If a continuous random variable has the following probability density function
$f(x) = \begin{cases} kx^3, & 0 \le x \le 1, \\ 0, & \text{otherwise;} \end{cases}$
then the value of k is __________________.
To find the value of the constant $k$ for the given probability density function (PDF), we use the property that the total probability over the entire range of the random variable must equal 1.
PDF Integration Property
For a continuous random variable, the integral of its PDF over all possible values must be equal to 1:
$ \int_{-\infty}^{\infty} f(x) dx = 1 $
In this case, the PDF is non-zero only between 0 and 1. Therefore, the integral becomes:
$ \int_{0}^{1} kx^3 dx = 1 $
Calculating the Constant k
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Factor out the constant $k$ from the integral:
$ k \int_{0}^{1} x^3 dx = 1 $
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Evaluate the integral of $x^3$:
$ \int x^3 dx = \frac{x^{3+1}}{3+1} = \frac{x^4}{4} $
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Apply the limits of integration (0 to 1):
$ k \left[ \frac{x^4}{4} \right]_{0}^{1} = 1 $
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Substitute the limits:
$ k \left( \frac{1^4}{4} - \frac{0^4}{4} \right) = 1 $
$ k \left( \frac{1}{4} - 0 \right) = 1 $
$ k \left( \frac{1}{4} \right) = 1 $
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Solve for $k$:
$ k = 4 $
Thus, the value of the constant $k$ 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
