Let the mean and the standard deviation of the probability distribution (X: α, 1, 0, -3; P(X): 1/3,

Q: Wait, why do we square the condition $\sigma - \mu = 2$?

Great question! We square it because our standard formulas for variance involve $\sigma^2$, not $\sigma$. By squaring $\sigma = \mu + 2$, we get $\sigma^2 = (\mu + 2)^2$, which allows us to directly substitute the variance formula $E[X^2] - \mu^2$ into the equation. It is a classic algebraic trick to bridge the gap between the given condition and the statistical definitions.

Q: Is there a faster way to find $\alpha$ without all this algebra?

While it might look intimidating, this algebraic path is the most robust method. Sometimes, in JEE Advanced, the 'long' way is the only way to ensure you don't miss a subtle constraint. Trust the process—the terms will simplify beautifully, and the constants will cancel out, leaving you with a clean quadratic equation.

Q: What if I get two values for $\alpha$? How do I know which one to pick?

Excellent intuition! If you get two values, check the context of the problem. Often, one value might make the probability distribution invalid (e.g., if a probability becomes negative) or it might be excluded by the problem statement. In this case, $\alpha=0$ would make the distribution degenerate, so we select the non-zero value.

Question

Let the mean and the standard deviation of the probability distribution $$\begin{array}{|c|c|c|c|c|} \hline X & \alpha & 1 & 0 & -3 \ \hline P(X) & \frac{1}{3} & K & \frac{1}{6} & \frac{1}{4} \ \hline \end{array}$$ be $\mu$ and $\sigma$, respectively. If $\sigma - \mu = 2$, then $\sigma + \mu$ is equal to________

Sigma Percentile · Accepted Answer

### The Foundation: Conservation of Probability

Every probability distribution obeys a sacred law: the sum of all probabilities must be exactly $1$. We write this as $\sum P(X) = 1$.

By summing our probabilities:
$$\frac{1}{3} + K + \frac{1}{6} + \frac{1}{4} = 1$$

Using a common denominator of $12$, this becomes:
$$\frac{4 + 2 + 3}{12} + K = 1 \Rightarrow \frac{9}{12} + K = 1 \Rightarrow \frac{3}{4} + K = 1$$

Thus, we find the missing piece: **$K = \frac{1}{4}$**.

### The Statistical Toolkit: Mean and Variance

The mean, or expected value $\mu = E[X]$, is the weighted average of all possible outcomes. We calculate it as:
$$\mu = \sum X \cdot P(X) = \left( \alpha \cdot \frac{1}{3} ight) + \left( 1 \cdot \frac{1}{4} ight) + \left( 0 \cdot \frac{1}{6} ight) + \left( -3 \cdot \frac{1}{4} ight)$$

Simplifying this expression, we obtain:
$$\mu = \frac{\alpha}{3} + \frac{1}{4} - \frac{3}{4} = \frac{\alpha}{3} - \frac{1}{2}$$

Next, we calculate the expected value of the square, $E[X^2] = \sum X^2 \cdot P(X)$:
$$E[X^2] = \left( \alpha^2 \cdot \frac{1}{3} ight) + \left( 1^2 \cdot \frac{1}{4} ight) + \left( 0^2 \cdot \frac{1}{6} ight) + \left( (-3)^2 \cdot \frac{1}{4} ight) = \frac{\alpha^2}{3} + \frac{1}{4} + \frac{9}{4} = \frac{\alpha^2}{3} + \frac{5}{2}$$

### The Algebraic Bridge

We are given the condition $\sigma - \mu = 2$, which we rewrite as $\sigma = \mu + 2$. Squaring both sides yields:
$$\sigma^2 = (\mu + 2)^2 = \mu^2 + 4\mu + 4$$

We know the fundamental identity $\sigma^2 = E[X^2] - \mu^2$. By equating these two expressions for $\sigma^2$:
$$E[X^2] - \mu^2 = \mu^2 + 4\mu + 4 \Rightarrow E[X^2] = 2\mu^2 + 4\mu + 4$$

This is the **master equation** that connects our statistical variables to the unknown $\alpha$.

### The Resolution

Substituting our expressions for $\mu$ and $E[X^2]$ into the master equation:
$$\frac{\alpha^2}{3} + \frac{5}{2} = 2\left( \frac{\alpha}{3} - \frac{1}{2} ight)^2 + 4\left( \frac{\alpha}{3} - \frac{1}{2} ight) + 4$$

Expanding the right side and simplifying the terms:
$$\frac{\alpha^2}{3} + \frac{5}{2} = 2\left( \frac{\alpha^2}{9} - \frac{\alpha}{3} + \frac{1}{4} ight) + \frac{4\alpha}{3} - 2 + 4$$
$$\frac{\alpha^2}{3} + \frac{5}{2} = \frac{2\alpha^2}{9} - \frac{2\alpha}{3} + \frac{1}{2} + \frac{4\alpha}{3} + 2$$

Rearranging the terms leads to:
$$\frac{\alpha^2}{9} - \frac{2\alpha}{3} = 0 \Rightarrow \alpha \left( \frac{\alpha}{9} - \frac{2}{3} ight) = 0$$

Assuming $\alpha 
eq 0$ for a distinct distribution, we conclude **$\alpha = 6$**.

With $\alpha = 6$, we find $\mu = \frac{6}{3} - \frac{1}{2} = 1.5$. Using our condition $\sigma = \mu + 2$, we get $\sigma = 3.5$.

Finally, the sum is:
$$\sigma + \mu = 3.5 + 1.5 = \mathbf{5}$$

Animated Solution for Mathematics - Probability: Let the mean and the standard deviation of the probability distribution $X P (X) α \frac{1}{3} 1 K 0 \frac{1}{6} - 3 \frac{1}{4}$ be $μ$ and $σ$ , respectively. If $σ - μ = 2$ , then $σ + μ$ is equal to________

Enter Numerical Value:

Visualized Solution

Sum of Probabilities Property

Solving for $K$

Defining the Mean $μ$

Calculating $μ$ in terms of $α$

Calculating $E [X^{2}]$

Simplifying $E [X^{2}]$

Using the Condition $σ - μ = 2$

Equating Variance Expressions

Substituting $α$ Expressions

Solving for $α$

Finding $μ$ and $σ$

Final Answer: $σ + μ$

The Sigma Insight: Random Variables and Probability Distributions

The Foundation

The Statistical Toolkit

The Algebraic Bridge

The Resolution

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Enter Numerical Value:

Visualized Solution

Sum of Probabilities Property

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K

Defining the Mean .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } μ

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } μ in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E[X2]

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E[X2]

Using the Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } σ−μ=2

Equating Variance Expressions

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α Expressions

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } μ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } σ

Final Answer: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } σ+μ