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
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?- Two fair dice (with faces labeled 1, 2, 3, 4, 5, and 6) are rolled. Let the random variable $X$ denote the sum of the outcomes obtained.
The expectation of $X$ is __________ (rounded off to two decimal places). - Let $X = aZ + b$, where $Z$ is a standard normal random variable, and $a, b$ are two unknown constants. It is given that
$E[X] = 1$, $E[(X – E[X])Z] = –2$, $E[(X – E[X])^2] = 4$,
where $E[X]$ denotes the expectation of random variable $X$. The values of $a, b$ are:
