Question

The random variable X has probability density function f(x)={ax+bx2 , 0<x<1 . If E(X)=0.6 , find (a)P(X<1/2) and (b)var(x).

Answer 1

Answer on Question #44950 – Math – Statistics and Probability

Question. The random variable $X$ has probability density function $f(x) = ax + bx^2$, $0 < x < 1$. If $E(X) = 0.6$, find

a) $P(X < 1/2)$

b) $Var(X)$

Solution. First of all we shall find $a$ and $b$. Since $f(x) = ax + bx^2$, $0 < x < 1$ is the probability density function, then $\int_0^1 f(x)dx = 1$. $\int_0^1 (ax + bx^2)dx = \frac{a}{2} x^2 +\frac{b}{3} x^3\big|_0^1 = \frac{a}{2} +\frac{b}{3}$. The first condition: $\frac{a}{2} +\frac{b}{3} = 1$.

$E(X) = \int_0^1 x(ax + bx^2)dx = \int_0^1 (ax^2 +bx^3)dx = \frac{a}{3} x^3 +\frac{b}{4} x^4\big|_0^1 = \frac{a}{3} +\frac{b}{4}.$ The second condition: $\frac{a}{3} +\frac{b}{4} = \frac{3}{5}$. We have the next linear system:

$$\left\{ \begin{array}{l} \frac {a}{2} + \frac {b}{3} = 1 \\ \frac {a}{3} + \frac {b}{4} = \frac {3}{5} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 3 a + 2 b = 6 \\ 4 a + 3 b = \frac {3 6}{5} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 9 a + 6 b = 1 8 \\ - 8 a - 6 a = - \frac {7 2}{5} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a = \frac {1 8}{5} \\ b = - \frac {1 2}{5} \end{array} \right.$$

The density function has the form: $f(x) = \frac{18}{5} x - \frac{12}{5} x^2, 0 < x < 1$.

a) $P(X < 1/2) = \int_0^{1/2} \left( \frac{18}{5} x - \frac{12}{5} x^2 \right) dx = \frac{9}{5} x^2 - \frac{4}{5} x^3 \big|_0^{1/2} = \frac{9}{20} - \frac{2}{20} = \frac{7}{20}$.

b) $E(X^2) = \int_0^1 x^2\left(\frac{18}{5} x - \frac{12}{5} x^2\right)dx = \int_0^1\left(\frac{18}{5} x^3 -\frac{12}{5} x^4\right)dx = \frac{9}{10} -\frac{12}{25} = \frac{21}{50}.$

$$Var(X) = E(X^2) - [E(X)]^2 = \frac{21}{50} - \frac{9}{25} = \frac{3}{50}.$$

Answer. a) $P(X < 1/2) = \frac{7}{20}$.

b) $Var(X) = \frac{3}{50}$.

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