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_____
Probability Density Function (PDF) Integration
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 $Calculating the Constant 'a'
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 $a$.
- Set up the integral:
$ \int_0^1 ax^2 dx = 1 $
- Perform the integration:
$ a \int_0^1 x^2 dx = 1 $
$ a \left[ \frac{x^3}{3} \right]_0^1 = 1 $
- Evaluate the definite integral:
$ a \left( \frac{1^3}{3} - \frac{0^3}{3} \right) = 1 $
$ a \left( \frac{1}{3} - 0 \right) = 1 $
$ \frac{a}{3} = 1 $
- Solve for 'a':
Multiplying both sides by 3 gives:
$ a = 3 $
Therefore, the value of $a$ is 3.
Important Questions from Random Variables
Consider the following probability density function for a random variable x.
(a) \(f_{1}(x)=1\ ;\ -1\le x\le 1\)
(b) \(f_{2}(x)=x\ ;\ 0\le x\le 1\)
(c) \(f_{3}(x)=(1-x)\ ;\ -1\le x\le 1\)
Arrange the above functions in terms of the increasing value of mean of random variable x.
Considering all the symbols with their usual meanings, match the following :
List - I List - II (a) \(f_{X}(x)\) (i) \(\displaystyle\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dx\) (b) \(\displaystyle\int_{-\infty}^{\infty}f_{X}(x)\,dx\) (ii) \(\displaystyle\int_{-\infty}^{a}f_{X}(x)\,dx\) (c) \(F_{X}(a)\) (iii) \(\displaystyle\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dy\) (d) \(f_{Y}(y)\) (iv) 1 Codes :
The density function of a random variable is given by \(p(x)=Ke^{-\frac{x^{2}}{2}}\) for \(-\infty \lt x \lt \infty\). The value of K should be :
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
