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Question

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 \(K\)

  1. 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 $

  2. 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 $

  3. Substitute the result back into the equation:

    $ \frac{K}{4} \times 1 = 1 $

  4. Solve for \(K\):

    $ \frac{K}{4} = 1 $

    $ K = 4 $

Therefore, the value of \(K\) is 4.

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Important Questions from Random Variables

  1. 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?

  2. 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).
  3. 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:
  4. Let $Y = Z^2$, $Z = \frac{X - \mu}{\sigma}$, where $X$ is a normal random variable with mean $\mu$ and variance $\sigma^2$. The variance of $Y$ is

  5. Consider a discrete random variable X whose probabilities are given below. The standard deviation of the random variable is ________ (round off to one decimal place).

    $x_1$1234
    $P(X = x_i)$0.30.10.30.3
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