The idea, step by step
Shifts and scales do different jobs
If , then . A fixed shift changes the location of every outcome by the same amount. It does not spread the outcomes farther apart.
Variance obeys and standard deviation obeys . The shift disappears from deviations about the new mean. The scale is squared for variance, so a negative scale still gives a positive variance.
The distribution changes along with the labels
To form a full transformed distribution, map each old value x to ax + b and carry its probability with it. If different old values map to the same new value, combine their probabilities. A non-zero linear scale is one-to-one, whereas a square need not be.
The simple linear rules do not apply to an arbitrary nonlinear rule such as Y = X². Also reverse an inequality when dividing by a negative number in a probability calculation.
A common trap
Var(3X − 2) = 3Var(X) − 2.
Instead: Multiply variance by 3² and do not subtract the constant. Variance is based on squared deviations.
See the thinking
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).
- E(Y) = 3(4) − 2 = 10.
- Var(Y) = 3²(9) = 81.
- SD(Y) = √81 = 9; subtracting 2 has not changed the spread.
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).
- The transformed values are 5, 3 and −1 with the same respective probabilities.
- The event Y < 4 contains transformed values 3 and −1.
- Add their probabilities: 0.5 + 0.3 = 0.8. Equivalently, 5 − 2X < 4 means X > 0.5.
Check your understanding
SD(X) = 2. What is SD(−4X + 7)?
Show answer
8.
Multiply standard deviation by |−4| = 4.
Your turn
6 questions · 14 marks · Work at your own pace.
Read the questions online
- E(X) = 5 and Var(X) = 4. Find E(2X + 3) and Var(2X + 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 = (X − 50)/10.2 marks
- X is −2 or 2 with equal probabilities. Find E(X²). 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 ≥ 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(2X + 3) and Var(2X + 3).2 marks
E(2X + 3) = 2(5) + 3.
Var(2X + 3) = 2²(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 = 100X.
E(Y) = 100(1.2) = 120; Var(Y) = 100²(0.04) = 400.
Mean 120 cm; variance 400 cm². - A score X has mean 50 and SD 10. Find mean and variance of Z = (X − 50)/10.2 marks
E(Z) = (50 − 50)/10 = 0.
Var(Z) = 100/100 = 1.
Mean 0; variance 1. - X is −2 or 2 with equal probabilities. Find E(X²). Explain why squaring E(X) gives the wrong answer.3 marks
X² always equals 4, so its mean is 4.
E(X) = 0, but a nonlinear transformation cannot be moved through expectation.
E(X²) = 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 ≥ 4).2 marks
6 − X ≥ 4 gives X ≤ 2.
Add probabilities at X = 1 and X = 2: 0.2 + 0.5.
0.7.
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 ↗