Q. 515.0( 1 Vote )

The probability d

Answer :

Given:


We know that,


Sum of the probabilities = 1



k + 4k + 9k + 8k + 10k + 12k = 1


44k = 1



(i) To find: E(X)


We know that, μ = E(X)



or


E(X) = ΣXP(X)


= 1 × k + 2 × 4k + 3 × 9k + 4 × 8k + 5 × 10k + 6 × 12k


= k + 8k + 27k + 32k + 50k + 72k


= 190k




= 4.32


(ii) To find: E(3X2)


Firstly, we find E(X2)


We know that,


E(X2) = ΣX2P(X)


= 12 × k + 22 × 4k + 32 × 9k + 42 × 8k + 52 × 10k + 62 × 12k


= k + 16k + 81k + 128k + 250k + 432k


= 908k



= 20.636


20.64


E(3X2) = 3 × 20.64 = 61.92


(iii) P(X ≥ 4) = P(X = 4) + P(X = 5) + P(X = 6)


= 8k + 10k + 12k


= 30k



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