If $X$ is a continuous random variable with the probability density function
$f(x) = \begin{cases} \frac{K}{4}, & 0 \le x \le 1 \\ 0, & \text{otherwise} \end{cases}$
then the value of $K$ is __________. (Answer in integer)
Finding the Constant \(K\) in a Probability Density Function
For a continuous random variable, the total area under its probability density function (PDF) curve must equal 1. This property is used to find unknown constants like \(K\).
The given probability density function is:
$f(x) = \begin{cases} \frac{K}{4}, & 0 \le x \le 1 \\ 0, & \text{otherwise} \end{cases}$
PDF Integral Property
The fundamental property of a PDF is that the integral over its domain must be 1:
$ \int_{-\infty}^{\infty} f(x) dx = 1 $
Calculating \(K\)
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Set up the integral using the given function, considering the range where $f(x)$ is non-zero:
$ \int_{0}^{1} \frac{K}{4} dx = 1 $
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Evaluate the integral. Since \(K\) is a constant, it can be taken out:
$ \frac{K}{4} \int_{0}^{1} 1 dx = 1 $
The integral of 1 with respect to $x$ is $x$. Evaluating this from 0 to 1:
$ \int_{0}^{1} 1 dx = [x]_{0}^{1} = 1 - 0 = 1 $
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Substitute the result back into the equation:
$ \frac{K}{4} \times 1 = 1 $
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Solve for \(K\):
$ \frac{K}{4} = 1 $
$ K = 4 $
Therefore, the value of \(K\) is 4.
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
