Why can't I just guess the values of $a$ and $b$?

In JEE Advanced, precision is everything. Guessing is a trap. You have two variables, so you need a system of two linear equations. It's the only way to guarantee accuracy.

Is the variance formula $\sigma^2 = E[X^2] - (E[X])^2$ always valid?

Yes, it is a universal identity for any random variable. It is often much faster than calculating deviations from the mean, especially when the mean is a decimal like 2.3.

What if I get a negative probability for $a$ or $b$?

That would be a red flag! Probabilities must always be between 0 and 1. If you get a negative value, re-check your arithmetic in the system of equations.

JEE Main 2021 (01 Sep Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Probability: 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 :

Enter Numerical Value:

Visualized Solution

Understanding the Probability Distribution

Given probability distribution table for random variable $X$ .
Values of $x$ : ${- 2, - 1, 3, 4, 6}$
Probabilities: ${\frac{1}{5}, a, \frac{1}{3}, \frac{1}{5}, b}$

The Sum of Probabilities Rule

Property: $\sum P (X = x) = 1$
The total probability must always equal exactly $1$ .

Setting up the Sum Equation

$\frac{1}{5} + a + \frac{1}{3} + \frac{1}{5} + b = 1$

Simplifying the First Equation

$(\frac{1}{5} + \frac{1}{3} + \frac{1}{5}) + a + b = 1$
$\frac{11}{15} + a + b = 1$
$a + b = \frac{4}{15}$ ... (Equation 1)

The Expected Value (Mean)

Mean $\overset{ˉ}{X} = \sum x_{i} P_{i} = 2.3$

Setting up the Mean Equation

$- 2 (\frac{1}{5}) + (- 1) (a) + 3 (\frac{1}{3}) + 4 (\frac{1}{5}) + 6 (b) = 2.3$

Simplifying the Mean Equation

$- 0.4 - a + 1 + 0.8 + 6 b = 2.3$
$1.4 - a + 6 b = 2.3$
$- a + 6 b = 0.9$ ... (Equation 2)

Solving for Variable $b$

Adding Eq (1) and Eq (2):
$(a + b) + (- a + 6 b) = \frac{4}{15} + 0.9$
$7 b = \frac{4}{15} + \frac{9}{10}$
$7 b = \frac{8 + 27}{30} = \frac{35}{30} = \frac{7}{6}$
$b = \frac{1}{6}$

Solving for Variable $a$

Substitute $b = \frac{1}{6}$ in Eq (1):
$a + \frac{1}{6} = \frac{4}{15}$
$a = \frac{4}{15} - \frac{1}{6} = \frac{8 - 5}{30}$
$a = \frac{3}{30} = \frac{1}{10}$

The Variance Formula

Variance $σ^{2} = E (X^{2}) - [E (X)]^{2}$
Where $E (X^{2}) = \sum x_{i}^{2} P_{i}$

Setting up $E (X^{2})$

$E (X^{2}) = (- 2)^{2} (\frac{1}{5}) + (- 1)^{2} (\frac{1}{10}) + 3^{2} (\frac{1}{3}) + 4^{2} (\frac{1}{5}) + 6^{2} (\frac{1}{6})$

Calculating $E (X^{2})$

$E (X^{2}) = 4 (0.2) + 1 (0.1) + 9 (\frac{1}{3}) + 16 (0.2) + 36 (\frac{1}{6})$
$E (X^{2}) = 0.8 + 0.1 + 3 + 3.2 + 6$
$E (X^{2}) = 13.1$

Final Variance Calculation

$σ^{2} = E (X^{2}) - (\overset{ˉ}{X})^{2}$
$σ^{2} = 13.1 - (2.3)^{2}$
$σ^{2} = 13.1 - 5.29$
$σ^{2} = 7.81$

The Final Answer

We need to find $100 σ^{2}$ .
$100 σ^{2} = 100 \times 7.81$
$100 σ^{2} = 781$

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The Sigma Insight: Random Variables and Probability Distributions

Solution Diagram

Analyzing the Probability Distribution

The fundamental law of probability distributions states that the sum of all probabilities must equal

1

. Given the values in our set, we establish the following equation:

\frac{1}{5} + a + \frac{1}{3} + \frac{1}{5} + b = 1

Simplifying this expression, we find our first anchor equation:

a + b = \frac{4}{15}

Determining the Mean

The mean, or the "center of mass" of the distribution, is defined by the expectation formula

E [X] = \sum x_{i} P_{i}

. Given that the mean is

2.3

, we set up the following equation:

- 2 (\frac{1}{5}) - 1 (a) + 3 (\frac{1}{3}) + 4 (\frac{1}{5}) + 6 (b) = 2.3

After performing the arithmetic, this simplifies to our second linear equation:

- a + 6 b = 0.9

Solving the System

We now have a system of two linear equations: 1)

a + b = \frac{4}{15}

- a + 6 b = 0.9

Adding these two equations eliminates

a

7 b = \frac{4}{15} + 0.9 = \frac{4}{15} + \frac{9}{10} = \frac{8 + 27}{30} = \frac{35}{30} = \frac{7}{6}

Solving for

b

gives

b = \frac{1}{6}

. Substituting this back into the first equation, we find

a = \frac{1}{10}

Calculating Variance

To find the variance, we use the identity

σ^{2} = E [X^{2}] - (E [X])^{2}

. First, we calculate the expected value of the square of the random variable:

E [X^{2}] = (- 2)^{2} (\frac{1}{5}) + (- 1)^{2} (\frac{1}{10}) + 3^{2} (\frac{1}{3}) + 4^{2} (\frac{1}{5}) + 6^{2} (\frac{1}{6})

E [X^{2}] = \frac{4}{5} + \frac{1}{10} + 3 + \frac{16}{5} + 6 = 13.1

Final Calculation

Now, we compute the variance

σ^{2}

σ^{2} = 13.1 - (2.3)^{2} = 13.1 - 5.29 = 7.81

The problem asks for the value of

100 σ^{2}

. Therefore:

100 \times 7.81 = 781

Animated Solution for Mathematics - Probability: 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 :

Enter Numerical Value:

Visualized Solution

Understanding the Probability Distribution

The Sum of Probabilities Rule

Setting up the Sum Equation

Simplifying the First Equation

The Expected Value (Mean)

Setting up the Mean Equation

Simplifying the Mean Equation

Solving for Variable $b$

Solving for Variable $a$

The Variance Formula

Setting up $E (X^{2})$

Calculating $E (X^{2})$

Final Variance Calculation

The Final Answer

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Probability Distribution

Determining the Mean

Solving the System

Calculating Variance

Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

Understanding the Probability Distribution

The Sum of Probabilities Rule

Setting up the Sum Equation

Simplifying the First Equation

The Expected Value (Mean)

Setting up the Mean Equation

Simplifying the Mean Equation

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

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

The Variance Formula

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E(X2)

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

Final Variance Calculation

The Final Answer

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Probability Distribution

Determining the Mean

Solving the System

Calculating Variance

Final Calculation

Similar Questions

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