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A Level H1 Mathematics Practice Paper 4
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TuitionGoWhere Practice Paper - Maths H1 A-Level
TuitionGoWhere Practice Paper (AI)
Subject: Mathematics H1 (8865) Level: A-Level Paper: Practice Paper 1 (Version 4) Duration: 3 hours Total Marks: 100
Name: _________________________ Class: _________________________ Date: _________________________
Instructions to Candidates
- This paper consists of two sections. Answer all questions.
- Section A (Pure Mathematics) carries 40 marks. Section B (Probability and Statistics) carries 60 marks.
- You are expected to use an approved graphing calculator (without CAS) where appropriate.
- Unless otherwise stated, give numerical answers to 3 significant figures.
- Show all necessary working. Marks will be awarded for method, not just final answers.
- The number of marks is given in brackets [ ] at the end of each question or part question.
- A formula sheet is provided separately.
Section A: Pure Mathematics (40 marks)
Answer all questions in this section.
1. Solve the inequality ( 3x^2 - 7x - 6 > 0 ).
[3 marks]
2. The curve ( C ) has equation ( y = x^3 - 6x^2 + 9x + 4 ).
(a) Find the coordinates of the stationary points of ( C ). [3 marks]
(b) Determine the nature of each stationary point. [2 marks]
3. Solve the simultaneous equations: [ \begin{aligned} y &= 3x - 1 \ y &= x^2 + x - 2 \end{aligned} ]
[4 marks]
4. The population of a colony of bacteria, ( P ) thousand, ( t ) hours after the start of an experiment, is modelled by the equation [ P = 20 e^{0.15t}. ]
(a) Find the population at the start of the experiment. [1 mark]
(b) Find the time taken for the population to double. [2 marks]
(c) Find the rate at which the population is increasing when ( t = 5 ). [2 marks]
5. A curve has equation ( y = \dfrac{4x + 1}{2x - 3} ), for ( x \neq \frac{3}{2} ).
(a) Find ( \dfrac{dy}{dx} ). Simplify your answer. [3 marks]
(b) Find the equation of the tangent to the curve at the point where ( x = 2 ). Give your answer in the form ( y = mx + c ). [3 marks]
6. The diagram shows part of the curve ( y = e^{2x} + 1 ) and the line ( y = 5 ).
The curve and the line intersect at point ( A ).
(a) Find the ( x )-coordinate of ( A ), giving your answer in exact form. [2 marks]
(b) Find the area of the region bounded by the curve, the line ( y = 5 ), the ( y )-axis, and the line through ( A ) parallel to the ( y )-axis. [5 marks]
7. A rectangular enclosure is to be built against an existing wall. The enclosure has width ( x ) metres and length ( y ) metres, as shown in the diagram. The total length of fencing available for the three sides is 60 metres.
(a) Show that the area ( A ) of the enclosure is given by ( A = 60x - 2x^2 ). [2 marks]
(b) Find the value of ( x ) that gives the maximum area. [3 marks]
(c) Verify that this value of ( x ) gives a maximum area. [1 mark]
(d) Find the maximum area. [1 mark]
Section B: Probability and Statistics (60 marks)
Answer all questions in this section.
8. A bag contains 5 red balls, 3 blue balls, and 2 green balls. Two balls are drawn at random from the bag without replacement.
(a) Draw a probability tree diagram to represent all possible outcomes. [2 marks]
(b) Find the probability that the two balls drawn are of the same colour. [2 marks]
(c) Find the probability that at least one red ball is drawn. [2 marks]
9. A random sample of 10 students recorded the time, in minutes, they spent on social media on a particular day. The times are summarised as follows:
[ \sum x = 850, \quad \sum x^2 = 74,800. ]
(a) Find the unbiased estimates of the population mean and variance of the time spent on social media. [3 marks]
(b) Explain what is meant by "unbiased estimate" in this context. [1 mark]
10. It is known that 25% of customers at a supermarket use the self-checkout counters. On a particular day, 20 customers are randomly selected.
(a) State two assumptions needed for the number of customers using self-checkout to be modelled by a binomial distribution. [2 marks]
(b) Find the probability that exactly 6 customers use the self-checkout counters. [2 marks]
(c) Find the probability that at least 8 customers use the self-checkout counters. [2 marks]
11. The mass of a packet of cereal is normally distributed with mean 500 g and standard deviation 8 g.
(a) Find the probability that a randomly selected packet has a mass less than 490 g. [2 marks]
(b) A packet is considered underweight if its mass is less than 485 g. Find the probability that a randomly selected packet is underweight. [2 marks]
(c) The cereal company wants to reduce the proportion of underweight packets to less than 1%. Find the minimum mean mass to which the machine should be set, assuming the standard deviation remains 8 g. [3 marks]
12. A company claims that the mean waiting time for calls to its customer service hotline is 4 minutes. A consumer group believes the mean waiting time is greater than 4 minutes. They take a random sample of 40 calls and find a mean waiting time of 4.5 minutes. The population standard deviation is known to be 1.8 minutes.
(a) State appropriate null and alternative hypotheses for a test. [2 marks]
(b) Carry out the test at the 5% significance level. [4 marks]
(c) State your conclusion in the context of the question. [1 mark]
13. A researcher is studying the relationship between the number of hours of revision (( x )) and the score in a test (( y )) for 8 students. The data are summarised as follows:
[ n = 8, \quad \sum x = 120, \quad \sum y = 560, \quad \sum x^2 = 2050, \quad \sum y^2 = 42,000, \quad \sum xy = 9200. ]
(a) Calculate the product moment correlation coefficient, ( r ). [2 marks]
(b) Comment on the relationship between hours of revision and test score. [1 mark]
(c) Find the equation of the least squares regression line of ( y ) on ( x ). Give the values of the coefficients to 3 significant figures. [3 marks]
(d) Use your regression line to estimate the test score of a student who revised for 18 hours. Comment on the reliability of this estimate. [2 marks]
14. A machine fills bottles with juice. The volume of juice in a bottle is normally distributed with mean ( \mu ) ml and standard deviation 5 ml. The machine is set so that ( \mu = 505 ) ml.
(a) Find the probability that a randomly selected bottle contains more than 510 ml. [2 marks]
(b) A random sample of 25 bottles is taken. Find the probability that the sample mean volume exceeds 507 ml. [3 marks]
(c) The machine setting is adjusted. A random sample of 36 bottles is taken and the sample mean is found to be 503.2 ml. Find a 95% confidence interval for the population mean ( \mu ). [3 marks]
15. A survey is to be conducted to estimate the proportion of residents in a town who support a new recycling scheme. The town has 12 000 residents.
(a) Describe how a simple random sample of 200 residents could be obtained. [2 marks]
(b) State one advantage and one disadvantage of using a simple random sample in this context. [2 marks]
16. The random variable ( X ) is normally distributed with mean ( \mu ) and variance ( \sigma^2 ). The random variable ( Y ) is independent of ( X ) and is normally distributed with mean ( 2\mu ) and variance ( 3\sigma^2 ).
(a) State the distribution of ( X + Y ). [2 marks]
(b) Given that ( \mu = 10 ) and ( \sigma^2 = 4 ), find ( \mathrm{P}(X + Y > 35) ). [3 marks]
17. A company produces light bulbs. The lifetime of a bulb, in hours, is normally distributed. A random sample of 8 bulbs is tested and the lifetimes are:
[ 1020, \quad 980, \quad 1050, \quad 1010, \quad 990, \quad 1030, \quad 1000, \quad 1040. ]
(a) Calculate the sample mean and the unbiased estimate of the population variance. [3 marks]
(b) The company claims the mean lifetime is 1020 hours. Test this claim at the 10% significance level, assuming the population variance is unknown but the sample size is small. State any necessary assumption. [5 marks]
18. A fair six-sided die is rolled 5 times. Find the probability of obtaining:
(a) exactly two sixes, [2 marks]
(b) at least one six. [2 marks]
19. The events ( A ) and ( B ) are such that ( \mathrm{P}(A) = 0.4 ), ( \mathrm{P}(B) = 0.5 ), and ( \mathrm{P}(A \cap B) = 0.2 ).
(a) Find ( \mathrm{P}(A \cup B) ). [1 mark]
(b) Find ( \mathrm{P}(A \mid B) ). [1 mark]
(c) Determine whether ( A ) and ( B ) are independent. Justify your answer. [2 marks]
20. A study is conducted to investigate the relationship between daily exercise time (( x ), in minutes) and resting heart rate (( y ), in beats per minute) for 10 individuals. The data are summarised as follows:
[ \sum x = 450, \quad \sum y = 720, \quad \sum x^2 = 24,500, \quad \sum y^2 = 52,600, \quad \sum xy = 31,200. ]
(a) Calculate the equation of the regression line of ( y ) on ( x ). [3 marks]
(b) Interpret the gradient of the regression line in the context of the question. [1 mark]
(c) An individual exercises for 60 minutes daily. Estimate their resting heart rate. [1 mark]
(d) Explain why it would be inappropriate to use the regression line to estimate the resting heart rate of an individual who exercises for 120 minutes daily. [1 mark]
END OF PAPER
TuitionGoWhere Practice Paper (AI) – Version 4 This paper is AI-generated for practice purposes and is not derived from past examination papers.
Answers
TuitionGoWhere Practice Paper - Maths H1 A-Level: Answer Key and Marking Scheme
TuitionGoWhere Practice Paper (AI)
Subject: Mathematics H1 (8865) Level: A-Level Paper: Practice Paper 1 (Version 4) Total Marks: 100
Section A: Pure Mathematics (40 marks)
1. Solve ( 3x^2 - 7x - 6 > 0 ). [3 marks]
Answer: [ 3x^2 - 7x - 6 = (3x + 2)(x - 3) = 0 \implies x = -\frac{2}{3},; x = 3. ] Since the coefficient of ( x^2 ) is positive, the parabola opens upward. [ \therefore ; x < -\frac{2}{3} ;; \text{or} ;; x > 3. ]
Marking:
- M1: Factorising or using quadratic formula correctly.
- M1: Identifying critical values ( x = -\frac{2}{3} ) and ( x = 3 ).
- A1: Correct solution ( x < -\frac{2}{3} ) or ( x > 3 ) (accept interval notation).
2. Curve ( y = x^3 - 6x^2 + 9x + 4 ). [5 marks]
(a) Stationary points [3 marks] [ \frac{dy}{dx} = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3). ] Set ( \frac{dy}{dx} = 0 ): ( x = 1 ) or ( x = 3 ). When ( x = 1 ): ( y = 1 - 6 + 9 + 4 = 8 ). Point: ( (1, 8) ). When ( x = 3 ): ( y = 27 - 54 + 27 + 4 = 4 ). Point: ( (3, 4) ).
Marking: M1 for differentiation; M1 for solving ( \frac{dy}{dx}=0 ); A1 for both coordinates.
(b) Nature [2 marks] [ \frac{d^2y}{dx^2} = 6x - 12. ] At ( x = 1 ): ( \frac{d^2y}{dx^2} = -6 < 0 ) → maximum point ( (1, 8) ). At ( x = 3 ): ( \frac{d^2y}{dx^2} = 6 > 0 ) → minimum point ( (3, 4) ).
Marking: M1 for second derivative or valid alternative method; A1 for correct nature of both points.
3. Simultaneous equations: ( y = 3x - 1 ), ( y = x^2 + x - 2 ). [4 marks]
Answer: [ x^2 + x - 2 = 3x - 1 \implies x^2 - 2x - 1 = 0. ] [ x = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}. ] When ( x = 1 + \sqrt{2} ): ( y = 3(1 + \sqrt{2}) - 1 = 2 + 3\sqrt{2} ). When ( x = 1 - \sqrt{2} ): ( y = 3(1 - \sqrt{2}) - 1 = 2 - 3\sqrt{2} ). Solutions: ( (1 + \sqrt{2},; 2 + 3\sqrt{2}) ) and ( (1 - \sqrt{2},; 2 - 3\sqrt{2}) ).
Marking: M1 for equating; M1 for solving quadratic; A1 for each correct pair (A2 total).
4. Bacteria population ( P = 20 e^{0.15t} ). [5 marks]
(a) At start, ( t = 0 ): ( P = 20 e^0 = 20 ) thousand. [1 mark]
(b) Double to 40: ( 40 = 20 e^{0.15t} \implies e^{0.15t} = 2 \implies 0.15t = \ln 2 \implies t = \frac{\ln 2}{0.15} \approx 4.62 ) hours. [2 marks]
(c) ( \frac{dP}{dt} = 20 \times 0.15 e^{0.15t} = 3 e^{0.15t} ). At ( t = 5 ): ( \frac{dP}{dt} = 3 e^{0.75} \approx 6.35 ) thousand per hour. [2 marks]
Marking: (a) A1; (b) M1 for setting up equation, A1 for correct time; (c) M1 for differentiation, A1 for correct rate.
5. Curve ( y = \dfrac{4x + 1}{2x - 3} ). [6 marks]
(a) Using quotient rule: [ \frac{dy}{dx} = \frac{(4)(2x - 3) - (4x + 1)(2)}{(2x - 3)^2} = \frac{8x - 12 - 8x - 2}{(2x - 3)^2} = \frac{-14}{(2x - 3)^2}. ] [3 marks]
(b) At ( x = 2 ): ( y = \frac{8 + 1}{4 - 3} = 9 ). Gradient ( m = \frac{-14}{(1)^2} = -14 ). Tangent: ( y - 9 = -14(x - 2) \implies y = -14x + 37 ). [3 marks]
Marking: (a) M1 for quotient rule, M1 for simplification, A1; (b) M1 for point, M1 for gradient, A1 for equation.
6. Curve ( y = e^{2x} + 1 ) and line ( y = 5 ). [7 marks]
(a) Intersection: ( e^{2x} + 1 = 5 \implies e^{2x} = 4 \implies 2x = \ln 4 \implies x = \frac{1}{2}\ln 4 = \ln 2 ). [2 marks]
(b) Area bounded by curve, ( y = 5 ), ( y )-axis (( x = 0 )), and ( x = \ln 2 ): [ \text{Area} = \int_0^{\ln 2} \bigl[5 - (e^{2x} + 1)\bigr] , dx = \int_0^{\ln 2} (4 - e^{2x}) , dx. ] [ = \left[ 4x - \frac{1}{2}e^{2x} \right]_0^{\ln 2} = \left(4\ln 2 - \frac{1}{2}e^{2\ln 2}\right) - \left(0 - \frac{1}{2}e^0\right) ] [ = 4\ln 2 - \frac{1}{2}(4) + \frac{1}{2} = 4\ln 2 - 2 + \frac{1}{2} = 4\ln 2 - \frac{3}{2}. ] [5 marks]
Marking: (a) M1 for setting up, A1; (b) M1 for correct integral setup, M1 for integration, M1 for limits, M1 for evaluation, A1 for exact answer.
7. Rectangular enclosure against wall. [7 marks]
(a) Fencing: ( 2x + y = 60 \implies y = 60 - 2x ). Area ( A = xy = x(60 - 2x) = 60x - 2x^2 ). [2 marks]
(b) ( \frac{dA}{dx} = 60 - 4x = 0 \implies x = 15 ). [3 marks]
(c) ( \frac{d^2A}{dx^2} = -4 < 0 ) → maximum. [1 mark]
(d) ( A_{\text{max}} = 60(15) - 2(15)^2 = 900 - 450 = 450 ) m². [1 mark]
Marking: (a) M1 for constraint, A1 for area expression; (b) M1 for differentiation, M1 for solving, A1 for ( x = 15 ); (c) A1; (d) A1.
Section B: Probability and Statistics (60 marks)
8. Balls: 5 red, 3 blue, 2 green. Total = 10. [6 marks]
(a) Tree diagram: First draw branches R (5/10), B (3/10), G (2/10). Second draw probabilities conditional on first draw (without replacement). [2 marks]
(b) Same colour: [ \mathrm{P}(RR) + \mathrm{P}(BB) + \mathrm{P}(GG) = \frac{5}{10} \cdot \frac{4}{9} + \frac{3}{10} \cdot \frac{2}{9} + \frac{2}{10} \cdot \frac{1}{9} = \frac{20}{90} + \frac{6}{90} + \frac{2}{90} = \frac{28}{90} = \frac{14}{45} \approx 0.311. ] [2 marks]
(c) At least one red: ( 1 - \mathrm{P}(\text{no red}) = 1 - \frac{5}{10} \cdot \frac{4}{9} = 1 - \frac{20}{90} = \frac{70}{90} = \frac{7}{9} \approx 0.778 ). [2 marks]
Marking: (a) M1 for correct structure, A1 for correct probabilities; (b) M1 for method, A1; (c) M1 for complement or direct, A1.
9. Unbiased estimates. [4 marks]
(a) ( n = 10 ), ( \sum x = 850 ), ( \sum x^2 = 74,800 ). [ \bar{x} = \frac{850}{10} = 85 \text{ minutes}. ] [ s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right) = \frac{1}{9}\left(74,800 - \frac{850^2}{10}\right) = \frac{1}{9}(74,800 - 72,250) = \frac{2550}{9} \approx 283.3 \text{ minutes}^2. ] [3 marks]
(b) An unbiased estimate means the expected value of the estimator equals the true population parameter. Using ( n-1 ) in the denominator for variance ensures the estimate is unbiased. [1 mark]
Marking: (a) M1 for mean, M1 for variance formula, A1 for both; (b) A1 for correct explanation.
10. Binomial distribution: ( n = 20 ), ( p = 0.25 ). [6 marks]
(a) Assumptions: (i) Each customer's choice is independent of others. (ii) The probability of using self-checkout is constant (0.25) for each customer. [2 marks]
(b) ( X \sim \mathrm{B}(20, 0.25) ). ( \mathrm{P}(X = 6) = \binom{20}{6}(0.25)^6(0.75)^{14} \approx 0.1686 \approx 0.169 ) (3 s.f.). [2 marks]
(c) ( \mathrm{P}(X \geq 8) = 1 - \mathrm{P}(X \leq 7) ). Using GC: ( \mathrm{P}(X \leq 7) \approx 0.8982 ). ( \mathrm{P}(X \geq 8) \approx 0.1018 \approx 0.102 ) (3 s.f.). [2 marks]
Marking: (a) A1 for each valid assumption; (b) M1 for binomial formula, A1; (c) M1 for complement, A1.
11. Normal distribution: ( \mu = 500 ), ( \sigma = 8 ). [7 marks]
(a) ( \mathrm{P}(X < 490) = \mathrm{P}!\left(Z < \frac{490 - 500}{8}\right) = \mathrm{P}(Z < -1.25) = 1 - \Phi(1.25) = 1 - 0.8944 = 0.1056 \approx 0.106 ). [2 marks]
(b) ( \mathrm{P}(X < 485) = \mathrm{P}!\left(Z < \frac{485 - 500}{8}\right) = \mathrm{P}(Z < -1.875) = 1 - \Phi(1.875) \approx 1 - 0.9696 = 0.0304 \approx 0.0304 ). [2 marks]
(c) Need ( \mathrm{P}(X < 485) < 0.01 ). Let new mean be ( \mu ). [ \mathrm{P}!\left(Z < \frac{485 - \mu}{8}\right) < 0.01 \implies \frac{485 - \mu}{8} < -2.326 \implies 485 - \mu < -18.608 \implies \mu > 503.608. ] Minimum mean = 504 g (to nearest gram). [3 marks]
Marking: (a) M1 for standardisation, A1; (b) M1, A1; (c) M1 for inverse normal, M1 for inequality, A1.
12. Hypothesis test for mean. [7 marks]
(a) ( H_0: \mu = 4 ); ( H_1: \mu > 4 ) (one-tail test). [2 marks]
(b) ( n = 40 ), ( \bar{x} = 4.5 ), ( \sigma = 1.8 ). Test statistic: ( Z = \frac{4.5 - 4}{1.8/\sqrt{40}} = \frac{0.5}{0.2846} \approx 1.757 ). Critical value at 5% (one-tail): ( z_{0.05} = 1.645 ). Since ( 1.757 > 1.645 ), reject ( H_0 ). [4 marks]
(c) There is sufficient evidence at the 5% significance level that the mean waiting time is greater than 4 minutes. [1 mark]
Marking: (a) A1 for each hypothesis; (b) M1 for test statistic, M1 for critical value or p-value, M1 for comparison, A1 for correct conclusion; (c) A1 for contextual conclusion.
13. Correlation and regression. [8 marks]
(a) ( r = \frac{n\sum xy - \sum x \sum y}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}} ) [ = \frac{8(9200) - 120(560)}{\sqrt{[8(2050) - 120^2][8(42,000) - 560^2]}} = \frac{73,600 - 67,200}{\sqrt{[16,400 - 14,400][336,000 - 313,600]}} = \frac{6400}{\sqrt{2000 \times 22,400}} = \frac{6400}{\sqrt{44,800,000}} \approx \frac{6400}{6693.3} \approx 0.956. ] [2 marks]
(b) There is a very strong positive linear correlation between hours of revision and test score. [1 mark]
(c) ( b = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2} = \frac{6400}{2000} = 3.2 ). ( a = \bar{y} - b\bar{x} = \frac{560}{8} - 3.2\left(\frac{120}{8}\right) = 70 - 3.2(15) = 70 - 48 = 22 ). Regression line: ( y = 22.0 + 3.20x ). [3 marks]
(d) When ( x = 18 ): ( y = 22 + 3.2(18) = 22 + 57.6 = 79.6 ). This is interpolation (18 is within the data range 10–20 approximately), so the estimate is reasonably reliable. [2 marks]
Marking: (a) M1 for formula, A1; (b) A1; (c) M1 for ( b ), M1 for ( a ), A1 for equation; (d) A1 for estimate, A1 for comment.
14. Normal distribution and sampling. [8 marks]
(a) ( \mu = 505 ), ( \sigma = 5 ). ( \mathrm{P}(X > 510) = \mathrm{P}!\left(Z > \frac{510 - 505}{5}\right) = \mathrm{P}(Z > 1) = 1 - 0.8413 = 0.1587 \approx 0.159 ). [2 marks]
(b) ( \bar{X} \sim \mathrm{N}!\left(505, \frac{5^2}{25}\right) = \mathrm{N}(505, 1) ). ( \mathrm{P}(\bar{X} > 507) = \mathrm{P}!\left(Z > \frac{507 - 505}{1}\right) = \mathrm{P}(Z > 2) = 1 - 0.9772 = 0.0228 \approx 0.0228 ). [3 marks]
(c) ( n = 36 ), ( \bar{x} = 503.2 ), ( \sigma = 5 ). 95% CI: ( \bar{x} \pm z_{0.025} \frac{\sigma}{\sqrt{n}} ). ( z_{0.025} = 1.96 ). Margin of error ( = 1.96 \times \frac{5}{\sqrt{36}} = 1.96 \times \frac{5}{6} \approx 1.633 ). CI: ( (503.2 - 1.633,; 503.2 + 1.633) = (501.6,; 504.8) ) (1 d.p.). [3 marks]
Marking: (a) M1, A1; (b) M1 for distribution of ( \bar{X} ), M1 for standardisation, A1; (c) M1 for ( z )-value, M1 for margin of error, A1 for interval.
15. Sampling methods. [4 marks]
(a) Assign each of the 12 000 residents a unique number from 1 to 12 000. Use a random number generator to select 200 distinct numbers. The residents corresponding to these numbers form the sample. [2 marks]
(b) Advantage: Every resident has an equal chance of being selected, reducing selection bias. Disadvantage: It may be time-consuming and expensive to contact residents who are geographically dispersed. [2 marks]
Marking: (a) M1 for numbering, M1 for random selection method; (b) A1 for valid advantage, A1 for valid disadvantage.
16. Sum of independent normal variables. [5 marks]
(a) ( X \sim \mathrm{N}(\mu, \sigma^2) ), ( Y \sim \mathrm{N}(2\mu, 3\sigma^2) ), independent. ( X + Y \sim \mathrm{N}(\mu + 2\mu,; \sigma^2 + 3\sigma^2) = \mathrm{N}(3\mu, 4\sigma^2) ). [2 marks]
(b) ( \mu = 10 ), ( \sigma^2 = 4 ). ( X + Y \sim \mathrm{N}(30, 16) ). ( \mathrm{P}(X + Y > 35) = \mathrm{P}!\left(Z > \frac{35 - 30}{4}\right) = \mathrm{P}(Z > 1.25) = 1 - 0.8944 = 0.1056 \approx 0.106 ). [3 marks]
Marking: (a) M1 for mean, A1 for variance; (b) M1 for distribution, M1 for standardisation, A1.
17. Small sample hypothesis test. [8 marks]
(a) Data: 1020, 980, 1050, 1010, 990, 1030, 1000, 1040. ( n = 8 ). ( \sum x = 8120 ), ( \bar{x} = \frac{8120}{8} = 1015 ). ( \sum x^2 = 1020^2 + 980^2 + \cdots + 1040^2 = 8,242,600 ). ( s^2 = \frac{1}{7}\left(8,242,600 - \frac{8120^2}{8}\right) = \frac{1}{7}(8,242,600 - 8,241,800) = \frac{800}{7} \approx 114.29 ). [3 marks]
(b) ( H_0: \mu = 1020 ); ( H_1: \mu \neq 1020 ) (two-tail test). Assumption: The lifetimes are normally distributed (since ( n ) is small). ( s = \sqrt{114.29} \approx 10.69 ). Test statistic: ( t = \frac{1015 - 1020}{10.69/\sqrt{8}} = \frac{-5}{3.779} \approx -1.323 ). Degrees of freedom = 7. Critical value at 10% (two-tail): ( t_{0.05, 7} = 1.895 ). Since ( |-1.323| < 1.895 ), do not reject ( H_0 ). There is insufficient evidence at the 10% level to reject the claim that the mean lifetime is 1020 hours. [5 marks]
Marking: (a) M1 for mean, M1 for variance formula, A1; (b) M1 for hypotheses, M1 for assumption, M1 for test statistic, M1 for critical value/comparison, A1 for conclusion in context.
18. Binomial probability. [4 marks]
(a) ( X \sim \mathrm{B}(5, \frac{1}{6}) ). ( \mathrm{P}(X = 2) = \binom{5}{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^3 = 10 \times \frac{1}{36} \times \frac{125}{216} = \frac{1250}{7776} \approx 0.1608 \approx 0.161 ). [2 marks]
(b) ( \mathrm{P}(X \geq 1) = 1 - \mathrm{P}(X = 0) = 1 - \left(\frac{5}{6}\right)^5 = 1 - \frac{3125}{7776} = \frac{4651}{7776} \approx 0.598 ). [2 marks]
Marking: (a) M1 for binomial formula, A1; (b) M1 for complement, A1.
19. Probability rules. [4 marks]
(a) ( \mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B) = 0.4 + 0.5 - 0.2 = 0.7 ). [1 mark]
(b) ( \mathrm{P}(A \mid B) = \frac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)} = \frac{0.2}{0.5} = 0.4 ). [1 mark]
(c) For independence: ( \mathrm{P}(A \cap B) = \mathrm{P}(A)\mathrm{P}(B) ). ( \mathrm{P}(A)\mathrm{P}(B) = 0.4 \times 0.5 = 0.2 ). Since ( \mathrm{P}(A \cap B) = 0.2 ), the events are independent. [2 marks]
Marking: (a) A1; (b) A1; (c) M1 for checking product, A1 for correct conclusion with justification.
20. Regression and interpretation. [6 marks]
(a) ( n = 10 ), ( \sum x = 450 ), ( \sum y = 720 ), ( \sum x^2 = 24,500 ), ( \sum y^2 = 52,600 ), ( \sum xy = 31,200 ). ( b = \frac{10(31,200) - 450(720)}{10(24,500) - 450^2} = \frac{312,000 - 324,000}{245,000 - 202,500} = \frac{-12,000}{42,500} \approx -0.28235 ). ( \bar{x} = 45 ), ( \bar{y} = 72 ). ( a = 72 - (-0.28235)(45) = 72 + 12.706 = 84.706 ). Regression line: ( y = 84.7 - 0.282x ) (3 s.f.). [3 marks]
(b) The gradient of −0.282 means that for each additional minute of daily exercise, the resting heart rate decreases by approximately 0.282 beats per minute, on average. [1 mark]
(c) When ( x = 60 ): ( y = 84.706 - 0.28235(60) = 84.706 - 16.941 = 67.765 \approx 67.8 ) bpm. [1 mark]
(d) 120 minutes is outside the range of the data (extrapolation), so the linear relationship may not hold and the estimate would be unreliable. [1 mark]
Marking: (a) M1 for ( b ), M1 for ( a ), A1 for equation; (b) A1 for correct interpretation; (c) A1; (d) A1 for extrapolation explanation.
END OF ANSWER KEY
TuitionGoWhere Practice Paper (AI) – Version 4 This answer key is AI-generated for practice purposes.