Q.The following function represents the p.d.f of a.r.v. X
$f(x) = \begin{cases} kx, & \text{for } 0 < x < 2 \\ 0, & \text{otherwise} \end{cases}$ then the value of K is ______

$P(X<1.5)=\int_0^{1.5}\frac{x}{8}\,dx = \left[\frac{x^2}{16}\right]_0^{1.5} = \frac{2.25}{16} = \frac{9}{64} \approx 0.1406$.

The probability density function is

$f(x) = \begin{cases} \dfrac{x}{8}, & 0 < x < 4 \\[4pt] 0, & \text{otherwise.} \end{cases}$

For a continuous random variable, the probability over an interval is the integral of the p.d.f.:

$P(X < 1.5) = \int_{0}^{1.5} f(x)\,dx = \int_{0}^{1.5} \frac{x}{8}\,dx.$

Integrating:

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For a continuous r.v., $P(a<x<b)=\int_a^b f(x)\,dx$. Here $f(x)=\frac{x}{8}$ on $(0,4)$, so integrating from $1$ to $2$ gives $\frac{3}{16}$.

Since $X$ is a continuous random variable with probability density function $f(x)=\dfrac{x}{8}$ for $0<x<4$, the required probability is the area under the curve between $x=1$ and $x=2$:

$P(1<x<2)=\displaystyle\int_1^2 f(x)\,dx=\int_1^2 \frac{x}{8}\,dx$

Integrating term by term,

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Because $f(x)=0$ for $x\ge 4$, $P(x>2)=\int_2^4 \frac{x}{8}\,dx=\frac{3}{4}$.

The density is $f(x)=\dfrac{x}{8}$ only on $0<x<4$ and is $0$ elsewhere, so the event $x>2$ contributes probability only over $2<x<4$:

$P(x>2)=\displaystyle\int_2^{4} f(x)\,dx=\int_2^{4}\frac{x}{8}\,dx$

Integrating,

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Apply $\int_{-\infty}^{\infty} f(x)\,dx = 1$: $\int_0^2 kx\,dx = 2k = 1$, so $k = \dfrac{1}{2}$.

For $f(x)$ to be a probability density function, it must be non-negative and satisfy

$\displaystyle \int_{-\infty}^{\infty} f(x)\,dx = 1$.

Since $f(x) = 0$ outside $(0,2)$,

$\displaystyle \int_0^2 kx\,dx = 1$.

Evaluating the integral,

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