If the probability density function of X is f(x)=1+ax2,−1≤x≤1,−1≤a≤1 then the expectation of X is? a) 6a b) a3 c) a2 d) 3a

Answer: E(X)=α3 Explanation: for a pdf f(x) the expectation of X E(X)=∫allxxf(x)dx E(X)=∫−11x(1+αx2)dx E(X)=12∫−11(x+αx2)dx E(X)=12[x22+αx33]−11 E(X)=12{[x22+αx33]1−[x22+αx33]0} E(X)=12{(⧸12+α3)−(⧸12−α3)} E(X)=12(α3+α3)=12×2α3 E(X)=α3

Explanation: By definition if f(x) is a continuous probability density function then: E(X)=∫−∞∞xf(x)dx So given that f(x)=1+ax2 for −1≤x≤1 then we have: E(X)=∫−11x(1+ax2)dx =12∫−11x(1+ax)dx =12∫−11x+ax2dx =12[x22+ax33]−11 =12{(12+a3)−(12−a3)} =12(12+a3−12+a3) =122a3 =a3

Answer: E(X)=α3 Explanation: for a pdf f(x) the expectation of X E(X)=∫allxxf(x)dx E(X)=∫−11x(1+αx2)dx E(X)=12∫−11(x+αx2)dx E(X)=12[x22+αx33]−11 E(X)=12{[x22+αx33]1−[x22+αx33]0} E(X)=12{(⧸12+α3)−(⧸12−α3)} E(X)=12(α3+α3)=12×2α3 E(X)=α3

Explanation: By definition if f(x) is a continuous probability density function then: E(X)=∫−∞∞xf(x)dx So given that f(x)=1+ax2 for −1≤x≤1 then we have: E(X)=∫−11x(1+ax2)dx =12∫−11x(1+ax)dx =12∫−11x+ax2dx =12[x22+ax33]−11 =12{(12+a3)−(12−a3)} =12(12+a3−12+a3) =122a3 =a3

Resolved: "If the probability density function of X is..."

Alexander Day

Answered question

2022-02-13

If the probability density function of X is

f (x) = \frac{1 + a x}{2}, - 1 \leq x \leq 1, - 1 \leq a \leq 1

then the expectation of X is?
a)

\frac{6}{a}

\frac{a}{3}

\frac{a}{2}

\frac{3}{a}

Answer & Explanation

Javon Ross

Beginner2022-02-14Added 13 answers

Answer: $E (X) = \frac{α}{3}$
Explanation: for a pdf f(x) the expectation of X
$E (X) = \int_{all x} x f (x) d x$
$E (X) = \int_{- 1}^{1} x (\frac{1 + α x}{2}) d x$
$E (X) = \frac{1}{2} \int_{- 1}^{1} (x + α x^{2}) d x$
$E (X) = \frac{1}{2} {[\frac{x^{2}}{2} + α \frac{x^{3}}{3}]}_{- 1}^{1}$
$E (X) = \frac{1}{2} {{[\frac{x^{2}}{2} + α \frac{x^{3}}{3}]}^{1} - {[\frac{x^{2}}{2} + α \frac{x^{3}}{3}]}_{0}}$
$E (X) = \frac{1}{2} {(⧸ \frac{1}{2} + \frac{α}{3}) - (⧸ \frac{1}{2} - \frac{α}{3})}$
$E (X) = \frac{1}{2} (\frac{α}{3} + \frac{α}{3}) = \frac{1}{2} \times \frac{2 α}{3}$
$E (X) = \frac{α}{3}$

Hannah Escobar

Beginner2022-02-15Added 9 answers

Explanation:
By definition if f(x) is a continuous probability density function then:

E (X) = \int_{- \infty}^{\infty} x f (x) d x

So given that

f (x) = \frac{1 + a x}{2} for - 1 \leq x \leq 1

then we have:

E (X) = \int_{- 1}^{1} x (\frac{1 + a x}{2}) d x

= \frac{1}{2} \int_{- 1}^{1} x (1 + a x) d x

= \frac{1}{2} \int_{- 1}^{1} x + a x^{2} d x

= \frac{1}{2} {[\frac{x^{2}}{2} + \frac{a x^{3}}{3}]}_{- 1}^{1}

= \frac{1}{2} {(\frac{1}{2} + \frac{a}{3}) - (\frac{1}{2} - \frac{a}{3})}

= \frac{1}{2} (\frac{1}{2} + \frac{a}{3} - \frac{1}{2} + \frac{a}{3})

= \frac{1}{2} \frac{2 a}{3}

= \frac{a}{3}

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