A Level H1 Mathematics Practice Paper 3

TuitionGoWhere Practice Paper — Answer Key

Subject: Mathematics H1 (A-Level)
Paper: Practice Paper — Version 3 of 5
Topic Focus: Statistics & Probability
Total Marks: 60

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Section A: Pure Mathematics Foundations [20 marks]

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Question 1 [2 marks]

Answer: $\mathrm{P}(X = 8) = 0.161$ (to 3 s.f.)

Working:

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

$\mathrm{P}(X = 8) = \binom{20}{8}(0.35)^8(0.65)^{12}$

$= 125970 \times (0.35)^8 \times (0.65)^{12}$

$= 0.16128...$

$\approx 0.161 \text{ (3 s.f.)}$

Marking notes:

• M1: Correct binomial probability formula with $n=20$, $p=0.35$, $r=8$
• A1: Correct answer 0.161 (3 s.f.)

Common mistake: Using $(0.35)^{12}(0.65)^8$ instead of $(0.35)^8(0.65)^{12}$ — the powers must correspond to the number of successes and failures respectively.

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Question 2 [3 marks]

(a) [1 mark]

Answer: $a = 0.25$

Working:

All probabilities must sum to 1:

$0.1 + 0.2 + a + 0.3 + 0.15 = 1$

$0.75 + a = 1$

$a = 0.25$

(b) [2 marks]

Answer: $\mathrm{E}(Y) = 3.15$

Working:

$\mathrm{E}(Y) = \sum y \cdot \mathrm{P}(Y=y)$

$= 1(0.1) + 2(0.2) + 3(0.25) + 4(0.3) + 5(0.15)$

$= 0.1 + 0.4 + 0.75 + 1.2 + 0.75$

$= 3.20$

Correction: Let me recalculate:

$= 0.1 + 0.4 + 0.75 + 1.2 + 0.75 = 3.20$

Answer: $\mathrm{E}(Y) = 3.20$

Marking notes:

• (a) A1: Correct value $a = 0.25$
• (b) M1: Correct formula for expected value applied; A1: Correct answer 3.20

Common mistake: Forgetting to multiply each value by its probability before summing.

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Question 3 [3 marks]

(a) [2 marks]

Answer: $\mathrm{P}(38 < X < 47) = 0.641$ (to 3 s.f.)

Working:

$X \sim \mathrm{N}(42, 5^2)$

Standardise:

$z_1 = \frac{38 - 42}{5} = -0.80, \quad z_2 = \frac{47 - 42}{5} = 1.00$

$\mathrm{P}(38 < X < 47) = \mathrm{P}(-0.80 < Z < 1.00)$

$= \Phi(1.00) - \Phi(-0.80)$

$= 0.8413 - 0.2119$

$= 0.6294$

$\approx 0.629 \text{ (3 s.f.)}$

Answer: $0.629$ (to 3 s.f.)

(b) [1 mark]

Answer: Approximately 11 plants.

Working:

$\mathrm{P}(X > 50) = \mathrm{P}\left(Z > \frac{50-42}{5}\right) = \mathrm{P}(Z > 1.60) = 1 - \Phi(1.60) = 1 - 0.9452 = 0.0548$

Expected number: $200 \times 0.0548 = 10.96 \approx 11$ plants.

Marking notes:

• (a) M1: Correct standardisation and use of normal tables; A1: Correct answer 0.629
• (b) A1: Correct answer 11 (accept 10.96 or 11)

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Question 4 [4 marks]

(a) [1 mark]

Answer: $\bar{x} = 5.11$ hours (to 3 s.f.)

Working:

$\bar{x} = \frac{4.2 + 5.8 + 3.1 + 6.5 + 4.9 + 7.2 + 3.8 + 5.1 + 6.0 + 4.5}{10}$

$= \frac{51.1}{10} = 5.11$

(b) [3 marks]

Answer: $s^2 = 1.57$ (to 3 s.f.)

Working:

$s^2 = \frac{1}{n-1}\sum(x_i - \bar{x})^2 = \frac{1}{9}\sum(x_i - 5.11)^2$

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
4.2	−0.91	0.8281
5.8	0.69	0.4761
3.1	−2.01	4.0401
6.5	1.39	1.9321
4.9	−0.21	0.0441
7.2	2.09	4.3681
3.8	−1.31	1.7161
5.1	−0.01	0.0001
6.0	0.89	0.7921
4.5	−0.61	0.3721

$\sum(x_i - \bar{x})^2 = 14.569$

$s^2 = \frac{14.569}{9} = 1.6188... \approx 1.62 \text{ (3 s.f.)}$

Answer: $s^2 = 1.62$ (to 3 s.f.)

Marking notes:

• (a) A1: Correct mean 5.11
• (b) M1: Correct formula for unbiased variance with $n-1$ denominator; M1: Correct computation of squared deviations; A1: Correct answer 1.62

Common mistake: Using $n = 10$ instead of $n - 1 = 9$ in the denominator. The unbiased estimate requires $n-1$.

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Question 5 [4 marks]

(a) [2 marks]

Answer: $\mathrm{P}(X < 49.5) = 0.0478$ (to 3 s.f.)

Working:

$X \sim \mathrm{N}(52, 1.5^2)$

$z = \frac{49.5 - 52}{1.5} = \frac{-2.5}{1.5} = -1.667$

$\mathrm{P}(X < 49.5) = \Phi(-1.667) = 1 - \Phi(1.667) = 1 - 0.9522 = 0.0478$

(b) [2 marks]

Answer: $0.0579$ (to 3 s.f.)

Working:

First find $\mathrm{P}(\text{one bar} > 53)$:

$z = \frac{53 - 52}{1.5} = 0.667$

$\mathrm{P}(X > 53) = 1 - \Phi(0.667) = 1 - 0.7476 = 0.2524$

Let $Y \sim \mathrm{B}(8, 0.2524)$. Then:

$\mathrm{P}(Y = 3) = \binom{8}{3}(0.2524)^3(0.7476)^5$

$= 56 \times 0.01608 \times 0.2337$

$= 0.2099$

$\approx 0.210 \text{ (3 s.f.)}$

Marking notes:

• (a) M1: Correct standardisation; A1: Correct answer 0.0478
• (b) M1: Correct probability for one bar > 53g, then correct binomial setup; A1: Correct answer 0.210

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Question 6 [4 marks]

(a) [2 marks]

Working:

For a valid PDF, $\int_{-\infty}^{\infty} f(x)\,dx = 1$:

$\int_0^6 kx(6-x)\,dx = 1$

$k\int_0^6 (6x - x^2)\,dx = 1$

$k\left[3x^2 - \frac{x^3}{3}\right]_0^6 = 1$

$k\left[\left(108 - 72\right) - 0\right] = 1$

$k \times 36 = 1$

$k = \frac{1}{36} \quad \text{(shown)}$

(b) [2 marks]

Answer: $\mathrm{E}(X) = 3$

Working:

$\mathrm{E}(X) = \int_0^6 x \cdot \frac{1}{36}x(6-x)\,dx = \frac{1}{36}\int_0^6 (6x^2 - x^3)\,dx$

$= \frac{1}{36}\left[2x^3 - \frac{x^4}{4}\right]_0^6$

$= \frac{1}{36}\left[\left(432 - 324\right) - 0\right]$

$= \frac{1}{36} \times 108 = 3$

Marking notes:

• (a) M1: Correct integration setup; A1: Correct derivation showing $k = \frac{1}{36}$
• (b) M1: Correct expectation integral; A1: Correct answer 3

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Section B: Statistics & Probability — Applied [40 marks]

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Question 7 [5 marks]

(a) [3 marks]

Answer: $\mathrm{P}(1000 < X < 1350) = 0.749$ (to 3 s.f.)

Working:

$X \sim \mathrm{N}(1200, 150^2)$

$z_1 = \frac{1000 - 1200}{150} = -1.333, \quad z_2 = \frac{1350 - 1200}{150} = 1.000$

$\mathrm{P}(1000 < X < 1350) = \Phi(1.000) - \Phi(-1.333)$

$= 0.8413 - (1 - 0.9088) = 0.8413 - 0.0912 = 0.7501$

$\approx 0.750 \text{ (3 s.f.)}$

(b) [2 marks]

Answer: Approximately 11 bulbs.

Working:

$\mathrm{P}(X < 900) = \mathrm{P}\left(Z < \frac{900-1200}{150}\right) = \mathrm{P}(Z < -2.00) = 1 - \Phi(2.00) = 1 - 0.9772 = 0.0228$

Expected number: $500 \times 0.0228 = 11.4 \approx 11$ bulbs.

Marking notes:

• (a) M1: Correct standardisation; M1: Correct use of normal tables; A1: Correct answer 0.750
• (b) M1: Correct probability calculation; A1: Correct answer 11 (accept 11.4)

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Question 8 [6 marks]

(a) [2 marks]

Answer: $\bar{x} = 83.17$, $s^2 = 54.52$ (to 3 s.f.)

Working:

$\bar{x} = \frac{78+85+92+71+88+76+95+82+90+74+87+80}{12} = \frac{998}{12} = 83.166... \approx 83.2$

$s^2 = \frac{1}{11}\sum(x_i - 83.167)^2$

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
78	−5.167	26.698
85	1.833	3.361
92	8.833	78.028
71	−12.167	148.036
88	4.833	23.361
76	−7.167	51.361
95	11.833	140.028
82	−1.167	1.361
90	6.833	46.694
74	−9.167	84.036
87	3.833	14.694
80	−3.167	10.028

$\sum(x_i - \bar{x})^2 = 627.686$

$s^2 = \frac{627.686}{11} = 57.062... \approx 57.1 \text{ (3 s.f.)}$

So $\bar{x} = 83.2$, $s^2 = 57.1$ (to 3 s.f.)

(b) [4 marks]

Answer: There is sufficient evidence at the 5% level to conclude that the new method produces a higher mean score.

Working:

Step 1: Hypotheses

$H_0: \mu = 79$ (the mean score equals the national mean) $H_1: \mu > 79$ (the new method gives a higher mean — one-tailed test)

Step 2: Test statistic

Using $t$-test (population variance unknown, small sample):

$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{83.167 - 79}{\sqrt{57.062}/\sqrt{12}} = \frac{4.167}{2.176} = 1.915$

Step 3: Critical value

Degrees of freedom $= n - 1 = 11$. At 5% significance (one-tailed), $t_{0.05, 11} = 1.796$.

Step 4: Conclusion

Since $t = 1.915 > 1.796$, we reject $H_0$.

There is sufficient evidence at the 5% significance level to conclude that the new teaching method produces a higher mean score than the traditional method.

Marking notes:

• (a) A1: Correct mean 83.2; A1: Correct variance 57.1
• (b) B1: Correct hypotheses (one-tailed); B1: Correct test statistic calculation; B1: Correct critical value or comparison; B1: Correct conclusion in context

Common mistake: Using a two-tailed test when the question asks whether the new method is higher (one-tailed). Also, using $z$-test instead of $t$-test when population variance is unknown.

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Question 9 [6 marks]

(a) [2 marks]

Answer: $0.114$ (to 3 s.f.)

Working:

For a 2-minute interval, $\lambda = 4.2 \times 2 = 8.4$.

$X \sim \mathrm{Po}(8.4)$

$\mathrm{P}(X = 6) = \frac{e^{-8.4}(8.4)^6}{6!} = \frac{e^{-8.4} \times 351298.032}{720}$

$= e^{-8.4} \times 487.914 = 0.0002246 \times 487.914$

$= 0.1096 \approx 0.110 \text{ (3 s.f.)}$

(b) [2 marks]

Answer: $0.350$ (to 3 s.f.)

Working:

For a 30-second interval, $\lambda = 4.2 \times 0.5 = 2.1$.

$Y \sim \mathrm{Po}(2.1)$

$\mathrm{P}(Y \ge 3) = 1 - \mathrm{P}(Y \le 2)$

$= 1 - \left[\frac{e^{-2.1}(2.1)^0}{0!} + \frac{e^{-2.1}(2.1)^1}{1!} + \frac{e^{-2.1}(2.1)^2}{2!}\right]$

$= 1 - e^{-2.1}\left[1 + 2.1 + \frac{4.41}{2}\right]$

$= 1 - e^{-2.1} \times 5.305$

$= 1 - 0.1225 \times 5.305$

$= 1 - 0.6498 = 0.3502$

$\approx 0.350 \text{ (3 s.f.)}$

(c) [2 marks]

Answer: Approximately 1 interval.

Working:

Total 1-minute intervals in a week: $5 \times 8 \times 60 = 2400$ intervals.

For a 1-minute interval, $\lambda = 4.2$.

$\mathrm{P}(X = 0) = e^{-4.2} = 0.0150$

Expected number: $2400 \times 0.0150 = 36$ intervals.

Answer: Approximately 36 intervals.

Marking notes:

• (a) M1: Correct $\lambda = 8.4$ for 2 minutes; A1: Correct answer 0.110
• (b) M1: Correct $\lambda = 2.1$ and complement method; A1: Correct answer 0.350
• (c) M1: Correct $\mathrm{P}(X=0) = e^{-4.2}$; A1: Correct answer 36

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Question 10 [5 marks]

(a) [2 marks]

Answer: $31.6$ minutes (to 3 s.f.)

Working:

Midpoints: 5, 15, 25, 40, 65

$\bar{x} = \frac{8(5) + 15(15) + 22(25) + 25(40) + 10(65)}{80}$

$= \frac{40 + 225 + 550 + 1000 + 650}{80} = \frac{2465}{80} = 30.8125$

$\approx 30.8 \text{ minutes (to 3 s.f.)}$

(b) [2 marks]

Answer: $s = 16.9$ minutes (to 3 s.f.)

Working:

$s^2 = \frac{8(5^2) + 15(15^2) + 22(25^2) + 25(40^2) + 10(65^2)}{80} - (30.8125)^2$

$= \frac{200 + 3375 + 13750 + 40000 + 42250}{80} - 949.410$

$= \frac{99575}{80} - 949.410 = 1244.688 - 949.410 = 295.278$

$s = \sqrt{295.278} = 17.184... \approx 17.2 \text{ (3 s.f.)}$

(c) [1 mark]

Answer: The class $0 \le t < 10$ has the highest frequency density.

Working:

Frequency density = frequency ÷ class width:

• $0 \le t < 10$: $8/10 = 0.80$
• $10 \le t < 20$: $15/10 = 1.50$
• $20 \le t < 30$: $22/10 = 2.20$
• $30 \le t < 50$: $25/20 = 1.25$
• $50 \le t < 80$: $10/30 = 0.33$

The highest frequency density is $2.20$ in the class $20 \le t < 30$.

Answer: $20 \le t < 30$

Marking notes:

• (a) M1: Correct midpoints used; A1: Correct answer 30.8
• (b) M1: Correct variance formula for grouped data; A1: Correct answer 17.2
• (c) A1: Correct class $20 \le t < 30$

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Question 11 [5 marks]

(a) [2 marks]

Answer: $\dfrac{1}{22}$

Working:

Total balls = 12. Drawing 3 without replacement.

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

(b) [2 marks]

Answer: $\dfrac{2}{11}$

Working:

$\mathrm{P}(\text{2 red, 1 blue}) = \frac{\binom{5}{2}\binom{4}{1}}{\binom{12}{3}} = \frac{10 \times 4}{220} = \frac{40}{220} = \frac{2}{11}$

(c) [1 mark]

Answer: $\dfrac{3}{11}$

Working:

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

Marking notes:

• (a) M1: Correct combination approach; A1: Correct answer $\frac{1}{22}$
• (b) M1: Correct numerator and denominator; A1: Correct answer $\frac{2}{11}$
• (c) M1: Correct product of combinations; A1: Correct answer $\frac{3}{11}$

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Question 12 [5 marks]

(a) [3 marks]

Answer: $0.749$ (to 3 s.f.)

Working:

$X \sim \mathrm{N}(4.8, 0.6^2)$

$z_1 = \frac{4.0 - 4.8}{0.6} = -1.333, \quad z_2 = \frac{5.5 - 4.8}{0.6} = 1.167$

$\mathrm{P}(4.0 < X < 5.5) = \Phi(1.167) - \Phi(-1.333)$

$= 0.8784 - (1 - 0.9088) = 0.8784 - 0.0912 = 0.7872$

$\approx 0.787 \text{ (3 s.f.)}$

(b) [2 marks]

Answer: $5.57$ kg (to 3 s.f.)

Working:

We need $w$ such that $\mathrm{P}(X > w) = 0.10$, i.e., $\mathrm{P}(X \le w) = 0.90$.

From tables, $\Phi(z) = 0.90$ gives $z \approx 1.282$.

$w = 4.8 + 1.282 \times 0.6 = 4.8 + 0.7692 = 5.5692$

$\approx 5.57 \text{ kg (to 3 s.f.)}$

Marking notes:

• (a) M1: Correct standardisation; M1: Correct use of normal tables; A1: Correct answer 0.787
• (b) M1: Correct $z$-value for 90th percentile; A1: Correct answer 5.57

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Question 13 [7 marks]

(a) [3 marks]

Answer: $y = 7.00 + 7.86x$ (to 3 s.f.)

Working:

$n = 8$

$\sum x = 1.2 + 2.0 + 2.8 + 3.5 + 4.0 + 4.8 + 5.5 + 6.2 = 30.0$

$\sum y = 15 + 22 + 28 + 33 + 38 + 42 + 48 + 55 = 281$

$\bar{x} = 30.0/8 = 3.75$, $\bar{y} = 281/8 = 35.125$

$\sum x^2 = 1.44 + 4.00 + 7.84 + 12.25 + 16.00 + 23.04 + 30.25 + 38.44 = 133.26$

$\sum xy = 18.0 + 44.0 + 78.4 + 115.5 + 152.0 + 201.6 + 264.0 + 341.0 = 1214.5$

$S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 133.26 - \frac{900}{8} = 133.26 - 112.5 = 20.76$

$S_{xy} = \sum xy - \frac{(\sum x)(\sum y)}{n} = 1214.5 - \frac{30.0 \times 281}{8} = 1214.5 - 1053.75 = 160.75$

$b = \frac{S_{xy}}{S_{xx}} = \frac{160.75}{20.76} = 7.743... \approx 7.74 \text{ (3 s.f.)}$

$a = \bar{y} - b\bar{x} = 35.125 - 7.743 \times 3.75 = 35.125 - 29.036 = 6.089 \approx 6.09$

Answer: $y = 6.09 + 7.74x$ (to 3 s.f.)

(b) [1 mark]

Answer: For every additional thousand dollars spent on advertising, the weekly sales revenue increases by approximately $7,740.

(c) [2 marks]

Answer: Estimated revenue = $29.3 thousand. This is an interpolation (since $x = 3.0$ lies within the data range), so the estimate is reliable.

Working:

$y = 6.09 + 7.74(3.0) = 6.09 + 23.22 = 29.31 \approx 29.3 \text{ (thousand dollars)}$

Since $x = 3.0$ lies within the range of the data ($1.2 \le x \le 6.2$), this is interpolation and the estimate is reasonably reliable.

(d) [1 mark]

Answer: $r = 0.997$ (to 3 s.f.)

Working:

$S_{yy} = \sum y^2 - \frac{(\sum y)^2}{n}$

$\sum y^2 = 225 + 484 + 784 + 1089 + 1444 + 1764 + 2304 + 3025 = 11119$

$S_{yy} = 11119 - \frac{281^2}{8} = 11119 - 9870.125 = 1248.875$

$r = \frac{S_{xy}}{\sqrt{S_{xx} \cdot S_{yy}}} = \frac{160.75}{\sqrt{20.76 \times 1248.875}} = \frac{160.75}{\sqrt{25926.6}} = \frac{160.75}{161.018} = 0.9983...$

$\approx 0.998 \text{ (3 s.f.)}$

Marking notes:

• (a) M1: Correct calculation of $S_{xx}$ and $S_{xy}$; M1: Correct $b$ and $a$ values; A1: Correct equation $y = 6.09 + 7.74x$
• (b) B1: Correct interpretation in context (mentioning units: thousand dollars)
• (c) M1: Correct substitution; A1: Correct estimate with valid reliability comment
• (d) A1: Correct answer 0.998

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Section Mark Summary

Section	Marks
Section A: Questions 1–6	20
Section B: Questions 7–13	40
Total	60