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The probability distribution of a random variable x is given as under:P(X = x) = kfor,,kfor,,otherwise{kx2 for x=1,2,32kx for x=4,5,60 otherwisewhere k is a constant. Calculate E(3X2) - Mathematics

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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(3X2)

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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(3X2) = 3[k + 4 × 4k + 9 × 9k + 16 × 8k + 25 × 10k + 36 × 12k]

= 3[k + 16k + 81k + 128k + 250k + 432k]

= 3[908k]

= `3 xx 908 xx 1/44`

= `2724/44`

= 61.9  ......(Approx)

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Chapter 13: Probability - Exercise [Page 278]

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NCERT Exemplar Mathematics [English] Class 12
Chapter 13 Probability
Exercise | Q 51.(ii) | Page 278

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