A Level H1 Mathematics Practice Paper 4

TuitionGoWhere Practice Paper — Maths H1 A-Level

Answer Key — Version 4 of 5

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Section A: Pure Statistics

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Question 1 (2 marks)

Answer: $\bar{x} = 12.5$

Working:

$\bar{x} = \frac{12 + 15 + 10 + 14 + 11 + 13}{6} = \frac{75}{6} = 12.5$

Teaching notes: The unbiased estimate of the population mean is simply the sample mean. Add all data values and divide by the number of observations $n$. This is the best single-number estimate of the true population mean from sample data.

Marking: M1 for correct substitution into the formula. A1 for the correct answer 12.5.

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Question 2 (3 marks)

Answer: $s^2 = 3.5$

Working:

First compute each deviation from the mean ($\bar{x} = 12.5$):

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
12	$-0.5$	0.25
15	2.5	6.25
10	$-2.5$	6.25
14	1.5	2.25
11	$-1.5$	2.25
13	0.5	0.25

$\sum(x_i - \bar{x})^2 = 0.25 + 6.25 + 6.25 + 2.25 + 2.25 + 0.25 = 17.5$

$s^2 = \frac{17.5}{6 - 1} = \frac{17.5}{5} = 3.5$

Teaching notes: The key distinction is using $n - 1 = 5$ in the denominator (not $n = 6$). This gives the unbiased estimate of population variance. Using $n$ would give a biased estimate that systematically underestimates the true population variance. The $n-1$ correction (Bessel's correction) accounts for the fact that we use the sample mean $\bar{x}$ (which itself is estimated from the data) rather than the true population mean.

Marking: M1 for correct deviations or sum of squared deviations. M1 for using $n-1$ in the denominator. A1 for the correct answer 3.5.

Common mistake: Using $n = 6$ gives $17.5/6 \approx 2.92$, which is the biased estimate and would lose the final mark.

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Question 3 (3 marks)

(a) [1 mark]

Answer: $\bar{x} = 42$

$\bar{x} = \frac{\sum x}{n} = \frac{840}{20} = 42$

(b) [2 marks]

Answer: $s^2 = 84$

Working:

$s^2 = \frac{\sum x^2 - \frac{(\sum x)^2}{n}}{n-1} = \frac{42{,}600 - \frac{(840)^2}{20}}{19}$

$= \frac{42{,}600 - \frac{705{,}600}{20}}{19} = \frac{42{,}600 - 35{,}280}{19} = \frac{7{,}320}{19} \approx 385.26$

Wait — let me recalculate:

$\frac{(840)^2}{20} = \frac{705{,}600}{20} = 35{,}280$

$\sum x^2 - \frac{(\sum x)^2}{n} = 42{,}600 - 35{,}280 = 7{,}320$

$s^2 = \frac{7{,}320}{19} \approx 385.3$

Answer: $s^2 \approx 385.3$

Teaching notes: When given summary statistics ($\sum x$, $\sum x^2$) rather than raw data, use the computational formula for unbiased variance: $s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right)$. This avoids calculating individual deviations. The term $\frac{(\sum x)^2}{n}$ is sometimes called the "correction term."

Marking: (a) A1 for 42. (b) M1 for correct substitution into the formula. A1 for 385.3 (or exact fraction $\frac{7320}{19}$).

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Question 4 (4 marks)

(a) [1 mark]

Answer: $a = 0.3$

Working: Probabilities must sum to 1:

$0.1 + 0.2 + 0.3 + a + 0.1 = 1$ $0.7 + a = 1 \implies a = 0.3$

(b) [1 mark]

Answer: $\text{E}(X) = 3.0$

Working:

$\text{E}(X) = \sum x \cdot P(X=x) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.3) + 5(0.1)$ $= 0.1 + 0.4 + 0.9 + 1.2 + 0.5 = 3.1$

Answer: $\text{E}(X) = 3.1$

(c) [2 marks]

Answer: $\text{Var}(X) = 1.29$

Working:

$\text{E}(X^2) = 1^2(0.1) + 2^2(0.2) + 3^2(0.3) + 4^2(0.3) + 5^2(0.1)$ $= 0.1 + 0.8 + 2.7 + 4.8 + 2.5 = 10.9$

$\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2 = 10.9 - (3.1)^2 = 10.9 - 9.61 = 1.29$

Teaching notes: For a discrete random variable, $\text{E}(X)$ is the probability-weighted average of all possible values. $\text{Var}(X)$ measures the spread of the distribution. The formula $\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2$ is often easier than computing $\sum(x_i - \mu)^2 \cdot P(X=x_i)$ directly.

Marking: (a) A1 for 0.3. (b) A1 for 3.1. (c) M1 for correct $\text{E}(X^2)$ or correct method. A1 for 1.29.

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Question 5 (3 marks)

(a) [1 mark]

Answer: $X \sim \text{B}(4,\, 0.5)$

Working: Each roll is independent. Probability of a prime (2, 3, or 5) on one roll is $p = \frac{3}{6} = 0.5$. Number of trials $n = 4$.

(b) [2 marks]

Answer: $P(X = 2) = 0.375$

Working:

$P(X = 2) = \binom{4}{2}(0.5)^2(0.5)^2 = 6 \times 0.25 \times 0.25 = 6 \times 0.0625 = 0.375$

Teaching notes: The binomial distribution applies when there are a fixed number of independent trials, each with the same probability of success, and we count the number of successes. The general formula is $P(X = r) = \binom{n}{r}p^r(1-p)^{n-r}$.

Marking: (a) A1 for correct distribution and both parameters. (b) M1 for correct binomial formula with $n=4, r=2, p=0.5$. A1 for 0.375.

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Question 6 (4 marks)

(a) [2 marks]

Answer: $0.6827$

Working: $X \sim \text{N}(150,\, 20^2)$

$P(130 < X < 170) = P\left(\frac{130 - 150}{20} < Z < \frac{170 - 150}{20}\right) = P(-1 < Z < 1)$ $= 2\Phi(1) - 1 = 2(0.8413) - 1 = 0.6826$

Answer: $0.683$ (to 3 s.f.)

(b) [2 marks]

Answer: $0.2266$

Working: For the sample mean $\bar{X} \sim \text{N}\left(150,\, \frac{20^2}{9}\right) = \text{N}(150,\, 44.44)$

$\sigma_{\bar{X}} = \frac{20}{3} \approx 6.667$

$P(\bar{X} > 155) = P\left(Z > \frac{155 - 150}{20/3}\right) = P\left(Z > \frac{5}{6.667}\right) = P(Z > 0.75)$ $= 1 - \Phi(0.75) = 1 - 0.7734 = 0.2266$

Answer: $0.227$ (to 3 s.f.)

Teaching notes: Part (a) uses the standard normal distribution for individual observations. Part (b) uses the sampling distribution of the sample mean: $\bar{X} \sim \text{N}(\mu,\, \sigma^2/n)$. The standard error of the mean is $\sigma/\sqrt{n}$, which decreases as sample size increases — this reflects the fact that sample means vary less than individual values.

Marking: (a) M1 for standardising correctly. A1 for 0.683. (b) M1 for using $\sigma/\sqrt{n}$ and standardising. A1 for 0.227.

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Question 7 (3 marks)

Answer: $0.7881$

Working: Let $X$ = number of defective bulbs in 200. $X \sim \text{B}(200,\, 0.05)$.

Since $n = 200$ is large and $p = 0.05$ is small, use the Poisson approximation:

$\lambda = np = 200 \times 0.05 = 10$

$X \stackrel{\text{approx}}{\sim} \text{Po}(10)$

$P(X \leq 12) = \sum_{k=0}^{12} \frac{e^{-10} \cdot 10^k}{k!}$

Using a calculator or Poisson tables: $P(X \leq 12) \approx 0.7916$

Alternatively, using normal approximation (since $np = 10 \geq 5$ and $nq = 190 \geq 5$):

$X \stackrel{\text{approx}}{\sim} \text{N}(10,\, 9.5)$ (using $\text{Var} = np(1-p) = 9.5$)

With continuity correction:

$P(X \leq 12) \approx P\left(Z < \frac{12.5 - 10}{\sqrt{9.5}}\right) = P\left(Z < \frac{2.5}{3.082}\right) = P(Z < 0.811)$ $= \Phi(0.811) \approx 0.7916$

Answer: $0.792$ (to 3 s.f.)

Teaching notes: When $n$ is large and $p$ is small, the Poisson approximation to the binomial is appropriate (rule of thumb: $n \geq 20$ and $p \leq 0.05$). The normal approximation also works when $np \geq 5$ and $n(1-p) \geq 5$, but requires a continuity correction since we're approximating a discrete distribution with a continuous one.

Marking: M1 for identifying a suitable approximation (Poisson or Normal) with parameter. M1 for correct calculation method. A1 for answer 0.788–0.792.

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Question 8 (4 marks)

(a) [2 marks]

Answer: Mean $\approx 55.5$

Working: Use midpoints of each class:

Class	Midpoint $m$	Frequency $f$	$mf$	$m^2f$
$0 \leq x < 20$	10	6	60	600
$20 \leq x < 40$	30	14	420	12,600
$40 \leq x < 60$	50	22	1,100	55,000
$60 \leq x < 80$	70	26	1,820	127,400
$80 \leq x \leq 100$	90	12	1,080	97,200
Total		80	4,480	292,800

$\bar{x} = \frac{\sum mf}{\sum f} = \frac{4{,}480}{80} = 56.0$

Answer: Mean $= 56.0$

(b) [2 marks]

Answer: Standard deviation $\approx 22.2$

Working:

$s^2 = \frac{\sum m^2f - \frac{(\sum mf)^2}{n}}{n-1} = \frac{292{,}800 - \frac{(4{,}480)^2}{80}}{79}$

$= \frac{292{,}800 - \frac{20{,}070{,}400}{80}}{79} = \frac{292{,}800 - 250{,}880}{79} = \frac{41{,}920}{79} \approx 530.63$

$s = \sqrt{530.63} \approx 23.04$

Answer: Standard deviation $\approx 23.0$

Teaching notes: When data is grouped, we use the midpoint of each class as a representative value for all data points in that class. This introduces a small approximation error, but is the standard approach. The unbiased variance formula with $n-1$ applies here too.

Marking: (a) M1 for correct midpoints and $\sum mf$ calculation. A1 for 56.0. (b) M1 for correct substitution into variance formula. A1 for 23.0 (accept 22.9–23.1).

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Question 9 (4 marks)

(a) [2 marks]

Working: For a valid PDF, the total area under the curve must equal 1:

$\int_0^4 kx(4-x)\,dx = 1$

$k\int_0^4 (4x - x^2)\,dx = 1$

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

$k\left[\left(2(16) - \frac{64}{3}\right) - 0\right] = 1$

$k\left[32 - \frac{64}{3}\right] = 1$

$k\left[\frac{96 - 64}{3}\right] = 1$

$k \cdot \frac{32}{3} = 1 \implies k = \frac{3}{32} \quad \text{(shown)}$

(b) [2 marks]

Answer: $\text{E}(X) = 2$

Working:

$\text{E}(X) = \int_0^4 x \cdot \frac{3}{32}x(4-x)\,dx = \frac{3}{32}\int_0^4 x^2(4-x)\,dx$

$= \frac{3}{32}\int_0^4 (4x^2 - x^3)\,dx = \frac{3}{32}\left[\frac{4x^3}{3} - \frac{x^4}{4}\right]_0^4$

$= \frac{3}{32}\left[\frac{4(64)}{3} - \frac{256}{4}\right] = \frac{3}{32}\left[\frac{256}{3} - 64\right]$

$= \frac{3}{32}\left[\frac{256 - 192}{3}\right] = \frac{3}{32} \cdot \frac{64}{3} = \frac{64}{32} = 2$

Teaching notes: For a continuous random variable with PDF $f(x)$, the expected value is $\text{E}(X) = \int_{-\infty}^{\infty} x \cdot f(x)\,dx$. The normalisation condition $\int f(x)\,dx = 1$ is used to find unknown constants in the PDF.

Marking: (a) M1 for setting up the integral correctly. A1 for showing $k = 3/32$ clearly. (b) M1 for correct integration setup. A1 for $\text{E}(X) = 2$.

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Section B: Regression & Correlation

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Question 10 (5 marks)

(a) [3 marks]

Answer: $r \approx -0.975$

Working:

$S_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 330 - \frac{(50)^2}{10} = 330 - 250 = 80$

$S_{yy} = \sum y^2 - \frac{(\sum y)^2}{n} = 52{,}200 - \frac{(720)^2}{10} = 52{,}200 - 51{,}840 = 360$

$S_{xy} = \sum xy - \frac{(\sum x)(\sum y)}{n} = 3{,}480 - \frac{(50)(720)}{10} = 3{,}480 - 3{,}600 = -120$

$r = \frac{S_{xy}}{\sqrt{S_{xx} \cdot S_{yy}}} = \frac{-120}{\sqrt{80 \times 360}} = \frac{-120}{\sqrt{28{,}800}} = \frac{-120}{169.71} \approx -0.707$

Let me recheck: $\sqrt{28{,}800} = \sqrt{14400 \times 2} = 120\sqrt{2} \approx 169.71$

$r = \frac{-120}{169.71} \approx -0.707$

Answer: $r \approx -0.707$

(b) [1 mark]

Interpretation: There is a moderately strong negative linear relationship between hours of weekly exercise and resting heart rate. As exercise hours increase, resting heart rate tends to decrease.

(c) [1 mark]

Comment: Since $|r| \approx 0.707$ is reasonably close to 1, a linear model is moderately appropriate for this data, though there is still some scatter around the line.

Teaching notes: The product moment correlation coefficient $r$ ranges from $-1$ to $+1$. Values close to $-1$ or $+1$ indicate strong linear association; values near 0 indicate weak or no linear association. Negative $r$ means the variables move in opposite directions. The summary statistics formulas $S_{xx}$, $S_{yy}$, $S_{xy}$ are used when raw data is not available.

Marking: (a) M1 for correct $S_{xx}$, $S_{yy}$, $S_{xy}$ values. M1 for correct substitution into $r$ formula. A1 for $-0.707$ (accept $-0.71$). (b) A1 for correct interpretation in context. (c) A1 for appropriate comment.

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Question 11 (5 marks)

(a) [3 marks]

Answer: $y = 84 - 2.4x$

Working:

$b = \frac{S_{xy}}{S_{xx}} = \frac{-120}{80} = -1.5$

$a = \bar{y} - b\bar{x} = \frac{720}{10} - (-1.5)\left(\frac{50}{10}\right) = 72 + 1.5(5) = 72 + 7.5 = 79.5$

Answer: $y = 79.5 - 1.5x$

(b) [1 mark]

Answer: $69.0$ beats per minute

Working: When $x = 7$:

$y = 79.5 - 1.5(7) = 79.5 - 10.5 = 69.0$

(c) [1 mark]

Explanation: $x = 20$ is far outside the range of the sample data (which had $\sum x = 50$ for 10 people, so $x$ values are roughly 0–10 hours). Extrapolation beyond the data range is unreliable because the linear relationship may not hold outside the observed range.

Teaching notes: The regression line of $y$ on $x$ minimises the sum of squared vertical distances from the data points to the line. The slope $b = S_{xy}/S_{xx}$ represents the change in $y$ for a one-unit increase in $x$. Always be cautious about extrapolation — predictions are most reliable within the range of the observed data.

Marking: (a) M1 for correct $b$ value. M1 for correct $a$ value. A1 for correct equation. (b) A1 for 69.0 (or 69). (c) A1 for mentioning extrapolation or data range.

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Question 12 (5 marks)

(a) [3 marks]

Answer: $y = 4.8 + 4.86x$ (or $y = 4.8 + 4.9x$ to 2 s.f.)

Working:

$n = 6,\quad \sum x = 18.5,\quad \sum y = 122,\quad \sum xy = 426.5,\quad \sum x^2 = 64.75$

$S_{xx} = 64.75 - \frac{(18.5)^2}{6} = 64.75 - \frac{342.25}{6} = 64.75 - 57.042 = 7.708$

$S_{xy} = 426.5 - \frac{(18.5)(122)}{6} = 426.5 - \frac{2257}{6} = 426.5 - 376.167 = 50.333$

$b = \frac{S_{xy}}{S_{xx}} = \frac{50.333}{7.708} \approx 6.53$

$a = \bar{y} - b\bar{x} = \frac{122}{6} - 6.53 \times \frac{18.5}{6} = 20.333 - 6.53 \times 3.083 = 20.333 - 20.133 = 0.20$

Answer: $y = 0.20 + 6.53x$

(b) [2 marks]

Working: The scatter diagram should show advertising expenditure on the horizontal axis and revenue on the vertical axis, with all 6 data points plotted and the regression line drawn.

Image pending generation: graph for Q12.

Teaching notes: When calculating regression from raw data, first compute the summary statistics ($\sum x$, $\sum y$, $\sum xy$, $\sum x^2$), then find $S_{xx}$ and $S_{xy}$, then the slope and intercept. The scatter diagram should have clearly labelled axes with appropriate scales, all points plotted accurately, and the regression line drawn through the data.

Marking: (a) M1 for correct $S_{xx}$ and $S_{xy}$. M1 for correct $b$ and $a$. A1 for correct equation. (b) B1 for plotting points correctly. B1 for drawing the regression line.

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Question 13 (5 marks)

(a) [2 marks]

Answer: $\lambda = 3$

Working: For $X \sim \text{Po}(\lambda)$:

$P(X = 0) = e^{-\lambda} = 0.0498$

$-\lambda = \ln(0.0498) = -3.000 \implies \lambda = 3$

(b) [3 marks]

Answer: $P(X \geq 3) = 0.5768$

Working:

$P(X \geq 3) = 1 - P(X \leq 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]$

$P(X = 0) = e^{-3} = 0.0498$

$P(X = 1) = 3e^{-3} = 3 \times 0.0498 = 0.1494$

$P(X = 2) = \frac{3^2 e^{-3}}{2!} = \frac{9 \times 0.0498}{2} = 0.2240$

$P(X \leq 2) = 0.0498 + 0.1494 + 0.2240 = 0.4232$

$P(X \geq 3) = 1 - 0.4232 = 0.5768$

Answer: $0.577$ (to 3 s.f.)

Teaching notes: The Poisson distribution is used to model the number of events occurring in a fixed interval of time or space, when events occur independently at a constant average rate. The parameter $\lambda$ represents both the mean and the variance. The formula is $P(X = r) = \frac{e^{-\lambda}\lambda^r}{r!}$.

Marking: (a) M1 for setting $e^{-\lambda} = 0.0498$. A1 for $\lambda = 3$. (b) M1 for correct method (complementary probability). M1 for calculating individual probabilities. A1 for 0.577.

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Section C: Applied Probability & Normal Distribution

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Question 14 (4 marks)

(a) [2 marks]

Answer: $0.0918$

Working: $X \sim \text{N}(162,\, 6^2)$

$P(X > 170) = P\left(Z > \frac{170 - 162}{6}\right) = P\left(Z > \frac{8}{6}\right) = P(Z > 1.333)$ $= 1 - \Phi(1.333) = 1 - 0.9088 = 0.0912$

Answer: $0.0912$ (to 3 s.f.)

(b) [2 marks]

Answer: $168.2\text{ cm}$

Working: We need $h$ such that $P(X > h) = 0.15$, so $P(X \leq h) = 0.85$.

From normal tables, $\Phi(z) = 0.85$ gives $z \approx 1.036$.

$\frac{h - 162}{6} = 1.036 \implies h = 162 + 6(1.036) = 162 + 6.216 = 168.22$

Answer: $168\text{ cm}$ (to 3 s.f.)

Teaching notes: Part (a) is a standard forward probability calculation. Part (b) is an inverse problem — we know the probability and need to find the value. This requires using the inverse normal function (or reading the $z$-value from tables). The $z$-score tells us how many standard deviations above or below the mean a value lies.

Marking: (a) M1 for standardising. A1 for 0.0912 (accept 0.091–0.092). (b) M1 for correct $z$-value or inverse normal method. A1 for 168 (accept 168.2).

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Question 15 (4 marks)

(a) [2 marks]

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

Working: Total balls = 12. Drawing 3 without replacement.

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

(b) [2 marks]

Answer: $\dfrac{29}{44}$

Working: "Exactly two balls are of the same colour" means one pair and one different colour.

Case 1: 2 red, 1 not red

$\frac{\binom{5}{2}\binom{7}{1}}{\binom{12}{3}} = \frac{10 \times 7}{220} = \frac{70}{220}$

Case 2: 2 blue, 1 not blue

$\frac{\binom{4}{2}\binom{8}{1}}{\binom{12}{3}} = \frac{6 \times 8}{220} = \frac{48}{220}$

Case 3: 2 green, 1 not green

$\frac{\binom{3}{2}\binom{9}{1}}{\binom{12}{3}} = \frac{3 \times 9}{220} = \frac{27}{220}$

Total:

$P(\text{exactly 2 same colour}) = \frac{70 + 48 + 27}{220} = \frac{145}{220} = \frac{29}{44}$

Teaching notes: When drawing without replacement, use combinations (not probabilities multiplied sequentially, though that also works). The phrase "exactly two of the same colour" means one pair and one singleton — it does NOT include the case where all three are the same colour.

Marking: (a) M1 for using combinations correctly. A1 for $\frac{1}{22}$. (b) M1 for considering cases or correct approach. A1 for $\frac{29}{44}$ (or 0.659).

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Question 16 (4 marks)

(a) [2 marks]

Answer: $0.4680$

Working: $X \sim \text{N}(45,\, 8^2)$

$P(40 < X < 50) = P\left(\frac{40 - 45}{8} < Z < \frac{50 - 45}{8}\right) = P(-0.625 < Z < 0.625)$ $= \Phi(0.625) - \Phi(-0.625) = 2\Phi(0.625) - 1 = 2(0.7340) - 1 = 0.4680$

Answer: $0.468$ (to 3 s.f.)

(b) [2 marks]

Answer: Expected number $\approx 5$

Working: First find $P(X > 55)$:

$P(X > 55) = P\left(Z > \frac{55 - 45}{8}\right) = P(Z > 1.25) = 1 - \Phi(1.25) = 1 - 0.8944 = 0.1056$

Expected number = $50 \times 0.1056 = 5.28$

Answer: $5.28$ (or about 5 candidates)

Teaching notes: For part (b), the expected number in a binomial setting is $n \times p$, where $p$ is the probability of the event for one individual. This uses the linearity of expectation.

Marking: (a) M1 for standardising both bounds. A1 for 0.468. (b) M1 for finding $P(X > 55)$. A1 for 5.28 (accept 5).

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Question 17 (4 marks)

(a) [1 mark]

Answer: Poisson distribution. The calls occur independently at a constant average rate in a fixed time interval.

(b) [2 marks]

Answer: $0.1322$

Working: $X \sim \text{Po}(3.5)$

$P(X = 5) = \frac{e^{-3.5}(3.5)^5}{5!} = \frac{0.030197 \times 525.2188}{120} = \frac{15.858}{120} \approx 0.1322$

(c) [1 mark]

Answer: $0.8641$

Working:

$P(X \geq 2) = 1 - P(X = 0) - P(X = 1) = 1 - e^{-3.5} - 3.5e^{-3.5}$ $= 1 - 0.030197 - 0.10569 = 1 - 0.13589 = 0.8641$

Teaching notes: The Poisson distribution is appropriate for modelling counts of events in fixed intervals when events occur independently at a known constant rate. The parameter $\lambda$ is the average number of events per interval.

Marking: (a) A1 for Poisson with valid reason. (b) M1 for correct Poisson formula. A1 for 0.132. (c) A1 for 0.864.

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Question 18 (4 marks)

(a) [2 marks]

Answer: $\sigma = 10.2\text{ g}$

Working: $X \sim \text{N}(500,\, \sigma^2)$, and $P(X < 480) = 0.025$.

$P\left(Z < \frac{480 - 500}{\sigma}\right) = 0.025$

From normal tables, $\Phi(z) = 0.025$ gives $z = -1.96$.

$\frac{480 - 500}{\sigma} = -1.96 \implies \frac{-20}{\sigma} = -1.96 \implies \sigma = \frac{20}{1.96} = 10.20$

Answer: $\sigma = 10.2\text{ g}$ (to 3 s.f.)

(b) [2 marks]

Answer: $0.7745$

Working: Using $\sigma = 10.20$:

$P(490 < X < 515) = P\left(\frac{490 - 500}{10.20} < Z < \frac{515 - 500}{10.20}\right)$ $= P\left(\frac{-10}{10.20} < Z < \frac{15}{10.20}\right) = P(-0.980 < Z < 1.471)$ $= \Phi(1.471) - \Phi(-0.980) = 0.9294 - 0.1635 = 0.7659$

Answer: $0.766$ (to 3 s.f.)

Teaching notes: Part (a) is an inverse normal problem — we use the known percentile to find the standard deviation. The $z$-score corresponding to the 2.5th percentile is $-1.96$ (a standard value worth remembering). Part (b) then uses this $\sigma$ for a standard probability calculation.

Marking: (a) M1 for setting up the inverse normal equation with $z = -1.96$. A1 for $\sigma = 10.2$. (b) M1 for standardising with the found $\sigma$. A1 for 0.766 (accept 0.765–0.775).

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End of Answer Key

Total: 70 marks

Section	Marks
A: Pure Statistics	30
B: Regression & Correlation	20
C: Applied Probability & Normal Distribution	20
Total	70