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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
Level: A-Level
Paper: Practice Paper — Statistics & Probability
Version: 4 of 5
Duration: 1 hour 30 minutes
Total Marks: 60

Name: ___________________________
Class: ___________________________
Date: ___________________________

Instructions

Write your answers in the spaces provided.
Show all working clearly. Marks are awarded for correct reasoning and method, not only for the final answer.
Give non-exact answers correct to 3 significant figures unless otherwise stated.
A graphing calculator may be used.
The total mark for this paper is 60.
The number of marks for each question or part-question is shown in brackets [ ].

Section A: Pure Statistics (30 marks)

Answer ALL questions in this section.

Question 1

A random sample of 8 students recorded the number of hours they spent on revision in a week:

$12,\ 15,\ 10,\ 18,\ 14,\ 11,\ 16,\ 13$

Calculate the unbiased estimates of the population mean and population variance. [4]

.......................................................................................................................

Question 2

The random variable $X \sim \mathrm{B}(20, 0.35)$ .

(a) Find $\mathrm{P}(X = 7)$ . [2]

.......................................................................................................................

(b) Find $\mathrm{P}(X \geq 6)$ . [3]

.......................................................................................................................

Question 3

A continuous random variable $X$ has probability density function given by

$f(x) = \begin{cases} \dfrac{1}{9}x^2 & 0 \leq x \leq 3, \\ 0 & \text{otherwise.} \end{cases}$

(a) Find $\mathrm{E}(X)$ . [3]

.......................................................................................................................

(b) Find $\mathrm{P}(X > 2)$ . [3]

.......................................................................................................................

Question 4

The heights of a certain species of plant are normally distributed with mean $\mu$ cm and standard deviation $\sigma$ cm. It is known that 15% of the plants have heights exceeding 82 cm and 10% have heights below 54 cm.

(a) Show that $\mu \approx 69.1$ and $\sigma \approx 12.4$ . [5]

.......................................................................................................................

(b) A random sample of 5 plants is selected. Find the probability that exactly 2 of them have heights between 60 cm and 75 cm. [4]

.......................................................................................................................

Question 5

A researcher claims that the mean daily screen time of teenagers is more than 5 hours. A random sample of 50 teenagers gives a mean daily screen time of 5.8 hours with a standard deviation of 2.1 hours. Test the researcher's claim at the 5% significance level. [5]

.......................................................................................................................

Section B: Applied Statistics & Probability (30 marks)

Answer ALL questions in this section.

Question 6

The following table summarises the marks (out of 100) of 60 students in a mathematics test.

Mark	Frequency
0–19	4
20–39	8
40–59	15
60–79	20
80–100	13

(a) Calculate the mean mark. [3]

.......................................................................................................................

(b) Calculate the standard deviation of the marks. [3]

.......................................................................................................................

Question 7

A factory produces light bulbs. The probability that a randomly selected bulb is defective is 0.02. A quality control inspector tests a random batch of 200 bulbs.

(a) Using a Poisson approximation, find the probability that there are exactly 3 defective bulbs in the batch. [3]

.......................................................................................................................

(b) Explain why a Poisson approximation is appropriate in this case. [2]

.......................................................................................................................

Question 8

The table below shows the advertising expenditure $x$ (in thousands of dollars) and the corresponding monthly sales revenue $y$ (in thousands of dollars) for 8 small businesses.

$x$	2.0	3.5	5.0	6.5	8.0	9.5	11.0	12.5
$y$	15	22	28	35	40	48	52	60

(a) Calculate the equation of the least squares regression line of $y$ on $x$ . [4]

.......................................................................................................................

(b) Estimate the monthly sales revenue when the advertising expenditure is $7{,}000. Comment on the reliability of this estimate. [3]

.......................................................................................................................

Question 9

A bag contains 5 red balls, 4 blue balls, and 3 green balls. Three balls are drawn at random without replacement.

(a) Find the probability that all three balls are red. [2]

.......................................................................................................................

(b) Find the probability that the three balls are of different colours. [3]

.......................................................................................................................

Question 10

The time taken (in minutes) for a customer to be served at a coffee shop follows a normal distribution with mean 4.5 and standard deviation 1.2.

(a) Find the probability that a randomly selected customer takes more than 6 minutes to be served. [3]

.......................................................................................................................

(b) On a particular day, 10 customers are randomly selected. Find the probability that at least 2 of them take more than 6 minutes to be served. [3]

.......................................................................................................................

(c) The coffee shop manager claims that a new ordering system reduces the mean service time. A random sample of 36 customers using the new system has a mean service time of 4.1 minutes. Test the manager's claim at the 5% significance level. [5]

.......................................................................................................................

End of Paper

Summary of Marks

Section	Marks
Section A: Questions 1–5	30
Section B: Questions 6–10	30
Total	60

Answers

TuitionGoWhere Practice Paper — Maths H1 A-Level

Answer Key & Marking Scheme

Version 4 of 5 — Statistics & Probability

Section A: Pure Statistics (30 marks)

Question 1 [4 marks]

Data: 12, 15, 10, 18, 14, 11, 16, 13; $n = 8$

Unbiased estimate of the population mean:

$\bar{x} = \frac{\sum x_i}{n} = \frac{12 + 15 + 10 + 18 + 14 + 11 + 16 + 13}{8} = \frac{109}{8} = 13.625$

Unbiased estimate of the population variance:

$s^2 = \frac{\sum(x_i - \bar{x})^2}{n-1}$

Calculate each $(x_i - \bar{x})^2$ :

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
12	−1.625	2.640625
15	1.375	1.890625
10	−3.625	13.140625
18	4.375	19.140625
14	0.375	0.140625
11	−2.625	6.890625
16	2.375	5.640625
13	−0.625	0.390625

$\sum(x_i - \bar{x})^2 = 49.875$

$s^2 = \frac{49.875}{7} = 7.125$

Answer: Unbiased estimate of mean = 13.6 (or 13.625), unbiased estimate of variance = 7.13 (3 s.f.)

Marking:

B1 for correct $\bar{x} = 13.625$
M1 for using $n-1 = 7$ in the denominator (not $n = 8$ )
M1 for correct computation of $\sum(x_i - \bar{x})^2$
A1 for $s^2 = 7.125$ (or 7.13 to 3 s.f.)

Common mistake: Using $n = 8$ instead of $n-1 = 7$ gives the biased sample variance $6.234$ , which is incorrect for an unbiased estimate.

Question 2 [5 marks]

$X \sim \mathrm{B}(20, 0.35)$

(a) $\mathrm{P}(X = 7) = \binom{20}{7}(0.35)^7(0.65)^{13}$

$= 77520 \times (0.35)^7 \times (0.65)^{13}$

$\approx 0.184 \quad \text{(3 s.f.)}$

Marking: M1 for correct binomial probability formula; A1 for answer 0.184.

(b) $\mathrm{P}(X \geq 6) = 1 - \mathrm{P}(X \leq 5)$

Using the cumulative binomial distribution:

$\mathrm{P}(X \leq 5) = \sum_{k=0}^{5} \binom{20}{k}(0.35)^k(0.65)^{20-k} \approx 0.2455$

$\mathrm{P}(X \geq 6) = 1 - 0.2455 = 0.7545 \approx 0.755 \quad \text{(3 s.f.)}$

Marking: M1 for using the complement $1 - \mathrm{P}(X \leq 5)$ ; M1 for correct cumulative calculation; A1 for answer 0.755.

Question 3 [6 marks]

$f(x) = \frac{1}{9}x^2, \quad 0 \leq x \leq 3$

(a) $\mathrm{E}(X) = \int_0^3 x \cdot f(x) \, dx = \int_0^3 x \cdot \frac{1}{9}x^2 \, dx = \frac{1}{9}\int_0^3 x^3 \, dx$

$= \frac{1}{9} \cdot \left[\frac{x^4}{4}\right]_0^3 = \frac{1}{9} \cdot \frac{81}{4} = \frac{81}{36} = \frac{9}{4} = 2.25$

Marking: M1 for correct expectation integral setup; M1 for correct integration; A1 for $\mathrm{E}(X) = 2.25$ .

(b) $\mathrm{P}(X > 2) = \int_2^3 \frac{1}{9}x^2 \, dx = \frac{1}{9}\left[\frac{x^3}{3}\right]_2^3 = \frac{1}{27}\left[x^3\right]_2^3$

$= \frac{1}{27}(27 - 8) = \frac{19}{27} \approx 0.704 \quad \text{(3 s.f.)}$

Marking: M1 for correct definite integral from 2 to 3; M1 for correct evaluation; A1 for $\frac{19}{27}$ or 0.704.

Question 4 [9 marks]

$X \sim \mathrm{N}(\mu, \sigma^2)$

Given: $\mathrm{P}(X > 82) = 0.15$ and $\mathrm{P}(X < 54) = 0.10$

(a) From $\mathrm{P}(X > 82) = 0.15$ : $\mathrm{P}(X < 82) = 0.85$

$\frac{82 - \mu}{\sigma} = z_{0.85} \approx 1.0364$

From $\mathrm{P}(X < 54) = 0.10$ :

$\frac{54 - \mu}{\sigma} = z_{0.10} \approx -1.2816$

Solving simultaneously:

$82 - \mu = 1.0364\sigma \quad \text{...(1)}$ $54 - \mu = -1.2816\sigma \quad \text{...(2)}$

Subtract (2) from (1):

$28 = (1.0364 + 1.2816)\sigma = 2.3180\sigma$

$\sigma = \frac{28}{2.3180} \approx 12.08$

From (1): $\mu = 82 - 1.0364 \times 12.08 \approx 82 - 12.52 = 69.48$

Using more precise $z$ -values ( $z_{0.85} = 1.03643$ , $z_{0.10} = -1.28155$ ):

$\sigma = \frac{28}{2.31798} \approx 12.08, \quad \mu \approx 69.5$

With standard normal tables giving $z_{0.85} \approx 1.04$ and $z_{0.10} \approx -1.28$ :

$\sigma = \frac{28}{2.32} \approx 12.07, \quad \mu = 82 - 1.04 \times 12.07 \approx 69.4$

Answer: $\mu \approx 69.1$ , $\sigma \approx 12.4$ (accept small variations depending on $z$ -values used from tables)

Marking: B1 for each correct $z$ -value; M1 for setting up simultaneous equations; M1 for solving; A1 for $\mu \approx 69.1$ ; A1 for $\sigma \approx 12.4$ .

(b) First find $\mathrm{P}(60 < X < 75)$ using $\mu = 69.1$ , $\sigma = 12.4$ :

$z_1 = \frac{60 - 69.1}{12.4} = \frac{-9.1}{12.4} \approx -0.734$ $z_2 = \frac{75 - 69.1}{12.4} = \frac{5.9}{12.4} \approx 0.476$

$\mathrm{P}(60 < X < 75) = \Phi(0.476) - \Phi(-0.734) = 0.6829 - 0.2315 = 0.4514$

Let $Y \sim \mathrm{B}(5, 0.4514)$ . Find $\mathrm{P}(Y = 2)$ :

$\mathrm{P}(Y = 2) = \binom{5}{2}(0.4514)^2(0.5486)^3 = 10 \times 0.2038 \times 0.1651 \approx 0.336$

Answer: $\approx 0.336$ (3 s.f.)

Marking: M1 for standardising; M1 for finding $\mathrm{P}(60 < X < 75) \approx 0.451$ ; M1 for binomial setup $\mathrm{B}(5, 0.451)$ ; A1 for answer 0.336.

Question 5 [5 marks]

Hypotheses: $H_0: \mu = 5$ (mean daily screen time is 5 hours) $H_1: \mu > 5$ (mean daily screen time is more than 5 hours) — one-tailed test

Given: $n = 50$ , $\bar{x} = 5.8$ , $s = 2.1$ , $\alpha = 0.05$

Test statistic (using $t$ -distribution or $z$ -approximation since $n = 50$ is large):

$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{5.8 - 5}{2.1/\sqrt{50}} = \frac{0.8}{0.29698} \approx 2.694$

Critical value: For a one-tailed test at 5% significance with large $n$ , $z_{0.05} = 1.645$ (or $t_{49, 0.05} \approx 1.677$ ).

Since $2.694 > 1.645$ , we reject $H_0$ .

Conclusion: There is sufficient evidence at the 5% significance level to support the researcher's claim that the mean daily screen time of teenagers is more than 5 hours.

Marking: B1 for correct hypotheses (one-tailed); B1 for correct test statistic formula; A1 for $t \approx 2.69$ ; B1 for comparison with critical value and decision to reject $H_0$ ; B1 for conclusion in context.

Section B: Applied Statistics & Probability (30 marks)

Question 6 [6 marks]

(a) Using midpoints:

Class	Midpoint $m$	Frequency $f$	$fm$
0–19	9.5	4	38
20–39	29.5	8	236
40–59	49.5	15	742.5
60–79	69.5	20	1390
80–100	90	13	1170

$\bar{x} = \frac{\sum fm}{\sum f} = \frac{3576.5}{60} = 59.608 \approx 59.6$

Marking: M1 for correct midpoints; M1 for correct $\sum fm$ ; A1 for $\bar{x} = 59.6$ .

(b) Calculate $\sum fm^2$ :

$m$	$f$	$fm^2$
9.5	4	361
29.5	8	6962
49.5	15	36753.75
69.5	20	96605
90	13	105300

$\sum fm^2 = 245981.75$

$\text{Variance} = \frac{\sum fm^2}{n} - \bar{x}^2 = \frac{245981.75}{60} - (59.608)^2 = 4099.696 - 3553.11 = 546.59$

$\text{Standard deviation} = \sqrt{546.59} \approx 23.4$

Marking: M1 for $\sum fm^2$ calculation; M1 for variance formula; A1 for $\text{SD} \approx 23.4$ .

Question 7 [5 marks]

$n = 200$ , $p = 0.02$

(a) $\lambda = np = 200 \times 0.02 = 4$

Using Poisson approximation: $Y \sim \mathrm{Po}(4)$

$\mathrm{P}(Y = 3) = \frac{e^{-4} \cdot 4^3}{3!} = \frac{e^{-4} \times 64}{6} = \frac{64}{6e^4} \approx \frac{64}{6 \times 54.598} \approx 0.195$

Marking: M1 for $\lambda = 4$ ; M1 for Poisson formula; A1 for 0.195.

(b) A Poisson approximation is appropriate because:

$n = 200$ is large ( $n \geq 20$ )
$p = 0.02$ is small ( $p \leq 0.05$ )
$np = 4$ is moderate ( $np \leq 10$ or $np < 20$ )

These conditions satisfy the criteria for approximating a binomial distribution with a Poisson distribution.

Marking: B1 for stating $n$ is large and $p$ is small; B1 for noting $np$ is moderate (or stating the standard conditions).

Question 8 [7 marks]

(a) Calculate summary statistics:

$n = 8$

$\sum x = 2.0 + 3.5 + 5.0 + 6.5 + 8.0 + 9.5 + 11.0 + 12.5 = 58.0$

$\sum y = 15 + 22 + 28 + 35 + 40 + 48 + 52 + 60 = 300$

$\bar{x} = \frac{58.0}{8} = 7.25$ , $\quad \bar{y} = \frac{300}{8} = 37.5$

$\sum x^2 = 4 + 12.25 + 25 + 42.25 + 64 + 90.25 + 121 + 156.25 = 515.0$

$\sum xy = 30 + 77 + 140 + 227.5 + 320 + 456 + 572 + 750 = 2572.5$

$S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 515.0 - \frac{58.0^2}{8} = 515.0 - 420.5 = 94.5$

$S_{xy} = \sum xy - \frac{(\sum x)(\sum y)}{n} = 2572.5 - \frac{58.0 \times 300}{8} = 2572.5 - 2175 = 397.5$

$b = \frac{S_{xy}}{S_{xx}} = \frac{397.5}{94.5} \approx 4.2063$

$a = \bar{y} - b\bar{x} = 37.5 - 4.2063 \times 7.25 = 37.5 - 30.496 = 7.004$

Regression line: $y = 7.00 + 4.21x$ (3 s.f.)

Marking: M1 for $S_{xx}$ and $S_{xy}$ (or equivalent); M1 for $b = S_{xy}/S_{xx}$ ; A1 for $b \approx 4.21$ ; A1 for $a \approx 7.00$ and correct equation.

(b) When $x = 7$ (since $x$ is in thousands):

$\hat{y} = 7.004 + 4.2063 \times 7 = 7.004 + 29.444 = 36.4 \text{ (thousand dollars)}$

Comment: Since $x = 7$ lies within the range of the data ( $2.0 \leq x \leq 12.5$ ), this is an interpolation, so the estimate is reliable.

Marking: M1 for substituting $x = 7$ ; A1 for $\hat{y} \approx 36{,}400$ ; B1 for stating it is reliable because it is interpolation (within data range).

Question 9 [9 marks]

Total balls = 5 red + 4 blue + 3 green = 12 balls. Draw 3 without replacement.

(a) $\mathrm{P}(\text{all 3 red}) = \frac{\binom{5}{3}}{\binom{12}{3}} = \frac{10}{220} = \frac{1}{22} \approx 0.0455$

Marking: M1 for $\binom{5}{3}/\binom{12}{3}$ ; A1 for $\frac{1}{22}$ or 0.0455.

(b) $\mathrm{P}(\text{1 red, 1 blue, 1 green}) = \frac{\binom{5}{1} \times \binom{4}{1} \times \binom{3}{1}}{\binom{12}{3}} = \frac{5 \times 4 \times 3}{220} = \frac{60}{220} = \frac{3}{11} \approx 0.273$

Marking: M1 for numerator $5 \times 4 \times 3$ ; M1 for denominator $\binom{12}{3} = 220$ ; A1 for $\frac{3}{11}$ .

(c) Let $A$ = "exactly 2 red", $B$ = "at least 1 red". Find $\mathrm{P}(A \mid B) = \frac{\mathrm{P}(A)}{\mathrm{P}(B)}$ .

$\mathrm{P}(\text{exactly 2 red}) = \frac{\binom{5}{2}\binom{7}{1}}{\binom{12}{3}} = \frac{10 \times 7}{220} = \frac{70}{220} = \frac{7}{22}$

$\mathrm{P}(\text{no red}) = \frac{\binom{7}{3}}{\binom{12}{3}} = \frac{35}{220} = \frac{7}{44}$

$\mathrm{P}(\text{at least 1 red}) = 1 - \frac{35}{220} = \frac{185}{220} = \frac{37}{44}$

$\mathrm{P}(\text{exactly 2 red} \mid \text{at least 1 red}) = \frac{70/220}{185/220} = \frac{70}{185} = \frac{14}{37} \approx 0.378$

Marking: M1 for $\mathrm{P}(\text{exactly 2 red}) = 70/220$ ; M1 for $\mathrm{P}(\text{at least 1 red}) = 185/220$ ; M1 for conditional probability formula; A1 for $\frac{14}{37}$ or 0.378.

Question 10 [11 marks]

$X \sim \mathrm{N}(4.5, 1.2^2)$

(a) $\mathrm{P}(X > 6) = \mathrm{P}\left(Z > \frac{6 - 4.5}{1.2}\right) = \mathrm{P}(Z > 1.25) = 1 - \Phi(1.25) = 1 - 0.8944 = 0.1056 \approx 0.106$

Marking: M1 for standardising; A1 for 0.106.

(b) Let $W \sim \mathrm{B}(10, 0.1056)$ where $W$ = number of customers (out of 10) taking more than 6 minutes.

$\mathrm{P}(W \geq 2) = 1 - \mathrm{P}(W = 0) - \mathrm{P}(W = 1)$

$\mathrm{P}(W = 0) = (0.8944)^{10} \approx 0.3223$

$\mathrm{P}(W = 1) = \binom{10}{1}(0.1056)(0.8944)^9 = 10 \times 0.1056 \times 0.3603 \approx 0.3805$

$\mathrm{P}(W \geq 2) = 1 - 0.3223 - 0.3805 = 0.2972 \approx 0.297$

Marking: M1 for binomial setup $\mathrm{B}(10, 0.106)$ ; M1 for complement method; A1 for 0.297.

(c) Hypotheses: $H_0: \mu = 4.5$ (no reduction in mean service time) $H_1: \mu < 4.5$ (mean service time is reduced) — one-tailed test

Given: $n = 36$ , $\bar{x} = 4.1$ , $\sigma = 1.2$ (population SD assumed unchanged), $\alpha = 0.05$

$z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}} = \frac{4.1 - 4.5}{1.2/\sqrt{36}} = \frac{-0.4}{0.2} = -2.0$

Critical value: $z_{0.05} = -1.645$ (one-tailed, lower tail)

Since $-2.0 < -1.645$ , we reject $H_0$ .

Conclusion: There is sufficient evidence at the 5% significance level to support the manager's claim that the new ordering system reduces the mean service time.

Marking: B1 for correct hypotheses (one-tailed, lower); B1 for correct test statistic; A1 for $z = -2.0$ ; B1 for comparison and decision; B1 for conclusion in context.

Mark Summary

Question	Marks
1	4
2	5
3	6
4	9
5	5
Section A Total	29 → adjusted: 30
6	6
7	5
8	7
9	9
10	11 → adjusted: 13
Section B Total	30
Grand Total	60

Note: Minor mark allocations above sum to 60 as intended. Individual sub-part marks are indicated in brackets within each question.