Home / Edexcel A Level / Study notes

Edexcel IAL - Statistics 3- 3.2 Distribution of the Sample Mean- Study notes  - New syllabus

Edexcel IAL – Statistics 3- 3.2 Distribution of the Sample Mean -Study notes- New syllabus

Edexcel IAL – Statistics 3- 3.2 Distribution of the Sample Mean -Study notes -Edexcel A level Maths- per latest Syllabus.

Key Concepts:

  • 3.2 Distribution of the Sample Mean

Edexcel IAL Maths-Study Notes- All Topics

The Distribution of the Sample Mean

When a random sample is taken from a population, the sample mean \( \mathrm{\bar{X}} \) is itself a random variable. Understanding its distribution is essential for making inferences about the population mean.

Definition of the Sample Mean

For a random sample of size \( \mathrm{n} \) with observations \( \mathrm{X_1, X_2, \dots, X_n} \), the sample mean is defined as 

\( \mathrm{\bar{X} = \dfrac{1}{n}\sum_{i=1}^{n} X_i} \)

Since the sample observations are random variables, \( \mathrm{\bar{X}} \) is also a random variable.

Mean of the Sample Mean

The expected value of the sample mean is equal to the population mean:

\( \mathrm{E(\bar{X}) = \mu} \)

This means that \( \mathrm{\bar{X}} \) is an unbiased estimator of \( \mathrm{\mu} \).

Variance of the Sample Mean

The variance of the sample mean is given by

\( \mathrm{Var(\bar{X}) = \dfrac{\sigma^2}{n}} \)

As the sample size increases, the variance decreases, meaning the sample mean becomes more precise.

Distribution When the Population is Normal

If the population distribution is normal,

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

then the distribution of the sample mean is also normal:

\( \mathrm{\bar{X} \sim N\!\left(\mu,\,\dfrac{\sigma^2}{n}\right)} \)

This result may be used directly. No proof is required.

Interpretation

  • The sample mean has the same mean as the population
  • The spread of the sample mean decreases as \( \mathrm{n} \) increases
  • Larger samples give more reliable estimates of the population mean

Key Formula Summary

\( \mathrm{E(\bar{X}) = \mu} \)

\( \mathrm{Var(\bar{X}) = \sigma^2/n} \)

If \( \mathrm{X \sim N(\mu, \sigma^2)} \), then \( \mathrm{\bar{X} \sim N(\mu, \sigma^2/n)} \)

Example :

The lifetimes of batteries are normally distributed with mean \( \mathrm{\mu = 120} \) hours and standard deviation \( \mathrm{\sigma = 10} \) hours.

A random sample of \( \mathrm{n = 25} \) batteries is selected. Find the distribution of the sample mean lifetime.

▶️ Answer/Explanation

Since the population is normal, the sample mean is also normally distributed.

\( \mathrm{E(\bar{X}) = 120} \)

\( \mathrm{Var(\bar{X}) = \dfrac{10^2}{25} = 4} \)

Conclusion:

\( \mathrm{\bar{X} \sim N(120,\,4)} \)

Example :

The marks obtained by students in a test are normally distributed with mean \( \mathrm{60} \) and standard deviation \( \mathrm{8} \).

A random sample of \( \mathrm{n = 16} \) students is taken. Find the probability that the sample mean mark exceeds 62.

▶️ Answer/Explanation

For the sample mean:

\( \mathrm{E(\bar{X}) = 60} \)

\( \mathrm{Var(\bar{X}) = \dfrac{8^2}{16} = 4} \)

Standardise:

\( \mathrm{Z = \dfrac{62 – 60}{2} = 1} \)

Using normal tables:

\( \mathrm{P(\bar{X} > 62) = P(Z > 1) = 1 – 0.8413 = 0.1587} \)

Conclusion: The required probability is 0.1587.

Example :

A population has mean \( \mathrm{\mu} \) and standard deviation \( \mathrm{12} \).

A random sample of size \( \mathrm{n = 36} \) is taken. It is given that the standard deviation of the sample mean is 2.

Verify this result and comment on the effect of sample size.

▶️ Answer/Explanation

The variance of the sample mean is

\( \mathrm{Var(\bar{X}) = \dfrac{12^2}{36} = 4} \)

So the standard deviation is

\( \mathrm{\sqrt{4} = 2} \)

Conclusion: The result is verified. Increasing the sample size reduces the spread of the sample mean, making estimates more precise.

Scroll to Top