The probability density function of a random variable X is given byfXx-8x3 for x-2Then the value of E - X3 - is equal to

E [ X3 ]=∫_2^∞x3f_X(x)dx =∫_2^∞x3.8x^3dx=43=1.33 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Complement of an Event

The probabili...

Question

The probability density function of a random variable $X$ is given by
$f_{X} (x) = \frac{8}{x^{3}}$ for $x > 2$
Then the value of $E [\frac{X}{3}]$ is equal to _____
1.333

Open in App

Solution

The correct option is A 1.333
$E [\frac{X}{3}] = \int_{2}^{\infty} \frac{x}{3} f_{X} (x) d x$
$= \int_{2}^{\infty} \frac{x}{3} . \frac{8}{x^{3}} d x = \frac{4}{3} = 1.33$

Suggest Corrections

Similar questions

Q. The probability density function of a ramdom variable X is given by

f_{x} (x) = {\begin{matrix} λ (x - 2) (x^{2} + 4); & 2 \leq x < 4 0 & o t h e r w i s e \end{matrix}

The value of the constant

λ

The probability density function of a random variable X is $P_{x} (x) = e^{- x} for x \geq 0 and 0 otherwise.$ The expected value of the function $g_{x} (x) = e^{3 x / 4} is$ .

A random variable X has probability density function f(x) as give below:
$f (x) = (\begin{matrix} a + bx for 0 < x < 1 0 otherwise \end{matrix}$

If the expected value E[X] = 2/3, then Pr[X < 0.5] is .

Q. The probability density function of a random variable

x

f (x) = \frac{x}{3} (9 - x^{2}) for 0 \leq x \leq 3

The mean,

μ_{x}

of the random variable is _____

Q. The probability density function of a random variable 'X' is given as

f_{x} (x)

. A random variabel Y is defined as Y = 2X. The PDF of random variable 'Y' is defined as

k f_{x} (\frac{y}{2})

, then the value of 'k' is equal to

Join BYJU'S Learning Program

The probability density function of a random variable X is given by fX(x)=8x3 for x>2 Then the value of E[X3] is equal to _____1.333

The correct option is A 1.333E[X3]=∫∞2x3fX(x)dx =∫∞2x3.8x3dx=43=1.33

The probability density function of a random variable $X$ is given by
$f_{X} (x) = \frac{8}{x^{3}}$ for $x > 2$
Then the value of $E [\frac{X}{3}]$ is equal to _____
1.333

The correct option is A 1.333
$E [\frac{X}{3}] = \int_{2}^{\infty} \frac{x}{3} f_{X} (x) d x$
$= \int_{2}^{\infty} \frac{x}{3} . \frac{8}{x^{3}} d x = \frac{4}{3} = 1.33$