# 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) $\dfrac{1}{4}$

Answer

Verified

277.2k+ views

**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

$\int\limits_{0}^{4}{\dfrac{k}{\sqrt{x}}dx}=1$.

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

$k\int\limits_{0}^{4}{{{x}^{-\dfrac{1}{2}}}dx}=1$.

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

$2k\left[ {{x}^{\dfrac{1}{2}}} \right]_{0}^{4}=1$.

Or,

$2k\left[ \sqrt{x} \right]_{0}^{4}=1$.

On putting the values of $x$, we get

$2k\left[ \sqrt{4}-0 \right]=1$.

Thus, we have

$4k=1$.

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.

Recently Updated Pages

Basicity of sulphurous acid and sulphuric acid are

Why should electric field lines never cross each other class 12 physics CBSE

An electrostatic field line is a continuous curve That class 12 physics CBSE

What are the measures one has to take to prevent contracting class 12 biology CBSE

Suggest some methods to assist infertile couples to class 12 biology CBSE

Amniocentesis for sex determination is banned in our class 12 biology CBSE

Trending doubts

The lightest gas is A nitrogen B helium C oxygen D class 11 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Which place is known as the tea garden of India class 8 social science CBSE

What is pollution? How many types of pollution? Define it

Write a letter to the principal requesting him to grant class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE