The probability distribution of a random variable x is given below: | X | P ||---|---|| 1 | 0.15 || 2 | 0.25 || 4 | 0.2 || 5 | 0.3 || 6 | 0.1 | What is the standard deviation of x ?

Option c: 1.69. Solution: Let's calculate the standard deviation of x systematically. 1. Compute the expected value (mean) E(X) : E(X) = sum xi P(xi) E(X) = 1(0.15) + 2(0.25) + 4(0.2) + 5(0.3) + 6(0.1) E(X) = 0.15 + 0.50 + 0.80 + 1.50 + 0.60 = 3.55 . 2. Compute the expected value of X2 , E(X2) : E(X2) = sum xi2 P(xi) E(X2) = 12(0.15) + 22(0.25) + 42(0.20) + 52(0.30) + 62(0.10) E(X2) = 1(0.15) + 4(0.25) + 16(0.20) + 25(0.30) + 36(0.10) E(X2) = 0.15 + 1.00 + 3.20 + 7.50 + 3.60 = 15.45 . 3. Compute the variance Var(X) : Var(X) = E(X2) - [E(X)]2 Var(X) = 15.45 - (3.55)2 = 15.45 - 12.6025 = 2.8475 . 4. Compute the standard deviation SD(X) : SD(X) = sqrt(Var)(X) = sqrt(2.8475) ≈ 1.68745 . Rounding to two decimal places, we get 1.69 . Hence, Option C is the correct answer.

The probability distribution of a random variable $x$ is given below: | X | P ||---|---|| 1 | $0.15$ || 2 | $0.25$ || 4 | $0.2$ || 5 | $0.3$ || 6 | $0.1$ | What is the standard deviation of $x$?

The correct answer is Option c: $1.69$. Let's calculate the standard deviation of $x$ systematically. 1. Compute the expected value (mean) $E(X)$: $E(X) = \sum x_i P(x_i)$ $E(X) = 1(0.15) + 2(0.25) + 4(0.2) + 5(0.3) + 6(0.1)$ $E(X) = 0.15 + 0.50 + 0.80 + 1.50 + 0.60 = 3.55$. 2. Compute the expected value of $X^2$, $E(X^2)$: $E(X^2) = \sum x_i^2 P(x_i)$ $E(X^2) = 1^2(0.15) + 2^2(0.25) + 4^2(0.20) + 5^2(0.30) + 6^2(0.10)$ $E(X^2) = 1(0.15) + 4(0.25) + 16(0.20) + 25(0.30) + 36(0.10)$ $E(X^2) = 0.15 + 1.00 + 3.20 + 7.50 + 3.60 = 15.45$. 3. Compute the variance $\text{Var}(X)$: $\text{Var}(X) = E(X^2) - [E(X)]^2$ $\text{Var}(X) = 15.45 - (3.55)^2 = 15.45 - 12.6025 = 2.8475$. 4. Compute the standard deviation $\text{SD}(X)$: $\text{SD}(X) = \sqrt{\text{Var}(X)} = \sqrt{2.8475} \approx 1.68745$. Rounding to two decimal places, we get $1.69$. Hence, **Option C** is the correct answer.

ProbabilityMCQPYQ Dec. 21Question 2836 of 295

All Questions

The probability distribution of a random variable $\displaystyle x$ is given below: | X | P ||---|---|| 1 | $\displaystyle 0.15$ || 2 | $\displaystyle 0.25$ || 4 | $\displaystyle 0.2$ || 5 | $\displaystyle 0.3$ || 6 | $\displaystyle 0.1$ | What is the standard deviation of $\displaystyle x$ ?

Options

\displaystyle 1.49

\displaystyle 1.56

\displaystyle 1.69

\displaystyle 1.72

For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option c — $\displaystyle 1.69$

All Options:

A $\displaystyle 1.49$
B $\displaystyle 1.56$
C $\displaystyle 1.69$
D $\displaystyle 1.72$

Detailed Solution & Explanation

Let's calculate the standard deviation of

\displaystyle x

systematically. 1. Compute the expected value (mean)

\displaystyle E(X)

\displaystyle E(X) = \sum x_i P(x_i)

\displaystyle E(X) = 1(0.15) + 2(0.25) + 4(0.2) + 5(0.3) + 6(0.1)

\displaystyle E(X) = 0.15 + 0.50 + 0.80 + 1.50 + 0.60 = 3.55

. 2. Compute the expected value of

\displaystyle X^2

\displaystyle E(X^2)

\displaystyle E(X^2) = \sum x_i^2 P(x_i)

\displaystyle E(X^2) = 1^2(0.15) + 2^2(0.25) + 4^2(0.20) + 5^2(0.30) + 6^2(0.10)

\displaystyle E(X^2) = 1(0.15) + 4(0.25) + 16(0.20) + 25(0.30) + 36(0.10)

\displaystyle E(X^2) = 0.15 + 1.00 + 3.20 + 7.50 + 3.60 = 15.45

. 3. Compute the variance

\displaystyle \text{Var}(X)

\displaystyle \text{Var}(X) = E(X^2) - [E(X)]^2

\displaystyle \text{Var}(X) = 15.45 - (3.55)^2 = 15.45 - 12.6025 = 2.8475

. 4. Compute the standard deviation

\displaystyle \text{SD}(X)

\displaystyle \text{SD}(X) = \sqrt{\text{Var}(X)} = \sqrt{2.8475} \approx 1.68745

. Rounding to two decimal places, we get

\displaystyle 1.69

. Hence, **Option C** is the correct answer.

About This Chapter: Probability

Paper

Paper 3: Quantitative Aptitude

Weightage

5-7 Marks

Key Topics

Probability Operations, Expected Value

A logic-heavy chapter dealing with random experiments, events (mutually exclusive, exhaustive), set theory probability, conditional probability, and Bayes' Theorem. It forms the basis for Theoretical Distributions.

View Official ICAI Syllabus

Exam Strategy Tip

Always draw a quick Venn Diagram or tree when faced with 'At least one' or 'Only A but not B' wording. It saves you from double-counting.

Key Concepts to Understand

Probability

A numerical measure of the likelihood of an event occurring, expressed as a number between 0 (impossible) and 1 (certain). Formula: P(A) = Number of favorable outcomes / Total number of equally likely outcomes.

Standard Deviation

A measure of the dispersion or spread of values in a data set around the mean. It is the square root of Variance. A lower standard deviation indicates data points are clustered close to the mean; a higher value indicates greater spread.

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