Q.A continuous random variable X has the following distribution function.

[!FORMULA] $F(x)=\begin{cases} 0 & \text{if } x\le 1 \\ k(x-1)^4 & \text{if } 1<x\le 3 \\ 1 & \text{if } x>3 \end{cases}$

Find

The total probability must equal $1$: $15k=1\Rightarrow k=\tfrac{1}{15}$.

For $P(x)$ to be a valid probability mass function, all probabilities are non-negative and their sum is $1$:

$\sum P(x) = k + 2k + 3k + 4k + 5k = 15k.$

…

$\sum p(x_i)=1$ for a discrete distribution.

A valid probability mass function satisfies $p(x_i)\ge 0$ for all $i$ and $\sum_i p(x_i)=1$. Hence the sum of all th …

$P(X = c) = 0$ for a continuous distribution.

In a continuous probability distribution, probabilities are areas under the probability density function $f(x)$:

$P(a \le X \le b) = \int_a^b f(x)\,dx.$

The probability that $X$ takes an exact single value $c$ is the area over an interval of zero width:

…

$k = \dfrac{1}{15}$.

For a discrete random variable, the total probability must equal $1$:

$\sum P(x) = 1.$

Adding the given probabilities:

$k + 2k + 3k + 4k + 5k = 15k.$

Set the sum equal to $1$:

$15k = 1 \;\Rightarrow\; k = \frac{1}{15}.$ …

A p.m.f. must satisfy $\sum P(x) = 1$. Adding the five probabilities gives $15k = 1$, hence $k = \frac{1}{15}$.

Step 1 — Total probability condition. For any discrete random variable, $\sum P(x) = 1$.

Step 2 — Add the given probabilities.

$k + 2k + 3k + 4k + 5k = 15k.$

Step 3 — Solve. …