If a continuous random variable X has probability density function $f(x) = \begin{cases} ax^2, & 0\le x\le 1 \\ 0, & \text{otherwise} \end{cases}$ then the value of a is_____
For a continuous random variable X, the probability density function (PDF) $f(x)$ must satisfy the condition that the total probability over its domain is equal to 1.
This means the integral of the PDF over its defined range must equal 1:
$ \int_{-\infty}^{\infty} f(x) dx = 1 $Given the PDF:
$ f(x) = \begin{cases} ax^2, & 0\le x\le 1 \\ 0, & \text{otherwise} \end{cases} $We need to integrate $f(x)$ from 0 to 1 and set the result equal to 1 to find the value of the constant $a$.
$ \int_0^1 ax^2 dx = 1 $
$ a \int_0^1 x^2 dx = 1 $
$ a \left[ \frac{x^3}{3} \right]_0^1 = 1 $
$ a \left( \frac{1^3}{3} - \frac{0^3}{3} \right) = 1 $
$ a \left( \frac{1}{3} - 0 \right) = 1 $
$ \frac{a}{3} = 1 $
Multiplying both sides by 3 gives:
$ a = 3 $
Therefore, the value of $a$ is 3.
If the odds in favour of any random event A are 5 ∶ 6, then the odds against the event are:
If random variable X follows binomial distribution with parameter n and p with mean 15 and variance 10, then the value of mode is
Let $X$ and $Y$ be continuous random variables with probability density functions $P_X(x)$ and $P_Y(y)$, respectively. Further, let $Y = X^2$ and $P_X(x) = \begin{cases} 1, & x\in (0,1] \\ 0, & \text{otherwise} \end{cases}$
Which one of the following options is correct?