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STA642 ASSIGNMENT NO. 1 FALL 2022 |
STA642 ASSIGNMENT NO. 1 FALL 2022 || 100% RIGHT SOLUTION || PROBABILITY DISTRIBUTIONS || BY VuTech
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Question 1
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
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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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