For the random variable X having PDF f(x) = 4x 3; 0

Question

For the random variable X having PDF f(x) = 4x 3; 0 < x < 1, the interquartile range is:

Prepp Team · Accepted Answer

Understanding the Interquartile Range (IQR) for a Random Variable

The interquartile range (IQR) is a measure of statistical dispersion, representing the spread of the middle 50% of the data. For a continuous random variable, it is the difference between the third quartile ($Q_3$) and the first quartile ($Q_1$).

The first quartile ($Q_1$) is the value below which 25% of the probability mass lies. If $X$ is the random variable, $P(X \le Q_1) = 0.25$.
The third quartile ($Q_3$) is the value below which 75% of the probability mass lies. $P(X \le Q_3) = 0.75$.
The interquartile range (IQR) is calculated as $IQR = Q_3 - Q_1$.

To find the quartiles for a continuous random variable with a given probability density function (PDF), we first need to find its cumulative distribution function (CDF).

Calculating the Cumulative Distribution Function (CDF)

The given probability density function (PDF) is $f(x) = 4x^3$ for $0 < x < 1$. The CDF, $F(x)$, is the integral of the PDF from the lower limit of the support to $x$.

For $0 < x < 1$, the CDF is:

$ F(x) = P(X \le x) = \int_{0}^{x} f(t) dt $

$ F(x) = \int_{0}^{x} 4t^3 dt $

Integrating $4t^3$ with respect to $t$, we get $t^4$. Evaluating this from 0 to $x$:

$ F(x) = \left[ t^4 \right]_{0}^{x} = x^4 - 0^4 = x^4 $

So, the CDF is $F(x) = x^4$ for $0 < x < 1$. The complete CDF is:

$ F(x) = \begin{cases} 0 & \text{for } x \le 0 \\ x^4 & \text{for } 0 < x < 1 \\ 1 & \text{for } x \ge 1 \end{cases} $

Finding the Quartiles ($Q_1$ and $Q_3$)

To find the quartiles, we set the CDF equal to the desired probability and solve for $x$.

Finding the First Quartile ($Q_1$)

We need to find $Q_1$ such that $F(Q_1) = 0.25$.

Using the CDF for $0 < x < 1$:

$ Q_1^4 = 0.25 $

$ Q_1^4 = \dfrac{1}{4} $

Taking the fourth root of both sides (and since $Q_1$ must be positive as per the domain):

$ Q_1 = \left( \dfrac{1}{4} \right)^{\dfrac{1}{4}} $

Finding the Third Quartile ($Q_3$)

We need to find $Q_3$ such that $F(Q_3) = 0.75$.

Using the CDF for $0 < x < 1$:

$ Q_3^4 = 0.75 $

$ Q_3^4 = \dfrac{3}{4} $

Taking the fourth root of both sides (and since $Q_3$ must be positive as per the domain):

$ Q_3 = \left( \dfrac{3}{4} \right)^{\dfrac{1}{4}} $

Calculating the Interquartile Range (IQR)

The interquartile range (IQR) is $Q_3 - Q_1$.

$ IQR = \left( \dfrac{3}{4} \right)^{\dfrac{1}{4}} - \left( \dfrac{1}{4} \right)^{\dfrac{1}{4}} $

This matches one of the given options.

Summary of Interquartile Range Calculation

Here's a quick summary of the steps:

Find the CDF, $F(x)$, from the given PDF, $f(x)$.
Set $F(x) = 0.25$ and solve for $x$ to find $Q_1$.
Set $F(x) = 0.75$ and solve for $x$ to find $Q_3$.
Calculate $IQR = Q_3 - Q_1$.

Revision Table: Key Concepts for IQR

Concept	Definition	How to Calculate for PDF
PDF ($f(x)$)	Describes the relative likelihood for a continuous random variable to take on a given value.	Given in the problem or derived. Must integrate to 1 over its support.
CDF ($F(x)$)	Gives the probability that a random variable $X$ is less than or equal to $x$. $F(x) = P(X \le x)$.	Integral of the PDF: $F(x) = \int_{-\infty}^{x} f(t) dt$.
Quartiles ($Q_1, Q_2, Q_3$)	Values that divide the probability distribution into four equal parts (25% each). $Q_1$ is 25th percentile, $Q_2$ is 50th (median), $Q_3$ is 75th.	Find $x$ such that $F(x) = p$, where $p$ is 0.25 for $Q_1$, 0.50 for $Q_2$, and 0.75 for $Q_3$.
Interquartile Range (IQR)	Difference between the third and first quartiles. Measures the spread of the middle 50%.	$IQR = Q_3 - Q_1$.

Additional Information on Percentiles and IQR

Quartiles are specific types of percentiles. The first quartile ($Q_1$) is the 25th percentile, the second quartile ($Q_2$, which is also the median) is the 50th percentile, and the third quartile ($Q_3$) is the 75th percentile. The IQR is a robust measure of spread because it is not affected by extreme outliers, unlike the range or standard deviation.

For any continuous random variable with PDF $f(x)$, if the support is $[a, b]$, the CDF is $F(x) = \int_{a}^{x} f(t) dt$. To find the percentile $P_p$ (where $p$ is a probability between 0 and 1), you solve $F(P_p) = p$. The quartiles are special cases where $p$ is 0.25, 0.50, and 0.75.

Prepp Team · Answer

$\left( \dfrac{3}{4} - \dfrac{1}{4} \right)^4$

Prepp Team · Answer

$\left( \dfrac{3}{4} - \dfrac{1}{4} \right)^\dfrac{1}{4}$

Prepp Team · Answer

$\left( \dfrac{1}{4} \right)^\dfrac{1}{4} - \left( \dfrac{3}{4} \right)^\dfrac{1}{4}$

Concept	Definition	How to Calculate for PDF
PDF (\(f(x)\))	Describes the relative likelihood for a continuous random variable to take on a given value.	Given in the problem or derived. Must integrate to 1 over its support.
CDF (\(F(x)\))	Gives the probability that a random variable \(X\) is less than or equal to \(x\). \(F(x) = P(X \le x)\).	Integral of the PDF: \(F(x) = \int_{-\infty}^{x} f(t) dt\).
Quartiles (\(Q_1, Q_2, Q_3\))	Values that divide the probability distribution into four equal parts (25% each). \(Q_1\) is 25th percentile, \(Q_2\) is 50th (median), \(Q_3\) is 75th.	Find \(x\) such that \(F(x) = p\), where \(p\) is 0.25 for \(Q_1\), 0.50 for \(Q_2\), and 0.75 for \(Q_3\).
Interquartile Range (IQR)	Difference between the third and first quartiles. Measures the spread of the middle 50%.	\(IQR = Q_3 - Q_1\).

X	0	1	2	3	4	5	6	7	8
f(x)	K	2k	3k	5k	5k	4k	3k	2k	k