E(aX+b) and Var(aX+b) for discrete random variables

A discrete random variable $X$ has $E(X) = 6$ and $E(X^2) = 40$. A student writes: "$\text{Var}(X) = E(X^2) = 40$." What is wrong with this, and what is the correct value of $\text{Var}(X)$?

• The student has confused $E(X^2)$ with $\text{Var}(X)$; correctly, $\text{Var}(X) = E(X^2) - [E(X)]^2 = 40 - 36 = 4$

  Correct, and this is a documented, named confusion — a real WST01 examiner report records exactly this: "some still think E(X²) = Var(X)" (Jan 2021, Q4(b–d)). E(X²) is a raw second moment; Var(X) is that moment with the square of the mean subtracted off.

• BNothing is wrong — $E(X^2)$ and $\text{Var}(X)$ are the same thing by definition

  This is the misconception itself, stated as fact. If they were the same, the formula booklet would only need to print one of them — instead it prints Var(X) = E(X²) − μ² specifically because the two differ by [E(X)]², which is nonzero here (6² = 36).

• CThe student needs to subtract $E(X)$ without squaring it: $\text{Var}(X) = E(X^2) - E(X) = 40 - 6 = 34$

  Half right: something does need subtracting, but it's [E(X)]², the square of the mean, not the mean itself. The correct formula is Var(X) = E(X²) − [E(X)]², and dropping the square here is the same family of omission as forgetting to square the coefficient in Var(aX+b) — a squaring step skipped, just attached to a different formula.

• D$\text{Var}(X) = [E(X)]^2 = 36$

  This drops E(X²) from the formula entirely and keeps only the squared mean — the reverse mistake from option C, and neither term alone is the answer. Both E(X²) and [E(X)]² are required, and it's their DIFFERENCE that gives Var(X).

Traps tested: E x squared confused with var x · E x not squared in variance formula · E x squared term dropped

A discrete random variable $X$ has $E(X) = 12$ and $\text{Var}(X) = 5$. Given that $Y = 3X - 4$, find $E(Y)$ and $\text{Var}(Y)$.

• $E(Y) = 32$, $\text{Var}(Y) = 45$

  Correct. E(Y) = 3(12) − 4 = 32. Var(Y) = 3² × 5 = 9 × 5 = 45 — the coefficient squared, and the −4 contributing nothing at all to the variance.

• B$E(Y) = 32$, $\text{Var}(Y) = 3(5) - 4 = 11$

  The mean's formula (which does carry a −b) has been applied to variance by mistake, and the coefficient wasn't squared either: 3(5) − 4 = 11, not the correct 45. Variance has no b term, present or absent, squared or not.

• C$E(Y) = 32$, $\text{Var}(Y) = 3(5) = 15$

  The coefficient was left unsquared: 3 × 5 = 15, not 3² × 5 = 45. This is the exact mechanism a WST01 examiner report names directly at Jun 2024 Q2(c) and Q3(d) — the single most common error attached to this formula.

• D$E(Y) = 32$, $\text{Var}(Y) = 9(5) - 4 = 41$

  The coefficient WAS squared correctly (9 × 5 = 45) — but then the −4 was subtracted anyway, as though variance carried a b term the way the mean does. It doesn't: Var(aX+b) = a²Var(X), full stop, with nothing left over to add or subtract.

Traps tested: Variance treated like mean transformation · Coefficient not squared in variance · Additive constant included in variance

A discrete random variable $X$ has $\text{Var}(X) = 9$. Student 1 finds $\text{Var}(3X)$; Student 2 finds $\text{Var}(-3X)$. Which statement is correct?

• Both get $81$ — variance depends on $a^2$, and $3^2 = (-3)^2 = 9$, so the sign of $a$ never matters for spread

  Correct. Var(3X) = 3² × 9 = 81 and Var(−3X) = (−3)² × 9 = 81, identically. Squaring erases the sign completely — a transformation and its exact mirror image always produce the same spread, whatever they do to the mean.

• BStudent 1 gets $27$, Student 2 gets $-27$

  Forgetting to square gives a × Var(X) = 3 × 9 = 27 for Student 1, and carrying the sign through unsquared for Student 2 gives −27 — a negative variance, which is impossible by definition (variance is an average of squared, hence non-negative, deviations). A negative result here is itself proof a step was skipped, not a valid second case.

• CStudent 2's variance really is negative, because the transformation reflects the distribution

  Reflecting a distribution can flip the sign of its mean's displacement from zero, but variance is defined by a SQUARED deviation and is never negative, for any transformation whatsoever. "Reflected" changes which direction values sit relative to the mean; it does not, and cannot, make spread itself negative.

• DThey can't be compared without knowing $E(X)$

  Var(aX+b) = a²Var(X) depends only on a and Var(X) — E(X) plays no role in it whatsoever, which is exactly why the additive constant b (which DOES need E(X) for the mean formula) drops out of the variance formula entirely.

Traps tested: Coefficient not squared in variance · Variance assumed signed · Unnecessary mean dependency assumed

A company pays each employee a monthly bonus, in dollars, $B = 50N - 20$, where $N$ is the random number of new clients that employee brings in that month. Records show $E(N) = 4$ and $\text{Var}(N) = 1.5$. Find $E(B)$ and $\text{Var}(B)$.

• $E(B) = 180$, $\text{Var}(B) = 3750$

  Correct. E(B) = 50(4) − 20 = 180. Var(B) = 50² × 1.5 = 2500 × 1.5 = 3750 — the $20 fixed adjustment changes every employee's bonus by the same fixed amount and so changes nothing about how much bonuses VARY between employees; only the $50-per-client rate, squared, does that.

• B$E(B) = 180$, $\text{Var}(B) = 50(1.5) - 20 = 55$

  The mean's formula has been reused for variance: 50 × 1.5 − 20 = 55. Var(aX+b) has neither an unsquared a nor any b in it — both features of this working are wrong at once.

• C$E(B) = 180$, $\text{Var}(B) = 50(1.5) = 75$

  The coefficient 50 was left unsquared: 50 × 1.5 = 75, not 50² × 1.5 = 3750 — a factor of 50 out, which on a real payroll figure is the difference between a rounding error and a spreadsheet somebody should have double-checked.

• D$E(B) = 180$, $\text{Var}(B) = 2500(1.5) - 20 = 3730$

  The coefficient was squared correctly (2500 × 1.5 = 3750) but the −20 was then subtracted anyway. Variance has no b term to subtract, present or absent — the correct value is 3750, full stop.

Traps tested: Variance treated like mean transformation · Coefficient not squared in variance · Additive constant included in variance

Which of the following would you need to recall from memory in a WST01 exam, rather than being able to look up in the Mathematical Formulae and Statistical Tables booklet?

• $\text{Var}(aX + b) = a^2\text{Var}(X)$

  Correct. This is explicitly NOT in the booklet — verified against the spec's own statement that formulae "students are expected to know... will not appear in the booklet," and confirmed directly by checking the booklet's own Statistics S1 section, which prints the general definitions but not this shortcut.

• B$\text{Var}(X) = \sum x_i^2 P(X=x_i) - \mu^2$

  This IS printed in the booklet's Statistics S1 section — it's a general, computational definition that applies to any discrete random variable, which is exactly the kind of formula the booklet supplies.

• C$E(X) = \sum x_i P(X=x_i)$

  This is also printed in the booklet, for the same reason: it's the general definition, not a special-case shortcut.

• D$E(g(X)) = \sum g(x_i) P(X=x_i)$

  This general form is printed in the booklet too — it's the formula the two aX+b shortcuts are actually derived FROM, which is precisely why it's provided and they aren't: Pearson gives you the general tool and expects the specific simplification to be your own knowledge.

Traps tested: Base var formula assumed unlisted · Base e formula assumed unlisted · General e g x formula assumed unlisted