Solve the following problem : Find the expected value and variance of the r. v. X if its probability distribution is as follows. x 1 2 3 ... n P(X = x) 1n 1n 1n ... 1n - Mathematics and Statistics

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Sum

Solve the following problem :

Find the expected value and variance of the r. v. X if its probability distribution is as follows.

x 1 2 3 ... n
P(X = x) `(1)/"n"` `(1)/"n"` `(1)/"n"` ... `(1)/"n"`
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Solution

E(X) = \[\sum\limits_{i=1}^{n} x_i\cdot\text{P}(x_i)\]

= `1(1/"n") + 2(1/"n") + 3(1/"n") + ... + "n"(1/"n")`

= `(1 + 2 + 3 + .... + "n")/"n"` 

= `(1)/"n" xx ("n"("n" + 1))/(2)`

= `("n" + 1)/(2)`

E(X2) = \[\sum\limits_{i=1}^{n} x_i^2\text{P}(x_i)\]

= `1^2(1/"n") + 2^2(1/"n") + 3^2(1/"n") + ... + "n"^2(1/"n")`

= `(1^2 + 2^2 + 3^2 + .... + "n"^2)/"n"` 

= `(1)/"n" xx ("n"("n" + 1)(2"n" + 1))/(6)`

= `(("n" + 1)(2"n" + 1))/(6)`

∴ Var(X) = E(X2) – [E(X)]2 

= `(("n" + 1)(2"n" + 1))/(2 xx 3) - ("n" + 1)^2/(4)`

= `("n" + 1)/(2) ((2"n" + 1)/(3) - ("n" + 1)/(2))`

= `("n" + 1)/(2) ((4"n" + 2 - 3"n" - 3)/6)`

= `(("n" + 1)("n" - 1))/(12)`

= `("n"^2 - 1)/(12)`.

Concept: Probability Distribution of Discrete Random Variables
  Is there an error in this question or solution?
Chapter 8: Probability Distributions - Part I [Page 156]

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