Effects of scale/shift | Year 12

01 / LEARN

Understand the idea.

Shifts and scales do different jobs

For $Y = a X + b$ , the mean transforms as $E (Y) = a E (X) + b$ . A fixed shift changes the centre but moves every outcome equally, so distances from the new mean are unaffected by the shift.

Scaling changes each deviation by $a$ . Therefore $Var (Y) = a^{2} Var (X)$ and $SD (Y) = ∣ a ∣ SD (X)$ . A negative scale reverses the order of outcomes, but spread remains nonnegative.

The distribution changes along with the labels

To transform a whole distribution, calculate $a x + b$ for each old value and carry its probability along. If several old values produce one new value, add their probabilities. A nonzero linear scale keeps distinct old values distinct.

For an event involving the transformed variable, substitute the rule and solve the inequality. Dividing by a negative number reverses its direction. The linear moment rules cannot be used unchanged for nonlinear transformations such as $Y = X^{2}$ .

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A score conversion

E(X) = 4 and Var(X) = 9. A converted score is Y = 3X − 2. Find E(Y), Var(Y) and SD(Y).

Worked solution

1
Transform the mean.
$E (Y) = 3 E (X) - 2 = 3 (4) - 2 = 10$
Expectation follows the same scale and shift as the outcomes.
2
Transform the variance.
$Var (Y) = 3^{2} Var (X) = 9 (9) = 81$
The scale acts on squared deviations; the fixed subtraction changes no spread.
3
Convert variance to standard deviation.
$SD (Y) = \sqrt{81} = 9$
Equivalently, $∣ 3 ∣ \sqrt{9} = 9$ , which checks the scale rule.

Answer

Mean 10; variance 81; standard deviation 9.

EXAMPLE 02

A negative scale and an event

X takes values 0, 1 and 3 with probabilities 0.2, 0.5 and 0.3. Set Y = 5 − 2X. Find P(Y < 4).

Worked solution

1
Map the support values.
$\begin{aligned} X = 0 & \Rightarrow Y = 5 (0.2) \\ X = 1 & \Rightarrow Y = 3 (0.5) \\ X = 3 & \Rightarrow Y = - 1 (0.3) \end{aligned}$
Each transformed value retains the probability of its originating outcome.
2
Translate the inequality.
$\begin{aligned} 5 - 2 X & < 4 \\ - 2 X & < - 1 \\ X & > \frac{1}{2} \end{aligned}$
Dividing by negative two reverses the inequality.
3
Add the included probabilities.
$P (Y < 4) = P (X = 1) + P (X = 3) = 0.5 + 0.3 = 0.8$
Both corresponding transformed values, $3$ and $- 1$ , are below $4$ .

Answer

P(Y < 4) = 0.8.

03 / YOUR TURN

Try it yourself.

Check your understanding

SD(X) = 2. What is SD(−4X + 7)?

Show answer & reasoning

The shift does not change spread. Multiply the original standard deviation by $∣ - 4 ∣$ : $SD (- 4 X + 7) = 4 (2) = 8$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

$E (X) = 5$ and $Var (X) = 4$ . Find $E (2 X + 3)$ and $Var (2 X + 3)$ .
2 marks
$E (X) = 8$ and $SD (X) = 3$ . Find the mean and SD of $Y = 10 - X$ .
2 marks
A random length $X$ in metres has mean 1.2 and variance 0.04. Convert both to centimetre-based quantities.
3 marks
A score $X$ has mean 50 and SD 10. Find mean and variance of $Z = \frac{X - 50}{10}$ .
2 marks
$X$ is −2 or 2 with equal probabilities. Find $E (X^{2})$ . Explain why squaring $E (X)$ gives the wrong answer.
3 marks
$X$ is 1, 2 or 4 with probabilities 0.2, 0.5, 0.3. For $Y = 6 - X$ , find $P (Y \geq 4)$ .
2 marks

Read the worked solutions

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

$E (X) = 5$ and $Var (X) = 4$ . Find $E (2 X + 3)$ and $Var (2 X + 3)$ .
2 marks
$E (2 X + 3) = 2 (5) + 3$ .
$Var (2 X + 3) = 2^{2} (4)$ .
13 and 16.
$E (X) = 8$ and $SD (X) = 3$ . Find the mean and SD of $Y = 10 - X$ .
2 marks
$E (Y) = 10 - 8$ .
The scale magnitude is 1, so the standard deviation stays 3.
Mean 2; SD 3.
A random length $X$ in metres has mean 1.2 and variance 0.04. Convert both to centimetre-based quantities.
3 marks
$Y = 100 X$ .
$E (Y) = 100 (1.2) = 120$ ; $Var (Y) = 100^{2} (0.04) = 400$ .
Mean 120 cm; variance 400 cm².
A score $X$ has mean 50 and SD 10. Find mean and variance of $Z = \frac{X - 50}{10}$ .
2 marks
$E (Z) = \frac{50 - 50}{10} = 0$ .
$Var (Z) = \frac{100}{100} = 1$ .
Mean 0; variance 1.
$X$ is −2 or 2 with equal probabilities. Find $E (X^{2})$ . Explain why squaring $E (X)$ gives the wrong answer.
3 marks
$X^{2}$ always equals 4, so its mean is 4.
$E (X) = 0$ , but a nonlinear transformation cannot be moved through expectation.
$E (X^{2}) = 4$ ; ${[E (X)]}^{2} = 0$ .
$X$ is 1, 2 or 4 with probabilities 0.2, 0.5, 0.3. For $Y = 6 - X$ , find $P (Y \geq 4)$ .
2 marks
$6 - X \geq 4$ gives $X \leq 2$ .
Add probabilities at $X = 1$ and $X = 2$ : $0.2 + 0.5$ .
0.7.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: discrete random variables and probability distributions, expected value and variance, transformations and independent sums, Bernoulli and binomial models. Original supporting lessons and exercises follow the approved Ascent chapter structure.

Official curriculum source ↗

In this lesson

Understand the idea.

Shifts and scales do different jobs

The distribution changes along with the labels

See the method in action.

A score conversion

Transform the mean.

Transform the variance.

Convert variance to standard deviation.

A negative scale and an event

Map the support values.

Translate the inequality.

Add the included probabilities.

Try it yourself.

Check your understanding

Keep practising