X, -2, -1, 0, 1, 2, P(X), 0.2, 0.1, 0.3, 0.2, 0.2, The variance of X will be : Option 1: 0.1 Option 2: 1.42 Option 3: 1.89 Option 4: 2.54

Answer: 1.89. Solution: Step 1: Calculate E(X): E(X) = x_i P(X=x_i) E(X) = (-2)(0.2) + (-1)(0.1) + (0)(0.3) + (1)(0.2) + (2)(0.2) E(X) = -0.4 - 0.1 + 0 + 0.2 + 0.4 = 0.1 Step 2: Calculate E(X²): E(X^2) = x_i^2 P(X=x_i) E(X^2) = (4)(0.2) + (1)(0.1) + (0)(0.3) + (1)(0.2) + (4)(0.2) E(X^2) = 0.8 + 0.1 + 0 + 0.2 + 0.8 = 1.9 Step 3: Find the variance: Var(X) = E(X^2) - [E(X)]^2 = 1.9 - (0.1)^2 = 1.9 - 0.01 = 1.89

Applied Mathematics

Probability Distribution

Medium

applied

$X$ -2 -1 0 1 2
$P(X)$ $0.2$ $0.1$ $0.3$ $0.2$ $0.2$

The variance of $X$ will be :

$X$	-2	-1	0	1	2
$P(X)$	$0.2$	$0.1$	$0.3$	$0.2$	$0.2$

$0.1$

$1.42$

$1.89$

$2.54$

✅ Correct Option: 3

Step 1: Calculate E(X):

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

$E(X) = (-2)(0.2) + (-1)(0.1) + (0)(0.3) + (1)(0.2) + (2)(0.2)$

$E(X) = -0.4 - 0.1 + 0 + 0.2 + 0.4 = 0.1$

Step 2: Calculate E(X²):

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

$E(X^2) = (4)(0.2) + (1)(0.1) + (0)(0.3) + (1)(0.2) + (4)(0.2)$

$E(X^2) = 0.8 + 0.1 + 0 + 0.2 + 0.8 = 1.9$

Step 3: Find the variance:

$\text{Var}(X) = E(X^2) - [E(X)]^2 = 1.9 - (0.1)^2 = 1.9 - 0.01 = 1.89$

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The probability distribution of a random variable X is given below :

X	0	1	2	3	4
P(X)	0.1	0.25	0.3	0.2	0.15

Then, $\text{Var}\left(\frac{X}{2}\right)$ is :

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2026: 18 May Shift 2

For 6 trials of an experiment, let X be a binomial variate which satisfies the relation $9P(X=4)=P(X=2)$ , then the probability of success is

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If $X$ is a random variable such that $X$ can take values 0,1, 2 or 3. Then the expectation of $X$ for the following data is :

$X$	0	1	2	3
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A box containing 10 wall clocks has three defective pieces in it. If random sample of two clocks is taken from the box, then the probability distribution of defective clocks is:

XXX-2-1012P(X)P(X)P(X)0.20.20.20.10.10.10.30.30.30.20.20.20.20.20.2 The variance of XXX will be :

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

$X$ -2 -1 0 1 2
$P(X)$ $0.2$ $0.1$ $0.3$ $0.2$ $0.2$

The variance of $X$ will be :