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Question
The probability distribution of a random variable x is given as under:
P(X = x) = `{{:("k"x^2, "for" x = 1"," 2"," 3),(2"k"x, "for" x = 4"," 5"," 6),(0, "otherwise"):}`
where k is a constant. Calculate E(X)
Solution
Given that: P(X = x) = `{{:("k"x^2, "for" x = 1"," 2"," 3),(2"k"x, "for" x = 4"," 5"," 6),(0, "otherwise"):}`
∴ Probability distribution of random variable X is
| X | 1 | 2 | 3 | 4 | 5 | 6 | otherwise |
| P(X) | k | 4k | 9k | 8k | 10k | 12k | 0 |
We know that `sum_("i" = 1)^"n" "P"("X"_"i")` = 1
∴ k + 4k + 9k + 8k + 10k + 12k = 1
⇒ 44k = 1
⇒ k = `1/44`
E(X) = `sum_("i" = 1)^"n" "P"_"i""X"_"i"`
= 1 × k + 2 × 4k + 3 × 9k + 4 × 8k + 5 × 10k + 6 × 12k
= k + 8k + 27k + 32k + 50k + 72k
= 190k
= `190 xx 1/44`
= `95/22`
= 4.32 ......(Approx)
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