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If a random variable $X$ has a Poisson distribution with mean $5$, then the expectation $E\left [ \left ( x+2 \right )^{2} \right ]$ equals ___.
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80 votes

In Poisson distribution :

Mean  =  Variance  as n is large and p is small

And we know:

$\text{Variance}  =  E\left(X^{2}\right)  -   [ E(X) ]^{2}$  

$\Rightarrow E(X^{2})     =  [ E(X) ]^{2}  + \text{Variance}$

$\Rightarrow E(X^{2})     =   5^{2}   + 5$

$\Rightarrow E(X^{2})     =   30$

So, by linearity of expectation,

$E[(X + 2)^{2} ]   =   E[ X^{2} + 4X + 4 ]$

$\quad = E(X^{2})   +   4 E(X)  +  4$

$\quad =30  +  (4 \times 5)  +  4$

$\quad = 54$

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E(X2+4X+4) = E(X2) + E(4X) + E(4)

we know that V(x) = E(x2) - (E(x))2   =    5 = E(x2) - 25        =>  E(x2) = 30

So      30 + 4*5 + 4  =  54

–2 votes
–2 votes

a discrete frequency distribution which gives the probability of a number of "independent events" occurring in a fixed time.

E(cX)=cE(x)

E(c)=c  where c is constant

E(XY)=E(X)E(Y)  iff X and Y are independent variable

so here it is given poison distribution and poison distribution is used when event are all distinct.

E(X2+4X+4) = E(X2) + E(4X) + E(4)

=E(X)E(X)+4E(X)+4

=5*5+4*5+4=49 is correct ans

Answer:

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