Suppose X has probability density function f(x)=6x(1-x) for 0&lt;x&lt;1 and 0 otherwise. What is E[X] and Var[X]?

E[X]=12&nbsp;and&nbsp;Var[X}=120&nbsp;Solution:&nbsp;E[X]=∫016x2(1−x)&nbsp;dx&nbsp;=12&nbsp;orE[X2]=∫016x3(1−x)&nbsp;dx&nbsp;=310&nbsp;So&nbsp;Var[X]=E[X2]−(E(X))2=310−14=120

To find EX, we use the formula, EX=∫01xP(x)dx That is, EX=∫01x×6x(1−x)dx =6∫01(x2−x3)dx Integrating, we get =6[x33−x44]01 Substituting the upper and lower limit, =6[(1)33−(1)44−((0)33−(0)44)] =6[13−14] =12 We have to find Var(X). For this, we use the formula, Var(X)=E(X2)−(EX)2 E(X2)=0.3 E(X)=0.5 Var(X)=E(X2)−(EX)2 =0.3−(0.5)2 =0.3-0.25 =0.05

E[X]=12&nbsp;and&nbsp;Var[X}=120&nbsp;Solution:&nbsp;E[X]=∫016x2(1−x)&nbsp;dx&nbsp;=12&nbsp;orE[X2]=∫016x3(1−x)&nbsp;dx&nbsp;=310&nbsp;So&nbsp;Var[X]=E[X2]−(E(X))2=310−14=120

To find EX, we use the formula, EX=∫01xP(x)dx That is, EX=∫01x×6x(1−x)dx =6∫01(x2−x3)dx Integrating, we get =6[x33−x44]01 Substituting the upper and lower limit, =6[(1)33−(1)44−((0)33−(0)44)] =6[13−14] =12 We have to find Var(X). For this, we use the formula, Var(X)=E(X2)−(EX)2 E(X2)=0.3 E(X)=0.5 Var(X)=E(X2)−(EX)2 =0.3−(0.5)2 =0.3-0.25 =0.05

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hunterofdeath63

Answered question

2022-01-17

Suppose X has probability density function f(x)=6x(1-x) for 0<x<1 and 0 otherwise. What is E[X] and Var[X]?

Answer & Explanation

Barbara Meeker

Beginner2022-01-17Added 38 answers

$E [X] = \frac{1}{2}$ and $V a r [X} = \frac{1}{20}$
Solution:
$E [X] = \int_{0}^{1} 6 x^{2} (1 - x) d x = \frac{1}{2}$
or
$E [X^{2}] = \int_{0}^{1} 6 x^{3} (1 - x) d x = \frac{3}{10}$
So $V a r [X] = E [X^{2}] - {(E (X))}^{2} = \frac{3}{10} - \frac{1}{4} = \frac{1}{20}$

kalupunangh

Beginner2022-01-18Added 29 answers

To find EX, we use the formula,

E X = \int_{0}^{1} x P (x) d x

That is,

E X = \int_{0}^{1} x \times 6 x (1 - x) d x

= 6 \int_{0}^{1} (x^{2} - x^{3}) d x

Integrating, we get

= 6 {[\frac{x^{3}}{3} - \frac{x^{4}}{4}]}_{0}^{1}

Substituting the upper and lower limit,

= 6 [\frac{{(1)}^{3}}{3} - \frac{{(1)}^{4}}{4} - (\frac{{(0)}^{3}}{3} - \frac{{(0)}^{4}}{4})]

= 6 [\frac{1}{3} - \frac{1}{4}]

= \frac{1}{2}

We have to find Var(X). For this, we use the formula,

V a r (X) = E (X^{2}) - {(E X)}^{2}

E (X^{2}) = 0.3

E(X)=0.5

V a r (X) = E (X^{2}) - {(E X)}^{2}

= 0.3 - {(0.5)}^{2}

=0.3-0.25
=0.05

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