Home / Edexcel A Level / Study notes

Edexcel IAL - Statistics 3- 1.1 Linear Combinations of Independent Normal Random Variables- Study notes  - New syllabus

Edexcel IAL – Statistics 3- 1.1 Linear Combinations of Independent Normal Random Variables -Study notes- New syllabus

Edexcel IAL – Statistics 3- 1.1 Linear Combinations of Independent Normal Random Variables -Study notes -Edexcel A level Maths- per latest Syllabus.

Key Concepts:

  • 1.1 Linear Combinations of Independent Normal Random Variables

Edexcel IAL Maths-Study Notes- All Topics

Distribution of Linear Combinations of Independent Normal Random Variables

If random variables are normally distributed, then any linear combination of those variables is also normally distributed, provided the variables are independent. This result is fundamental and is widely used in probability and statistics.

Statement of the Result

Let

\( \mathrm{X \sim N(\mu_X, \sigma_X^2)} \)

\( \mathrm{Y \sim N(\mu_Y, \sigma_Y^2)} \)

where \( X \) and \( Y \) are independent random variables.

Then, for constants \( a \) and \( b \), the linear combination

\( \mathrm{Z = aX + bY} \)

is also normally distributed.

Mean of the Linear Combination

Using the properties of expectation:

\( \mathrm{E(Z) = aE(X) + bE(Y)} \)

\( \mathrm{E(Z) = a\mu_X + b\mu_Y} \)

Variance of the Linear Combination

Because \( X \) and \( Y \) are independent, the variance is

\( \mathrm{Var(Z) = a^2 Var(X) + b^2 Var(Y)} \)

\( \mathrm{Var(Z) = a^2 \sigma_X^2 + b^2 \sigma_Y^2} \)

There are no cross terms because of independence.

Generalisation

If \( X_1, X_2, \dots, X_n \) are independent normal random variables and

\( \mathrm{Z = a_1 X_1 + a_2 X_2 + \dots + a_n X_n} \)

then

\( \mathrm{Z \sim N\!\left(\sum a_i \mu_i,\; \sum a_i^2 \sigma_i^2 \right)} \)

This result may be used directly without proof.

Important Conditions

  • Each random variable must be normally distributed
  • The variables must be independent

If the variables are not independent, this result does not apply in this simple form.

Interpretation

  • Sums and differences of independent normal variables are normal
  • Scaling a normal variable changes the mean and variance but preserves normality

Example :

Let

\( \mathrm{X \sim N(10,\,4)} \), \( \mathrm{Y \sim N(6,\,1)} \)

where \( X \) and \( Y \) are independent.

Find the distribution of \( \mathrm{Z = X + Y} \).

▶️ Answer/Explanation

Mean:

\( \mathrm{E(Z) = E(X) + E(Y) = 10 + 6 = 16} \)

Variance:

\( \mathrm{Var(Z) = Var(X) + Var(Y) = 4 + 1 = 5} \)

Conclusion:

\( \mathrm{Z \sim N(16,\,5)} \)

Example :

Let

\( \mathrm{X \sim N(50,\,9)} \), \( \mathrm{Y \sim N(20,\,4)} \)

where \( X \) and \( Y \) are independent.

Find the distribution of \( \mathrm{W = 2X – Y} \).

▶️ Answer/Explanation

Mean:

\( \mathrm{E(W) = 2E(X) – E(Y) = 2(50) – 20 = 80} \)

Variance:

\( \mathrm{Var(W) = 2^2 Var(X) + (-1)^2 Var(Y)} \)

\( \mathrm{Var(W) = 4(9) + 1(4) = 40} \)

Conclusion:

\( \mathrm{W \sim N(80,\,40)} \)

Example :

The masses of apples and oranges are independent and normally distributed.

\( \mathrm{A \sim N(180,\,25)} \) g, \( \mathrm{O \sim N(120,\,16)} \) g

One apple and one orange are selected at random.

Find the distribution of the total mass \( \mathrm{T} \).

▶️ Answer/Explanation

Let \( \mathrm{T = A + O} \).

Mean:

\( \mathrm{E(T) = 180 + 120 = 300} \)

Variance:

\( \mathrm{Var(T) = 25 + 16 = 41} \)

Conclusion:

\( \mathrm{T \sim N(300,\,41)} \)

Scroll to Top