The probability density function of a random variable x is fx - x39-x2 for 0 - x - 3The mean- -x of the random variable is

F(x) = {x3(9 x^2), 0 ≤ x ≤ 3 nbsp; 0, otherwise. μ_x = ∫_0^3 x f(x) dx mean (μ_x)= ∫_0^3 x x3(9 x^2)dx = ∫_0^3 ( 3x^2 x^43 ) dx nbsp; = [ x^3 x^515 ]_0^3=2 ...

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Standard XII

Mathematics

Coefficient of Variation

The probabili...

Question

The probability density function of a random variable $x$ is $f (x) = \frac{x}{3} (9 - x^{2}) for 0 \leq x \leq 3$
The mean, $μ_{x}$ of the random variable is _____
10.8

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Solution

The correct option is A 10.8
$f (x) = ⎧ ⎨ ⎩ \begin{matrix} \frac{x}{3} (9 - x^{2}), 0 \leq x \leq 3 0, otherwise \end{matrix}$
$μ_{x} = 3 \int 0 x f (x) d x$
mean $(μ_{x}) = 3 \int 0 x \frac{x}{3} (9 - x^{2}) d x = 3 \int 0 (3 x^{2} - \frac{x^{4}}{3}) d x$
$= {[x^{3} - \frac{x^{5}}{15}]}_{0}^{3} = 27 - \frac{243}{15} = 10.8$

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The probability density function of a random variable x is f(x)=x3(9−x2) for 0≤x≤3 The mean, μx of the random variable is _____10.8

The correct option is A 10.8f(x)=⎧⎨⎩x3(9−x2), 0≤x≤30, otherwise μx=3∫0x f(x) dx mean (μx)=3∫0x x3(9−x2)dx=3∫0(3x2−x43) dx =[x3−x515]30=27−24315=10.8

The probability density function of a random variable $x$ is $f (x) = \frac{x}{3} (9 - x^{2}) for 0 \leq x \leq 3$
The mean, $μ_{x}$ of the random variable is _____
10.8