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Probability distribution of random variable X is

X-2-1012
P(X)2/111/114/113/111/11

Then the value of E(X) is

Solution

✅ Correct Option: 1

The expected value of a random variable is calculated by multiplying each value by its probability and summing the results.

E(X)=∑[X×P(X)]E(X) = \sum [X \times P(X)]


Creating a table to organize the calculations:

XP(X)X × P(X)
-22/11(-2) × (2/11) = -4/11
-11/11(-1) × (1/11) = -1/11
04/11(0) × (4/11) = 0
13/11(1) × (3/11) = 3/11
21/11(2) × (1/11) = 2/11

Adding all values from the last column:

E(X)=−411+−111+0+311+211E(X) = \frac{-4}{11} + \frac{-1}{11} + 0 + \frac{3}{11} + \frac{2}{11}

E(X)=−4−1+0+3+211E(X) = \frac{-4 - 1 + 0 + 3 + 2}{11}

E(X)=011E(X) = \frac{0}{11}

E(X)=0E(X) = 0


Therefore, the value of E(X)=0E(X) = 0

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