ISS Previous year 2016 paper-01 Q. No. 2 Solution

ISS Previous year 2016 paper-01 Q. No. 2 Solution

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ISS Previous Year Question – Bernoulli Distribution

Previous year questions are very helpful for understanding the level and pattern of the UPSC Indian Statistical Service (ISS) examination. The following question tests the concept of Bernoulli distribution and variance of a linear transformation.

Question

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

Options:

(a) 0.24(b) 0.48(c) 0.60(d) 0.96

Solution

Since X follows a Bernoulli distribution,

P(X = 1) = p, P(X = 0) = 1 − p

For a Bernoulli random variable,

E(X) = p

Given in the question,

E(X) = 0.4

Therefore,

p = 0.4

Step 1: Variance of X

For a Bernoulli distribution,

Var(X) = p(1 − p)

Substitute p = 0.4

Var(X) = 0.4(1 − 0.4)

Var(X) = 0.4 × 0.6

Var(X) = 0.24

Step 2: Variance of (2X − 3)

We use the variance transformation rule:

Var(aX + b) = a² Var(X)

Here

a = 2b = −3

Therefore,

Var(2X − 3) = 2² Var(X)

Var(2X − 3) = 4 × 0.24

Var(2X − 3) = 0.96

Final Answer

Var(2X − 3) = 0.96

Correct option: (d) 0.96

Exam Insight for ISS Aspirants

In many UPSC ISS questions, variance under linear transformation is frequently asked. Always remember:

Var(aX + b) = a² Var(X)

The constant term b does not affect variance. Only the square of the coefficient of X matters.

ISS Previous year 2016 paper-01 Q. No. 2 Solution Download pdf Here