Solve the following problem : Let the p. m. f. of the r. v. X be P(x)={3-x10, for x=-1, 0, 1, 2.0otherwise.Calculate E(X) and Var(X). - Mathematics and Statistics

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Solve the following problem :

Let the p. m. f. of the r. v. X be

`"P"(x) = {((3 - x)/(10)", ","for"  x = -1", "0", "1", "2.),(0,"otherwise".):}`
Calculate E(X) and Var(X).

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Solution 1

P(X)=`(3-x) /10`

X takes values -1, 0, 1, 2

P(X = -1)= P(-1) = `(3+1) /10 =4/ 10`

P(X = 0)= P(0) = `(3-0) /10 =3/ 10`

P(X = 1)= P(1) = `(3-1) /10 =2/ 10`

P(X = 2)= P(2) = `(3-2) /10 =1/ 10`

We construct the following table to calculate the mean and variance of X :

xi P (xi) xi P (xi) xi2 P (xi)
-1 `4/10` -`4/10` `4/10`
0 `3/10` 0 0
1 `2/ 10` `2/10` `2/10`
2 `1/10` `2/10` `4/10`
Total 1 0 1

From the table

∑xi P(xi)0 and ∑xi2 ·P(xi)= 1

E(X) = xi P(xi) = 0

Var (X) = ∑xi2 · P(xi) - [E(X)]2

= 1 - 0 = 1

Hence, E(X) = 0, Var (X) = 1.

Solution 2

E(X) = `sum_("i" = 1)^4x_"i""P"(x_"i")`

= `-1 xx ((3 - (-1))/10) + 0 xx ((3 - 0)/10) + 1 xx ((3 - 1)/10) + 2 xx ((3 - 2)/10)`

= `(-4 + 0 + 2 + 2)/10`

= 0

E(X2) = `sum_("i" = 1)^4x_"i"^2"P"(x_"i")`

= `(-1)^2 xx 4/10 + (0)^2 xx 3/10 + (1)^2 xx 2/10 + (2)^2 xx 1/10`

= `(4 + 0 + 2 + 4)/10`

= 1

Var(X) = E(X2) – [E(X)]2

= 1 – (0)2

= 1

Concept: Probability Distribution of Discrete Random Variables
  Is there an error in this question or solution?
Chapter 8: Probability Distributions - Part I [Page 156]

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