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MathsMediumMCQ2024 · 04 Apr Shift 2

Q69.If the mean of the following probability distribution of a random variable X : X 0 2 4 6 8 46 is , then the variance of the distribution is P(X) a 2a a + b 2b 3b 9 (1) 173 (2) 566 27 81 (3) 151 (4) 581 27 81

What This Question Tests

This question requires applying the definitions of mean and variance for a discrete probability distribution. It involves solving for unknown probabilities (a and b) and then calculating the variance.

Concepts Tested

Probability distributionMean of a discrete random variableVariance of a discrete random variable

Formulas Used

ΣP(X) = 1

Mean E(X) = Σx_i P(X=x_i)

Variance Var(X) = E(X²) - [E(X)]²

E(X²) = Σx_i² P(X=x_i)

📚 NCERT Sections This Tests

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