Question

Probability distribution of a random variable X is as follows:
X=x
-2
-1
0
1
2
P(x)
0.2
0.1
0.3
0.3
0.1
Find (1) E(X) (2) V(X) (3) E(3X+2) (4) v(3x+2)

Answer 1

Answer:

E(x) = 0

E(3X+2) = 2

V(x) = 1.6

V(3x + 2) = 14.4

Step-by-step explanation:

X=x   P(x)        x * P(x)     X²*P(x)

-2     0.2         -0.4          0.8

-1      0.1          -0.1           0.1

0      0.3          0              0

1       0.3           0.3           0.3

2      0.1           0.2           0.4

Total   1          0              1.6

E(x) = -0.4  - 0.1 + 0 + 0.3 + 0.2  =  0

E(x)  = 0

V(x) = E(X²) - E(X)²

= 1.6 - 0²

= 1.6

V(x) = 1.6

E[aX + b] = aE[X] + b

=> E(3X+2) = 3E[X] + 2  = 3 * 0 + 2 = 2

E(3X+2) = 2

V(ax + b) = a²V(x)

=> V(3x + 2) = 3²V(x)

=> V(3x + 2) = 9 * 1.6

=> V(3x + 2) = 14.4

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

For the given Probability distribution of a random variable X

(1) E(X)

(2) V(X)

(3) E(3X+2)

(4) v(3x+2)

CALCULATION

(1)

$\sf{E(X)}$

$= \sf{ \sum xp(x)}$

$= \sf{( - 2 \times 0.2) + ( - 1 \times 0.1) + (0 \times 0.3) + (1 \times 0.3) + (2 \times 0.1) \: }$

$= \sf{ - 0.4 - 0.1 + 0 + 0.3 + 0.2 \: \: }$

$\sf{ = 0}$

(2)

Here

$\sf{E({X}^{2} )}$

$= \sf{ \sum {x}^{2}p(x) \: \: }$

$= \sf{ \{ {( - 2)}^{2} \times 0.2 + {( - 1)}^{2} \times 0.1 + {( 0)}^{2} \times 0.3 + {( 1)}^{2} \times 0.3 + {( 2)}^{2} \times 0.1 \: \}\: \: }$

$= \sf{ 0.8 + 0.1 + 0 + 0.3 + 0.4\: \: }$

$= \sf{ 1.6\: }$

Hence

$\sf{V(X)}$

$= \sf{Var(X)}$

$= \sf{E({X}^{2} ) - {( \sf{E({X}) )} }^{2} }$

$= \sf{1.6 - 0}$

$= \sf{1.6}$

(3)

$\sf{E(3X+2)}$

$= \sf{E(3X \: )+E(2)}$

$= \sf{3E(X \: )+2}$

$\sf{ = 0 + 2\: }$

$= \sf{2}$

(4)

$\sf{V(3X + 2)}$

$= \sf{Var(3X + 2)}$

$= \sf{ {3}^{2} Var(X)}$

$= \sf{9 \times 1.6}$

$= \sf{11.4}$

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