The sampling distribution of the mean and the central limit theorem: the mean and variance of a sample mean, and probabilities that it lies within given bounds (new in the 2024 Extension 1 syllabus)

Because the sample size is $50 \geq 30$, the central limit theorem applies: the mean $\bar{X}$ of the $50$ amounts is approximately normal with mean $\mu = 8$ and standard deviation $\frac{6.5}{\sqrt{50}} \approx 0.919$.

Standardise: $z = \frac{10 - 8}{6.5/\sqrt{50}} \approx 2.18$ (to two decimal places).

From the table, $P(Z \leq 2.18) = 0.9854$, so $P(\bar{X} > 10) \approx 1 - 0.9854 = 0.0146$, about $1.46\%$.

NESA's marking guidelines award 3 marks for the correct solution; 2 marks for recognising the standard normal distribution and finding the $z$-score (or equivalent merit); 1 mark for recognising that $n \geq 30$ allows use of the central limit theorem, or for finding the standard deviation of $\bar{X}$ in terms of $n$ (or equivalent merit).

Source: NESA HSC Mathematics Extension 1 annotated sample examination materials (Mathematics Extension 1 11-12 Syllabus (2024)), Question 14(c), and marking guidelines.

With $n = 36 \geq 30$, the central limit theorem gives $\bar{X}$ approximately normal with mean $400$ and standard deviation $\frac{60}{\sqrt{36}} = 10$.

$z = \frac{385 - 400}{10} = -1.5$, so $P(\bar{X} < 385) \approx P(Z < -1.5) = 1 - 0.9332 = 0.0668$.

Markers expect the central limit theorem to be named or justified ($n \geq 30$), the standard deviation of $\bar{X}$ (not of one battery), and a correct use of symmetry to handle the negative $z$-score.

Mean of the sample mean. $E(\bar{X}) = \mu = 50$.

Standard deviation of the sample mean. $\operatorname{Var}(\bar{X}) = \frac{\sigma^2}{n}$, so

Marker's note: one mark for each value. Dividing by $n$ instead of $\sqrt{n}$ (giving $\frac{1}{3}$) is the usual error.

Variance. $\operatorname{Var}(\bar{X}) = \frac{\sigma^2}{n} = \frac{81}{49} \approx 1.65$.

Standard deviation. Take the square root: $\sqrt{\frac{81}{49}} = \frac{9}{7} \approx 1.29$.

Marker's note: one mark for the variance, one for the standard deviation. Keep the exact $\frac{9}{7}$ if you will use it later.

Apply the central limit theorem

Since $n = 64 \geq 30$, $\bar{X}$ is approximately normal with mean $170$ and standard deviation $\frac{8}{\sqrt{64}} = 1$.

Standardise

$z = \frac{172 - 170}{1} = 2$.

Find the probability

Marker's note: one mark for the standard deviation $1$ (with the CLT justified by $n \geq 30$), one for $z = 2$, one for $0.0228$.

Why normal

The population is not normal, but the central limit theorem says that for $n \geq 30$ the sampling distribution of the mean is approximately normal whatever the population's shape. Here $n = 36$.

Parameters

$E(\bar{X}) = 12$ and $\sigma_{\bar{X}} = \frac{12}{\sqrt{36}} = 2$.

Standardise both bounds

$z_1 = \frac{10 - 12}{2} = -1$ and $z_2 = \frac{15 - 12}{2} = 1.5$.

Marker's note: one mark for the CLT explanation, one for both $z$-scores, one for $0.7745$. Using $P(Z < -1) = 0.8413$ instead of $0.1587$ is the common slip.

Compute both. For $n = 25$: $\frac{20}{\sqrt{25}} = 4$. For $n = 100$: $\frac{20}{\sqrt{100}} = 2$.

Interpret. Multiplying the sample size by $4$ halves the standard deviation of $\bar{X}$, because it depends on $\frac{1}{\sqrt{n}}$. Larger samples give sample means that cluster more tightly around $\mu$.

Marker's note: one mark for both values, one for the $\frac{1}{\sqrt{n}}$ explanation (quadruple $n$, halve the spread).

Set up the condition. By the central limit theorem $\bar{X}$ is approximately $N\left( 500, \frac{16}{n} \right)$, with standard deviation $\frac{4}{\sqrt{n}}$. We need

Symmetric bounds. $P(-z < Z < z) = 0.95$ when $z = 1.96$, since $P(Z \leq 1.96) = 0.975$ leaves $0.025$ in each tail. So we need

Smallest whole number. $n = 62$.

Marker's note: one mark for $\sigma_{\bar{X}} = \frac{4}{\sqrt{n}}$, one for linking $0.95$ to $z = 1.96$, one for the inequality in $n$, one for rounding up to $62$ (not down to $61$).

Find the $z$-score

$P(Z > z) = 0.0228$ means $P(Z \leq z) = 0.9772$, so $z = 2$.

Standard deviation of $\bar{X}$

$\frac{10}{\sqrt{40}} \approx 1.5811$.

Solve for $\mu$

Marker's note: one mark for $z = 2$, one for $\frac{10}{\sqrt{40}}$, one for the equation, one for $\mu \approx 48.84$.

(a) Mean and variance of one roll. $E(X) = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{21}{6} = 3.5$.

$E(X^2) = \frac{1 + 4 + 9 + 16 + 25 + 36}{6} = \frac{91}{6}$, so

(b) Sample mean of 50 rolls. $n = 50 \geq 30$, so $\bar{X}$ is approximately normal with mean $3.5$ and standard deviation

$z = \frac{4 - 3.5}{0.2415} \approx 2.07$, so

Marker's note: two marks for (a) (the mean, then the variance via $E(X^2) - \mu^2$); in (b), one for the standard deviation of $\bar{X}$, one for $z \approx 2.07$, one for $0.0192$.