The Binomial Distribution and Sampling Distribution of the Mean

This focus area was worth 7 marks on NESA's sample paper for the 2024 syllabus, including 3 marks on the central limit theorem, and the binomial distribution alone was worth 3–9 marks in recent HSC papers. School trial papers often give it less space than the HSC does, so make sure you practise it.

• Bernoulli distributions: one trial with success probability $p$; $E(X)=p$ and $\text{Var}(X)=p(1-p)$.
• Binomial distributions: $X\sim \text{Bin}(n,p)$, the probability formula, expected frequencies, $E(X)=np$ and $\text{Var}(X)=np(1-p)$.
• Sampling distribution of the mean: populations and samples, the sample mean $\bar{X}$ as a random variable, $E(\bar{X})=\mu$ and $\text{Var}(\bar{X})=\frac{{\sigma }^2}{n}$.
• The central limit theorem: for large samples, $\bar{X}$ is approximately normal even when the population is not, and using it to find probabilities.

The 2024 syllabus does not include the normal approximation to the binomial distribution or the distribution of sample proportions $\hat{p}$. Those were in the old syllabus, so ignore them in older practice papers.

A Bernoulli trial has two outcomes, success or failure. The random variable $X$ takes the value 1 for success (probability $p$) and 0 for failure (probability $q=1-p$).

For example, if 30% of a factory's bolts are a particular size and one is chosen at random, the Bernoulli variable "bolt is that size" has mean 0.3 and variance $0.3\times 0.7=0.21$.

A binomial random variable $X$ counts the number of successes in $n$ independent Bernoulli trials, each with the same probability $p$. Write $X\sim \text{Bin}(n,p)$. Check the conditions in context: a fixed number of trials, two outcomes each time, independence and constant $p$.

for $r=0,1,\dots ,n$. The $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ counts the number of orders in which the $r$ successes can occur.

If the whole experiment of $n$ trials is repeated $N$ times, the expected frequency of exactly $r$ successes is $N\times P(X=r)$.

For "at least" questions, use the complement: $P(X\ge 1)=1-P(X=0)=1-(1-p{)}^n$.

Worked example 1: seeds in a tray

85% of a variety of seeds germinate, independently. A tray holds 12 seeds. Let $X$ be the number that germinate. Find (a) $P(X=10)$, (b) $P(X\ge 11)$, (c) the mean and standard deviation of $X$, and (d) the expected number of trays, out of 200, with exactly 10 seeds germinating.

$X\sim \text{Bin}(12,0.85)$.

(a)

(b)

(Adding the unrounded values gives 0.4435; adding the rounded ones gives 0.4434, so carry extra decimal places until the end.)

(c) $E(X)=12\times 0.85=10.2$ and $\text{Var}(X)=12\times 0.85\times 0.15=1.53$, so the standard deviation is $\sqrt{1.53}\approx 1.24$.

(d) The expected number is $200\times P(X=10)\approx 58.5$ trays (an expected frequency need not be a whole number).

A population is the whole group of interest; a sample is the part you actually measure. The population mean $\mu$ and variance ${\sigma }^2$ are usually unknown, so we estimate them from a random sample. Larger random samples (as a rule of thumb $n\ge 30$) give better estimates.

For a random sample $X_1,X_2,\dots ,X_n$, the sample mean is

Different samples give different values of $\bar{X}$, so $\bar{X}$ is itself a random variable. Its distribution over all possible samples of size $n$ is the sampling distribution of the mean. It has

so its standard deviation is $\frac{\sigma }{\sqrt{n}}$. The centre stays at $\mu$, but the spread shrinks as $n$ grows. To halve the standard deviation of $\bar{X}$, you need four times the sample size.

The central limit theorem

If the population has mean $\mu$ and variance ${\sigma }^2$, then for large enough $n$ (typically $n\ge 30$)

The striking part is that this holds whatever the shape of the population, even if it is skewed or discrete. That is why sample means are so useful for prediction in areas like polling, finance, agriculture and biology. If the population is itself normal, $\bar{X}$ is exactly normal for any $n$.

To find a probability, convert to a $z$-score using the standard deviation of $\bar{X}$, not of the population:

Worked example 2: battery life

The lifetime of a type of battery has mean 40 hours and standard deviation 8 hours, and its distribution is skewed. A random sample of 64 batteries is tested. Find the probability that the sample mean lifetime is between 39 and 42 hours.

Since $n=64\ge 30$, by the central limit theorem $\bar{X}$ is approximately normal with

The $z$-scores are $\frac{39-40}{1}=-1$ and $\frac{42-40}{1}=2$. From the standard normal table:

Using the empirical rule instead gives about $34\%+47.5\%=81.5\%$, a good check. Note that a single battery lasting between 39 and 42 hours is much less likely: the population is spread out far more than the sample means are.

• Using $\sigma$ instead of $\frac{\sigma }{\sqrt{n}}$ when finding a probability about a sample mean.
• Confusing variance and standard deviation: $\text{Var}(\bar{X})=\frac{{\sigma }^2}{n}$, but the $z$-score uses $\frac{\sigma }{\sqrt{n}}$.
• Saying the central limit theorem makes the population normal. It is the distribution of sample means that becomes approximately normal.
• Forgetting the $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ factor, or swapping the powers of $p$ and $1-p$.
• Calculating "at least 2" as $1-P(X=0)$, which leaves out $P(X=1)$.
• Using a binomial model when trials are not independent (for example, sampling without replacement from a small group).
• Rounding intermediate probabilities and then adding them, as noted in worked example 1.

• Always state the distribution: "$X\sim \text{Bin}(12,0.85)$" or "by the CLT, $\bar{X}\approx N(40,1^2)$". Markers often give a mark for identifying the model and its parameters.
• Justify a binomial model by naming the conditions in the context of the question.
• If you know $E(X)$ and $\text{Var}(X)$, divide to get $1-p=\frac{\text{Var}(X)}{E(X)}$, then find $p$ and $n$.
• "How many trials are needed so that the probability of at least one success exceeds..." leads to an inequality with logs: solve $1-(1-p{)}^n>c$ and round $n$ up.
• Give probabilities to 4 decimal places unless told otherwise, and show the substituted formula before the calculator answer.
• For CLT questions, say why it applies (sample size $n\ge 30$, random sample) before you use it.

• Permutations and combinations (ME11-PC): $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ counts the arrangements of successes.
• The binomial theorem (ME11-BT): the binomial probabilities are the terms of $(q+p{)}^n$, which is why they add to 1.
• Advanced, Random Variables (MA12-RV): discrete distributions, expected value, variance, the normal distribution, $z$-scores and the empirical rule are all assumed here.
• Advanced, Probability and Data (MA11-PD): independence and the complement rule.