Suppose that X is a continuous random variable whose probability density function is given by: $f (x) = k (2 x - x^{2}) f or 0 < x < 2$ $f(x)= k(2x - x^2) for 0 < x < 2$ ; 0 for all other x. What is the value of k, P(X>1), E(X) and Var(X)?

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Answer 1 · 2016-11-13T08:31:00

$k = \frac{3}{4}$ $k=3/4$
$P (x > 1) = \frac{1}{2}$ $P(x>1)=1/2$
$E (X) = 1$ $E(X)=1$
$V (X) = \frac{1}{5}$ $V(X)=1/5$

To find $k$ $k$ , we use $\int_{0}^{2} f (x) d x = \int_{0}^{2} k (2 x - x^{2}) d x = 1$ $int_0^2f(x)dx=int_0^2k(2x-x^2)dx=1$

$:. k[2x^2/2-x^3/3]_0^2=1$

$k(4-8/3)=1$ $=>$ $4/3k=1$ $=>$ $k=3/4$

To calculate $P(x>1)$ , we use $P(X>1)=1-P(0< x <1)$

$=1-int_0^1(3/4)(2x-x^2)=1-3/4[2x^2/2-x^3/3]_0^1$

$=1-3/4(1-1/3)=1-1/2=1/2$

To calculate $E(X)$

$E(X)=int_0^2xf(x)dx=int_0^2(3/4)(2x^2-x^3)dx$

$=3/4[2x^3/3-x^4/4]_0^2=3/4(16/3-16/4)=3/4*16/12=1$

To calculate $V(X)$

$V(X)=E(X^2)-(E(X))^2=E(X^2)-1$

$E(X^2)=int_0^2x^2f(x)dx=int_0^2(3/4)(2x^3-x^4)dx$

$=3/4[2x^4/4-x^5/5]_0^2=3/4(8-32/5)=6/5$

$:.V(X)=6/5-1=1/5$

Suppose that X is a continuous random variable whose probability density function is given by: f(x)= k(2x - x^2) for 0 < x < 2f(x)=k(2x−x2)for0<x<2f(x)= k(2x - x^2) for 0 < x < 2; 0 for all other x. What is the value of k, P(X>1), E(X) and Var(X)?