

A-Level H1 Mathematics (8865) - Practice Paper (Version 4)

Topic: Statistics & Probability

Total Marks: 60

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Section A: Probability and Counting Principles

1. (a) Total people = $8 + 6 = 14$. Select 5. $^{14}C_5 = \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} = 2002$ [1]

(b) Select 3 men from 8 and 2 women from 6. $^8C_3 \times ^6C_2 = 56 \times 15 = 840$ [2]

(c) At least 4 women means 4 women and 1 man, OR 5 women and 0 men. Case 1 (4W, 1M): $^6C_4 \times ^8C_1 = 15 \times 8 = 120$ Case 2 (5W, 0M): $^6C_5 \times ^8C_0 = 6 \times 1 = 6$ Total = $120 + 6 = 126$ [2]

2. (a) Let $X$ be the number of defective items in the sample. $X \sim B(15, 0.1)$ [1]

(b) $P(X = 2) = ^{15}C_2 (0.1)^2 (0.9)^{13}$ $= 105 \times 0.01 \times 0.25418... \approx 0.2669$ Answer: 0.267 (3 s.f.) [2]

(c) $P(X > 1) = 1 - P(X \le 1) = 1 - [P(X=0) + P(X=1)]$ $P(X=0) = (0.9)^{15} \approx 0.2059$ $P(X=1) = 15(0.1)(0.9)^{14} \approx 0.3432$ $P(X > 1) = 1 - (0.2059 + 0.3432) = 1 - 0.5491 = 0.4509$ Answer: 0.451 (3 s.f.) [2]

3. (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 $P(A \cap B) = P(A)P(B)$, events $A$ and $B$ are independent. [2] (1 for calculation, 1 for conclusion with reason)

(b) $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ $= 0.4 + 0.5 - 0.2 = 0.7$ [1]

(c) $P(A | B') = \frac{P(A \cap B')}{P(B')}$ $P(B') = 1 - P(B) = 1 - 0.5 = 0.5$ $P(A \cap B') = P(A) - P(A \cap B) = 0.4 - 0.2 = 0.2$ (Since A and B are independent, $A$ and $B'$ are also independent, so $0.4 \times 0.5 = 0.2$) $P(A | B') = \frac{0.2}{0.5} = 0.4$ [2]

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Section B: Descriptive Statistics and Estimation

4. (a) Mean $\bar{t} = \frac{\sum t}{n} = \frac{120}{10} = 12$ minutes. [1]

(b) Unbiased estimate of variance $s^2 = \frac{1}{n-1} \left[ \sum t^2 - \frac{(\sum t)^2}{n} \right]$ $s^2 = \frac{1}{9} \left[ 1550 - \frac{120^2}{10} \right]$ $s^2 = \frac{1}{9} [ 1550 - 1440 ] = \frac{110}{9} \approx 12.22$ Answer: 12.2 (3 s.f.) [3] (1 for formula/setup, 1 for substitution, 1 for answer)

5. (a) Mean of coded data $\bar{h} = \frac{\sum h}{n} = \frac{40}{80} = 0.5$. Since $h = \frac{x - 170}{5} \implies x = 5h + 170$. $\bar{x} = 5\bar{h} + 170 = 5(0.5) + 170 = 2.5 + 170 = 172.5$ cm. [3]

(b) Variance of coded data $s_h^2 = \frac{1}{n-1} \left[ \sum h^2 - \frac{(\sum h)^2}{n} \right]$ $s_h^2 = \frac{1}{79} \left[ 120 - \frac{40^2}{80} \right] = \frac{1}{79} [ 120 - 20 ] = \frac{100}{79}$ Variance is invariant under change of origin but scales by square of multiplier for change of scale. $s_x^2 = 5^2 \times s_h^2 = 25 \times \frac{100}{79} = \frac{2500}{79} \approx 31.645$ Answer: 31.6 (3 s.f.) [3]

6. (a) $E(\bar{X}) = \mu$ $Var(\bar{X}) = \frac{\sigma^2}{n}$ [2]

(b) Dividing by $n-1$ makes the estimator unbiased. Dividing by $n$ tends to underestimate the population variance. [1]

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Section C: Normal Distribution and Sampling

7. Let $M$ be the mass of a bag. $M \sim N(5.0, 0.1^2)$. (a) $P(M < 4.9)$. Using GC: normalcdf(-E99, 4.9, 5.0, 0.1) $P(M < 4.9) \approx 0.1587$ Answer: 0.159 (3 s.f.) [2]

(b) We want $P(M > m) = 0.95$, which implies $P(M < m) = 0.05$. Using GC: invNorm(0.05, 5.0, 0.1) $m \approx 4.8355$ Answer: 4.84 kg (3 s.f.) [3]

8. Let $X$ be expenditure. $X \sim N(150, 40^2)$. Sample size $n=16$. Sample mean $\bar{X} \sim N(150, \frac{40^2}{16}) = N(150, 100)$. Standard deviation of $\bar{X}$ is $\sqrt{100} = 10$. (a) $P(\bar{X} < 140)$. Using GC: normalcdf(-E99, 140, 150, 10) $P(\bar{X} < 140) \approx 0.1587$ Answer: 0.159 (3 s.f.) [3]

(b) The Central Limit Theorem is not required because the population distribution is already stated to be normal. Therefore, the sampling distribution of the mean is normal for any sample size. [1]

(c) Standard error $SE = \frac{\sigma}{\sqrt{n}}$. If $n$ increases from 16 to 50, the denominator $\sqrt{n}$ increases. Therefore, the standard error decreases. Specifically, it changes from $\frac{40}{\sqrt{16}}=10$ to $\frac{40}{\sqrt{50}} \approx 5.66$. [2]

9. $X \sim N(10, 4) \implies E(X)=10, Var(X)=4$. $Y \sim N(5, 9) \implies E(Y)=5, Var(Y)=9$. (a) $E(2X - Y) = 2E(X) - E(Y) = 2(10) - 5 = 15$. [1]

(b) Since $X$ and $Y$ are independent: $Var(2X - Y) = 2^2 Var(X) + (-1)^2 Var(Y) = 4(4) + 1(9) = 16 + 9 = 25$. [2]

(c) Linear combination of independent normal variables is normal. $2X - Y \sim N(15, 25)$. [1]

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Section D: Hypothesis Testing and Regression

10. (a) $H_0: \mu = 1000$ $H_1: \mu < 1000$ [2]

(b) Test statistic $Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$ $Z = \frac{980 - 1000}{100/\sqrt{50}} = \frac{-20}{14.142} \approx -1.414$ Critical value for one-tail test at 5%: $z_{0.05} = -1.645$. Since $-1.414 > -1.645$ (or $P(Z < -1.414) \approx 0.0787 > 0.05$), 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 1000 hours. [4] (1 for Z calc, 1 for critical value/p-value, 1 for comparison, 1 for conclusion in context)

11. (a) Scatter diagram:

• Axes labeled "Age (years)" and "Price ($100s)".
• Points plotted correctly: (2,15), (3,14), (5,11), (7,9), (8,8), (10,5).
• Reasonable scale. [2]

(b) Using GC: $r \approx -0.986$ Answer: -0.986 (3 s.f.) [2]

(c) Using GC for regression $y$ on $x$: $a \approx 17.57$, $b \approx -1.29$ Equation: $y = 17.6 - 1.29x$ (coefficients to 3 s.f.) [3]

(d) For every additional year of age, the selling price of the car decreases by approximately \129$(since$y$ is in hundreds). [1]

(e) Age 20 is outside the range of the observed data (2 to 10 years). This is extrapolation, and the linear relationship may not hold for older cars (e.g., price cannot go below zero, or classic car value might increase). [1]