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The value of k when $f\left( x \right)=\dfrac{k}{\sqrt{x}},0 < x < 4$ is the pdf of RV $x$ is
(a) $-4$
(b) $-\dfrac{1}{4}$
(c) 4
(d) $\dfrac{1}{4}$

Last updated date: 15th Jul 2024
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Hint: We know that the area under a probability density function is always equal to unity. Hence, we can integrate the given function over the given range and equate it to unity. We can solve this integration using the formula $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}$, and find the required value of k.

Complete step by step answer:
We know that the probability density function, also represented as a pdf, at any given space or point, gives the probability of the random variable to achieve that particular value in that space or point.
We also know that the area under the probability density function is equal to 1.
So, if the probability density function is represented as $f\left( x \right)$, where $x$ is the random variable, then we can say that the integration of $f\left( x \right)$ with respect to $x$ over the entire range, will be equal to unity.
We can write this mathematically as
$\int\limits_{-\infty }^{+\infty }{f\left( x \right)dx}=1$.
Here, in this question, the probability density function is defined as $f\left( x \right)=\dfrac{k}{\sqrt{x}}$ in the range $0 < x < 4$.
Hence, we can write
We know that $\dfrac{1}{\sqrt{x}}$ can also be written as ${{x}^{-\dfrac{1}{2}}}$ . And since k is a constant, we can write the integral as
We know that $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}$.
So, by using this formula, we can write,
$k\left[ \dfrac{{{x}^{\dfrac{-1}{2}+1}}}{-\dfrac{1}{2}+1} \right]_{0}^{4}=1$.
On simplification, we can write
$k\left[ \dfrac{{{x}^{\dfrac{1}{2}}}}{\dfrac{1}{2}} \right]_{0}^{4}=1$.
We can also write this as
$2k\left[ {{x}^{\dfrac{1}{2}}} \right]_{0}^{4}=1$.
$2k\left[ \sqrt{x} \right]_{0}^{4}=1$.
On putting the values of $x$, we get
$2k\left[ \sqrt{4}-0 \right]=1$.
Thus, we have
Hence, we get $k=\dfrac{1}{4}$.

So, the correct answer is “Option d”.

Note: We can see that the probability density function is defined only for the range $0 < x < 4$, and this is why we have integrated the pdf in this range and not from $-\infty \text{ to }+\infty $. We must also not confuse thinking that pdf means probability distribution function.