A random variable X has the probability distribution , E(x), E(X2) Also show that E(3x+2) = 3E(x) + 2 | Query Point Official

A Random Variable X Has a Probability Distribution | Find E(X), E(X²) and Show E(3X+2) = 3E(X)+2

A random variable X has the probability distribution 

x	-2	3	1
f(x)	1/3	1/2	1/6

Find 

(i)    E(x) 

(ii)    E(X2)

 Also show that  

(iii)    E(3x+2) = 3E(x) + 2

Solution

(i)     E(x) 

`E( x ) = \sum xf( x )`

`E( x ) =  - 2(\frac{1}{3}) + 3(\frac{1}{2}) + 1(\frac{1}{6})`

`E( x ) = \frac{ - 4 + 9 + 1}{6} = \frac{6}{6} = 1`

(ii)  E(X2)

 `E(X^2) = \sum x^2f( x )`

`E(X^2) = 4(\frac{1}{3}) + 9(\frac{1}{2}) + 1(\frac{1}{6})`

`E(X^2) = \frac{8+27+1}{6}=\frac{36}{6}=6`

(iii) E(3x+2) = 3E(x) + 2

Lets first solve L.H.S

`L.H.S = E(3x + 2)f( x )`

`L.H.S = E(3x + 2)(\frac{1}{3} + \frac{1}{2} + \frac{1}{6})`

`L.H.S = (3(- 2) + 2)(\frac{1}{3}) + (3( 3 ) + 2)(\frac{1}{2}) + (3( 1 ) + 2)(\frac{1}{6})`

`L.H.S = ( - 6 + 2)(\frac{1}{3}) + (9 + 2)(\frac{1}{2}) + (3 + 2)(\frac{1}{6})`

`L.H.S = (- 4)(\frac{1}{3}) + (11)(\frac{1}{2}) + ( 5 )(\frac{1}{6})`

`L.H.S = \frac{-8+33+5}{6}`

`L.H.S = \frac{30}{6} = 5`

Now R.H.S

`R.H.S = 3E( x right) + 2`

`R.H.S = 3( 1 ) + 2`

`R.H.S = 3 + 2`

`R.H.S = 5`

Hence,

R.H.S = L.H.S

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Frequently Asked Questions (FAQs)

What is the expected value E(X)?

The expected value E(X) of a random variable is the probability-weighted average of all possible values.

How do you calculate E(X²)?

To find E(X²), square each value of X, multiply by its probability, and then sum.

Why is E(3X + 2) = 3E(X) + 2?

This follows from the linearity of expectation — the expected value operator distributes over addition and scalar multiplication.

Can expected value be negative?

Yes — if the weighted average of values is negative.

Where are such problems used?

Expected value and probability distributions are used in statistics, economics, risk analysis, and decision making.

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