1Let x be a random variable with the following probability d

1-Let x be a random variable with the following probability distribution

x

-3

6

9

F(x)

1/6

1/2

1/3

find E( g(x) where g(x)- (2x+1)^2

Calculate the variance for g(x)= (2x+1)^2

2- For a laboratory assignment, if the equipment is working, the probability densty function of the observed outcome X is:

f(x)=            2 (1-x) 0 <x<1

Otherwise

1-find the E(X)

2-Calculate the variance of X

Solution

A)Given

X        -3                    6                       9

P(X)       1/6                1/2                    1/3

and

Y=g(X)=(2x+1)²

then probability distribution of g(X) is given by

g(X)                  25             169                361

P(g(x))              1/6                1/2             1/3

we have to find E(g(X)) and Var(g(X))

so E(g(X))=(25*1)/6 +(169*1)/2 + (361*1)/3

               =4.17+84.5+120.33=209

var(g(X))=E((g(X))²)-[E(g(X))]²

      E[(g(X))²)=(25)²/6 + (169)²/2     + (361)²/3

                   =104.17 +14280.5 +43440.33=57825

then var(g(X))=57825 - (209)²

                    =57825 - 43681 =14144 answer

B) given

f(x)=2(1-x)    0<x<1

we have to find E(X) and var (X)

hence

E(X) =integrate (x*2(1-x)) from 0 to 1

             on integration we got

    E(X)=    [x² - 2x³/3 ] _{at 1}    -     [x²   -2x³/3]_{at 0}

            =1-    2/3 =1/3    answer

var(X)= E(X²) - [E(X)]²

E(X²) = integrate (x²*2(1-x)) from 0 to 1

            on integration we got

E(X²)= [(2x³/3 ) - (x⁴/2)] _{at 1} -   [(2x³/3) - (x⁴/2)] at 0

          =(2/3) -(1/2) = 1/6

then var(X) =(1/6) - (1/3)²       = (1/6) - (1/9) =1/18 answer