ISS 2016 Statistics Paper-1 Solution: Question 2 (Variance of Bernoulli Transformation)

Continuing our question-by-question walk through the ISS Statistics Paper-1 previous year papers, today we move to the second item of the 2016 paper. If you followed along with Question 1 on the joint distribution of X and Y, this one is a nice change of pace — shorter, but it hides a very common mistake that trips up a lot of first-time solvers.

Quick Summary

• Topic: Bernoulli distribution and variance of a linear transformation

• Question Reference: ISS 2016, Statistics Paper-1, Question 2

• Correct Answer: Option (d) — 0.96

The Question, Exactly As Asked

Let X have a Bernoulli distribution with mean 0.4. What is the variance of (2X − 3)?

• (a) 0.24

• (b) 0.48

• (c) 0.6

• (d) 0.96

Step 1: What Does "Bernoulli Distribution with Mean 0.4" Actually Mean?

A Bernoulli random variable is the simplest possible random variable — it only ever takes two values, 0 or 1. Think of it as the outcome of a single coin toss where "success" is recorded as 1 and "failure" as 0. If the probability of success is p, then:

• P(X = 1) = p

• P(X = 0) = 1 − p

The mean (expected value) of a Bernoulli variable is simply E(X) = p. This is easy to check directly: E(X) = 1 × p + 0 × (1 − p) = p. So when the question tells us the mean is 0.4, it is really just telling us that p = 0.4 directly — no extra algebra needed to extract it.

Step 2: Find the Variance of X Itself

For any Bernoulli variable, there is a standard formula for variance:

Var(X) = p(1 − p)

Let's quickly see where this comes from, since it's worth understanding rather than memorizing blindly. Variance is defined as Var(X) = E(X²) − [E(X)]². Since X only takes values 0 and 1, X² is exactly the same as X (because 0² = 0 and 1² = 1). So E(X²) = E(X) = p. Substituting:

Var(X) = p − p² = p(1 − p)

Now plug in p = 0.4:

Var(X) = 0.4 × (1 − 0.4) = 0.4 × 0.6 = 0.24

So the variance of X on its own is 0.24. Notice this matches option (a) — which is exactly why (a) is placed there as a distractor. If you stop at this step and pick an answer, you've answered the wrong question; the paper is asking for the variance of (2X − 3), not the variance of X.

Step 3: The Rule for Variance Under a Linear Transformation

This is the real concept being tested here. Whenever a random variable is transformed linearly — that is, written in the form aX + b, where a and b are constants — there is a clean rule connecting Var(aX + b) to Var(X):

Var(aX + b) = a² × Var(X)

Let's derive this from scratch instead of just quoting it, so it actually sticks. By definition:

Var(aX + b) = E[(aX + b)²] − [E(aX + b)]²

Expand (aX + b)² = a²X² + 2abX + b². Taking expectation term by term:

E[(aX + b)²] = a²E(X²) + 2abE(X) + b²

Also, E(aX + b) = aE(X) + b, so squaring this gives:

[E(aX + b)]² = a²[E(X)]² + 2abE(X) + b²

Now subtract the second expression from the first. The 2abE(X) terms cancel exactly, and the b² terms cancel exactly, leaving:

Var(aX + b) = a²E(X²) − a²[E(X)]² = a²{E(X²) − [E(X)]²} = a² × Var(X)

This is a very important structural fact:adding or subtracting a constant never changes the spread (variance) of a distribution — it only shifts the whole distribution left or right on the number line, without stretching or compressing it. Multiplying by a constant a, on the other hand, stretches the spread by a factor of a, and since variance is a squared quantity, that stretch factor gets squared too.

Pro Tip: The single most common error on this type of question is forgetting to square the multiplier. Students often compute Var(2X − 3) as 2 × Var(X) − 3, dragging both the coefficient and the constant into the variance calculation exactly as they appear in the original expression. But variance only cares about "how spread out" the values are — a constant shift doesn't change spread at all, and a multiplicative stretch changes spread by the

Step 4: Apply the Rule to Our Specific Case

Here, our transformation is 2X − 3, so comparing with aX + b, we have a = 2 and b = −3. Using the rule:

Var(2X − 3) = 2² × Var(X) = 4 × Var(X)

We already found Var(X) = 0.24 in Step 2, so:

Var(2X − 3) = 4 × 0.24 = 0.96

Step 5: Match With the Options and Confirm Against the Official Key

Our computed value of 0.96 matches option (d)exactly. Checking against the official ISS 2016 Statistics Paper-1 answer key (Series A), the recorded correct response for Question 2 is also D. Our independently derived answer agrees fully with the official key — there is no discrepancy to flag here.

Why This Question Matters

This question looks tiny, but the underlying idea —how variance behaves under linear transformations — shows up constantly throughout statistical inference: standardizing a variable (subtracting the mean and dividing by the standard deviation), scaling data before regression, working with sums of independent variables, and deriving standard errors of estimators. Getting comfortable with Var(aX + b) = a²Var(X) at the Bernoulli level, where the numbers are simple, builds the muscle memory needed for much messier derivations later in the paper — with binomial, Poisson, or continuous distributions where the algebra gets heavier but the underlying logic is identical.

Frequently Asked Questions

Why is the variance of a Bernoulli variable given by p(1 − p)?

Because X only takes the values 0 and 1, X² equals X for every possible outcome, so E(X²) equals E(X), which is p. Substituting into the general variance formula Var(X) = E(X²) − [E(X)]² gives p − p², which factors neatly into p(1 − p).

Why doesn't subtracting 3 from 2X change the variance?

Variance measures how spread out a distribution is around its mean, not where the distribution is located on the number line. Subtracting a constant shifts every possible value down by exactly the same amount, so the distances between values — and hence the spread — stay identical. Only the mean shifts; the variance is untouched.

Why does the coefficient 2 get squared when computing variance?

Multiplying a random variable by 2 doubles the distance of every value from the mean, and variance is defined using squared distances. Doubling a distance and then squaring it multiplies the squared distance by four, which is exactly 2². This is why the general rule uses a², not a.

Is p(1 − p) at its maximum when p = 0.4?

No — p(1 − p) is maximized at p = 0.5, where it equals 0.25, the largest possible variance a Bernoulli variable can have. At p = 0.4 the variance is 0.24, slightly below that maximum, which makes sense because 0.4 is close to 0.5.

Does the same Var(aX + b) = a²Var(X) rule apply to other distributions, like Binomial or Normal?

Yes, this rule is completely general — it holds for any random variable with a finite variance, regardless of its distribution. The Bernoulli case here is just the simplest setting to practice it in before applying it to Binomial, Poisson, Normal, or any other distribution encountered later in the syllabus.

What is the standard deviation of (2X − 3) in this problem?

Since variance is 0.96, the standard deviation is simply its square root: √0.96, which is approximately 0.98. Standard deviation is always the non-negative square root of variance and is expressed in the same units as the original variable.

That wraps up Question 2 of the ISS 2016 Statistics Paper-1 series. If any step felt unclear or you spotted something worth double-checking, drop a comment below — and if this helped, consider sharing it with a fellow ISS aspirant who's working through the same paper.