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Find the expected value, variance and standard deviation of the random variable whose p.m.f.’s are given below : x = x 1 2 3 ... n P (X = x) 1n 1n 1n ... 1n - Mathematics and Statistics

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Sum

Find the expected value, variance and standard deviation of the random variable whose p.m.f.’s are given below :

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

We construct the following table to find the expected value, variance, and standard deviation:

xi P (xi) xi·P (xi) xi 2·P (xi) = xi × xi·P (xi)
1 `1/n` `1/n` `1/n`
2 `1/n` `2/n` `2^2/n`
3 `1/n` `3/n` `3^2/n`
... ... ... ...
n `1/n` `n/n` `n^2/n`

From the table,

∑xi . P(xi) = `1/n + 2/n + 3/n + ... + n/n `

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

= `1/n sum _( r=1)^n r = 1/n xx (n(n+1 ))/2`

= `(n+1)/2`

∑xi2 . P(xi) = `1/n + 2^2/n + 3^2/n + ... + n^2/n `

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

= `1/n sum _( r=1)^n r^2 = 1/n xx (n(n+1 )(2n +1))/6`

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

∴ expected value = E(X) = ∑ xi · P(xi)

= `(n+1)/2`

Variance = V(X) = ∑ xi2 . P (xi) - [ ∑ xi·P (xi) ]2

= `((n+1)(2n + 1))/6  - ((n + 1)/2)^2`

= `(n+1)/2 [(2n + 1)/ 3 -(n + 1) /2 ]`

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

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

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

Standard deviation = `sqrt (V(X)`

= `sqrt ((n^2-1)/12)`

= `(sqrt (n^2-1))/(2sqrt 3)`

Concept: Variance of a Random Variable
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APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 7 Probability Distributions
Miscellaneous Exercise 2 | Q 10.3 | Page 244
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