If the mean of the following probability distribution of a random variable X: X: 0, 2, 4, 6, 8; P(X)

Q: Wait, why do we need to solve for $a$ and $b$ first? Can't we just find the variance directly?

That is a great question! While it would be wonderful to find a shortcut, the variance formula $\sigma^2 = E(X^2) - [E(X)]^2$ explicitly depends on the probabilities $P(X_i)$. Since those probabilities are defined in terms of $a$ and $b$, you cannot calculate the expected value of $X^2$ without knowing the specific values of $a$ and $b$. Think of $a$ and $b$ as the 'DNA' of this distribution; you must decode them before you can understand the spread of the data.

Q: I keep getting a different value for $E(X^2)$. Is there a common mistake I might be making?

The most common trap is forgetting to square the $X$ values! Remember, $E(X^2)$ is the sum of $x_i^2 \cdot P(x_i)$, not $(E(X))^2$. Students often calculate the mean, square it, and think they are done. Always write out the term $x_i^2$ explicitly for each entry in the table before multiplying by the probability. Precision is your best friend here.

Q: Why is the variance formula $\sigma^2 = E(X^2) - [E(X)]^2$ better than the definition $\sum (x_i - \mu)^2 P(x_i)$?

Both are mathematically equivalent, but the computational formula $\sigma^2 = E(X^2) - [E(X)]^2$ is almost always faster in JEE problems. Calculating $(x_i - \mu)^2$ for every term when $\mu$ is a fraction like $\frac{46}{9}$ will lead to messy arithmetic and a high chance of error. Using the computational formula keeps the fractions manageable until the very last step.

Question

If the mean of the following probability distribution of a random variable $X$: $$\begin{array}{|c|c|c|c|c|c|} \hline X & 0 & 2 & 4 & 6 & 8 \ \hline P(X) & a & 2a & a+b & 2b & 3b \ \hline \end{array}$$ is $46/9$, then the variance of the distribution is

Options:

(A) $\frac{173}{27}$

(B) $\frac{566}{81}$

(C) $\frac{151}{27}$

(D) $\frac{581}{81}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of a probability distribution. In the world of JEE Advanced, probability is about understanding the 'DNA' of a random variable.

We have a random variable $X$ that takes values $\{0, 2, 4, 6, 8\}$, and its behavior is governed by two unknown parameters, $a$ and $b$. Our mission is to find the variance, but to get there, we must first uncover the identity of these variables.

### The Normalization Constraint

Before we can calculate anything, we must respect the fundamental law of probability: the total probability of all possible outcomes must sum to exactly $1$. This is the normalization condition, the bedrock upon which all statistical models are built.

We look at the bottom row of our table: $a, 2a, (a+b), 2b, 3b$. Summing these up, we get:

$$a + 2a + (a+b) + 2b + 3b = 1$$

Grouping the terms, we find the following linear relationship:

$$4a + 6b = 1$$

This is our first anchor. It tells us that $a$ and $b$ are locked in a linear relationship.

### The Center of Mass (The Mean)

Nature provides us with the second piece of the puzzle: the mean, or the expected value $E(X)$. The problem states that $E(X) = \frac{46}{9}$. The formula for the mean is the weighted average of all outcomes:

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

Substituting our values, we get:

$$0(a) + 2(2a) + 4(a+b) + 6(2b) + 8(3b) = \frac{46}{9}$$

Expanding this carefully, we obtain $8a + 40b = \frac{46}{9}$. Dividing by $2$ to simplify, we arrive at:

$$4a + 20b = \frac{23}{9}$$

Now, we solve the system of equations: $4a + 6b = 1$ and $4a + 20b = \frac{23}{9}$. Subtracting the first from the second eliminates $a$ instantly:

$$14b = \frac{23}{9} - 1 = \frac{14}{9} \Rightarrow b = \frac{1}{9}$$

Substituting $b = \frac{1}{9}$ back into $4a + 6b = 1$, we find $4a + \frac{6}{9} = 1$, which simplifies to $4a = \frac{1}{3}$, or $a = \frac{1}{12}$. We have successfully decoded the distribution.

### The Spread (Variance)

Now that we know $a$ and $b$, we can tackle the variance. Variance, denoted by $\sigma^2$, measures the 'spread' of the random variable around its mean. We use the standard computational formula:

$$\sigma^2 = E(X^2) - [E(X)]^2$$

We already know $E(X) = \frac{46}{9}$. Now, we calculate $E(X^2)$, the expected value of the squares:

$$E(X^2) = 0^2(a) + 2^2(2a) + 4^2(a+b) + 6^2(2b) + 8^2(3b)$$

Substituting our values for $a$ and $b$ into this expression, we calculate:

$$E(X^2) = 24a + 280b = 24\left(\frac{1}{12}ight) + 280\left(\frac{1}{9}ight) = 2 + \frac{280}{9} = \frac{298}{9}$$

### Final Calculation

We are at the finish line. Using $E(X^2) = \frac{298}{9}$ and $E(X) = \frac{46}{9}$, the variance is:

$$\sigma^2 = \frac{298}{9} - \left(\frac{46}{9}ight)^2$$

$$\sigma^2 = \frac{298 	imes 9}{81} - \frac{2116}{81} = \frac{2682 - 2116}{81} = \frac{566}{81}$$

The final result is **$\frac{566}{81}$**. This value represents the quantification of the uncertainty in this specific system.

Animated Solution for Mathematics - Probability: If the mean of the following probability distribution of a random variable $X$ : $X P (X) 0 a 2 2 a 4 a + b 6 2 b 8 3 b$ is $46/9$ , then the variance of the distribution is

Select Answer:

Visualized Solution

Understanding the Probability Distribution

The Sum of Probabilities

Setting up the First Equation

Simplifying the First Equation

Defining the Mean $E (X)$

Setting up the Mean Equation

Simplifying the Mean Equation

Dividing the Mean Equation

Solving for $b$

Solving for $a$

The Variance Formula

Calculating $E (X^{2})$

Simplifying $E (X^{2})$

Numerical Value of $E (X^{2})$

Final Calculation of Variance

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The Normalization Constraint

The Center of Mass (The Mean)

The Spread (Variance)

Final Calculation

Similar Questions

Let $X$ be a random variable with distribution. $x P (X = x) - 2 \frac{1}{5} - 1 a 3 \frac{1}{3} 4 \frac{1}{5} 6 b$ If the mean of $X$ is 2.3 and variance of $X$ is $σ^{2}$ , then $100 σ^{2}$ is equal to :

From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. If the variance of $X$ is $σ^{2}$ , then $96 σ^{2}$ is equal to

Three rotten apples are accidently mixed with fifteen good apples. Assuming the random variable $x$ to be the number of rotten apples in a draw of two apples, the variance of $x$ is

Let $A = [a_{ij}]$ be a $2 \times 2$ matrix such that $a_{ij} \in {0, 1}$ for all i and j. Let the random variable X denote the possible values of the determinant of the matrix A. Then, the variance of X is :

A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let $X$ denote the number of defective pens. Then the variance of $X$ is

A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let $X$ be the number of white balls, among the drawn balls. If $σ^{2}$ is the variance of $X$ , then $100 σ^{2}$ is equal to

Three defective oranges are accidently mixed with seven good ones and on looking at them, it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If x denote the number of defective oranges, then the variance of x is