Independent sums and sample means | Year 12

01 / LEARN

Understand the idea.

Combine independent measurements

For a linear combination, combine means using the signed coefficients. For independent variables, combine variances using squared coefficients. Thus subtracting an independent measurement still adds its variance. Standard deviations cannot be added in place of variances.

A linear combination of independent normal variables is normal. Independence is a property of the joint process, not something guaranteed by normal marginal distributions. If observations are dependent, covariance information is needed for the variance calculation.

\begin{aligned} E (a X + b Y) & = a μ_{X} + b μ_{Y} \\ Var (a X + b Y) & = a^{2} σ_{X}^{2} + b^{2} σ_{Y}^{2} \end{aligned}

Average independent observations

For $n$ independent observations from the same population, their sum has mean $n μ$ and variance $n σ^{2}$ . Dividing the sum by $n$ gives $E (\bar{X}) = μ$ and $Var (\bar{X}) = σ^{2} / n$ . Averaging reduces spread without shifting the centre.

If the population is normal, the sample mean is exactly normal for every positive $n$ . Otherwise normality may be an approximation for a sufficiently large sample. Repeating one value $n$ times is not the same experiment as obtaining $n$ independent observations.

\begin{array}{c} \bar{X} \sim N (μ, \frac{σ^{2}}{n}) \\ for independent normal observations \end{array}

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A normal difference

Independent $X \sim N (10, 4)$ and $Y \sim N (5, 9)$ . Find the distribution of X-Y.

Worked solution

1
Combine the means for the difference.
$E (X - Y) = 10 - 5 = 5$
Expectation follows the signed coefficients.
2
Combine variances using independence.
$Var (X - Y) = 1^{2} (4) + {(- 1)}^{2} (9) = 13$
The negative sign disappears when its coefficient is squared.
3
State the full distribution.
$X - Y \sim N (5, 13)$
Independent normal variables have a normal difference; the second parameter is its variance.

Answer

$X - Y \sim N (5, 13)$ .

EXAMPLE 02

A small sample from a normal population

Four independent values each follow $N (10, 4)$ . Find the distribution of their mean.

Worked solution

1
Define the mean and calculate its expectation.
$\bar{X} = \frac{X_{1} + X_{2} + X_{3} + X_{4}}{4}, E (\bar{X}) = \frac{4 (10)}{4} = 10$
Each observation comes from the same population, so their average has the same mean.
2
Scale the variance of the independent sum.
$Var (\bar{X}) = \frac{1}{4^{2}} (4 + 4 + 4 + 4) = 1$
The sum has four variance contributions, then division by four squares the scale factor.
3
State the exact sampling distribution.
$\bar{X} \sim N (10, 1)$
The original observations are normal, so no large-sample approximation is needed.

Answer

$\bar{X} \sim N (10, 1)$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Independent $X \sim N (10, 4)$ and $Y \sim N (5, 9)$ . Find Var(2X-Y).

Show answer & reasoning

25.

Use squared coefficients and independent variances: $Var (2 X - Y) = 2^{2} (4) + {(- 1)}^{2} (9) = 16 + 9 = 25$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Independent $X \sim N (10, 4)$ , $Y \sim N (5, 9)$ . Find the distribution of $X + Y$ .
2 marks
A normal population has mean 80 and SD 12. Find the distribution of a mean from 9 independent observations.
2 marks
For that sample mean, find $P (mean > 84)$ .
2 marks
If $X$ has variance 4, compare $Var (2 X)$ with the variance of $X_{1} + X_{2}$ for independent copies.
2 marks
Two normally distributed processing times come from the same machine cycle and are strongly related. Can you automatically add their variances?
2 marks
Independent normal $X, Y$ have means 20,12 and variances 9,4. Find $P (X - Y < 5)$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Independent $X \sim N (10, 4)$ , $Y \sim N (5, 9)$ . Find the distribution of $X + Y$ .
2 marks
Means add and independent variances add.
$N (15, 13)$ .
A normal population has mean 80 and SD 12. Find the distribution of a mean from 9 independent observations.
2 marks
The variance is $\frac{144}{9} = 16$ .
$N (80, 16)$ .
For that sample mean, find $P (mean > 84)$ .
2 marks
Standard error is 4; $z = \frac{84 - 80}{4} = 1$ .
0.1587.
If $X$ has variance 4, compare $Var (2 X)$ with the variance of $X_{1} + X_{2}$ for independent copies.
2 marks
$Var (2 X) = 4 (4) = 16$ .
Independent sum variance is $4 + 4 = 8$ .
16 versus 8.
Two normally distributed processing times come from the same machine cycle and are strongly related. Can you automatically add their variances?
2 marks
Dependence may produce nonzero covariance; independence was not established.
No.
Independent normal $X, Y$ have means 20,12 and variances 9,4. Find $P (X - Y < 5)$ .
3 marks
$X - Y \sim N (8, 13)$ .
Standardise 5 using SD $\sqrt{13}$ .
0.2027.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: continuous random variables, normal distributions and sampling distributions of sample means. Original lessons and practice.

Official curriculum source ↗

In this lesson

Understand the idea.

Combine independent measurements

Average independent observations

See the method in action.

A normal difference

Combine the means for the difference.

Combine variances using independence.

State the full distribution.

A small sample from a normal population

Define the mean and calculate its expectation.

Scale the variance of the independent sum.

State the exact sampling distribution.

Try it yourself.

Check your understanding

Keep practising