Find the 50th percentile (median) the 25th percentile (first quartile) the 75th percentile (third quartile) and the 90th percentile for the following densities
a) $f(x) = 4x^3 , 0 \leq X < 1$
b) $f(x) = e^-x , 0 \leq X < \infty$
Solution
(a)
$$ \begin{equation*} f(x) = 4x^3, 0\leq x < 1 \end{equation*} $$
Distribution function of $X$ is
$$ \begin{eqnarray*} F(x) &=& P(X\leq x) \\ &=&\int_0^x f(x) \; dx \\ &=& \int_0^x 4x^3\;dx\\ & =& \bigg(\frac{4x^4}{4}\bigg)_{x=0}^x\\ & = & x^4. \end{eqnarray*} $$
- 25th percentile: $25^{th}$ percentile which is same as $1^{st}$ quartile is given by
$$ \begin{eqnarray*} F(p_{25}) &=& 0.25 \\ p_{25}^4 &=& 0.25 \\ p_{25} &=& (0.25)^{1/4}= 0.707. \end{eqnarray*} $$
- 50th percentile: $50^{th}$ percentile which is same as $2^{nd}$ quartile is given by
$$ \begin{eqnarray*} F(p_{50}) &=& 0.50 \\ p_{50}^4 &=& 0.50 \\ p_{50} &=& (0.50)^{1/4}= 0.8409. \end{eqnarray*} $$
- 75th percentile: $75^{th}$ percentile which is same as $3^{rd}$ quartile is given by
$$ \begin{eqnarray*} F(p_{75}) &=& 0.75 \\ p_{75}^4 &=& 0.75 \\ p_{75} &=& (0.75)^{1/4}= 0.9147. \end{eqnarray*} $$
- 90th percentile: $90^{th}$ percentile is given by
$$ \begin{eqnarray*} F(p_{90}) &=& 0.90 \\ p_{90}^4 &=& 0.90 \\ p_{90} &=& (0.90)^{1/4}= 0.974. \end{eqnarray*} $$
(b)
The given probability density function is
$$ \begin{equation*} f(x) = e^{-x}, 0\leq x < \infty \end{equation*} $$
Distribution function of $X$ is
$$ \begin{eqnarray*} F(x) &=& P(X\leq x) \\ &=&\int_0^x f(x) \; dx \\ &=& \int_0^x e^{-x}\;dx\\ & =& \bigg(-e^{-x}\bigg)_{x=0}^x\\ & = & 1-e^{-x}. \end{eqnarray*} $$
- 25th percentile: $25^{th}$ percentile which is same as $1^{st}$ quartile is given by
$$ \begin{eqnarray*} F(p_{25}) &=& 0.25 \\ 1-e^{-p_{25}} &=& 0.25 \\ e^{-p_{25}} &=& 1-0.25\\ p_{25} & = & -\ln(1-0.25)\\ & = & 0.2877. \end{eqnarray*} $$
- 50th percentile: $25^{th}$ percentile which is same as $1^{st}$ quartile is given by
$$ \begin{eqnarray*} F(p_{50}) &=& 0.50 \\ 1-e^{-p_{50}} &=& 0.50 \\ e^{-p_{50}} &=& 1-0.50\\ p_{50} & = & -\ln(1-0.50)\\ & = & 0.6931. \end{eqnarray*} $$
- 75th percentile: $25^{th}$ percentile which is same as $1^{st}$ quartile is given by
$$ \begin{eqnarray*} F(p_{75}) &=& 0.75 \\ 1-e^{-p_{75}} &=& 0.75 \\ e^{-p_{75}} &=& 1-0.75\\ p_{75} & = & -\ln(1-0.75)\\ & = & 1.3863. \end{eqnarray*} $$
- 90th percentile: $90^{th}$ percentile is given by
$$ \begin{eqnarray*} F(p_{90}) &=& 0.90 \\ 1-e^{-p_{90}} &=& 0.90 \\ e^{-p_{90}} &=& 1-0.95\\ p_{90} & = & -\ln(1-0.95)\\ & = & 2.3026. \end{eqnarray*} $$
Do read my step by step tutorial on Problem 105 - Distribution of square of standard normal variate.
Let me know in the comments if you have any questions on Problem 106 - Finding percentile from the probability density function and your thought on this article.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
