Calculate ${{D}_{4}}$ and ${{P}_{48}}$ from the following data:$$

Mid Value	2.5	7.5	12.5	17.5	22.5	Total
Frequency	7	18	25	30	20	100

Hint: To solve the problem we need to have the knowledge of statistics. The first step is to change the mid value into the class interval. We will find the class interval by calculating the difference of the mid value. We will calculate the cumulative frequency, and will apply the formula to find ${{D}_{4}}$ and ${{P}_{48}}$.

Complete step-by-step solution:
The question asks us to find ${{D}_{4}}$ and ${{P}_{48}}$ from the following data given in the question. The first step is to change mid- value into the class interval having certain number of frequency so the table thus formed is given below:


Now we are given the total number of frequencies which is denoted by “N= 100”. ${{D}_{4}}$ basically indicate the class containing ${{\left( \dfrac{4N}{10} \right)}^{th}}$ observation. So mathematically it is represented by:
${{D}_{4}}$class = class containing ${{\left( \dfrac{4N}{10} \right)}^{th}}$ observation
On putting the values in the formula we get:
$\Rightarrow {{\left( \dfrac{4N}{10} \right)}^{th}}={{\left( \dfrac{4\times 100}{10} \right)}^{th}}$
$\Rightarrow {{40}^{th}}$ observation
For knowing the class in which ${{40}^{th}}$ observation lies we will see the cumulative frequency after $40$. Now the Cumulative frequency above $40$ is $50$. This shows ${{D}_{4}}$lies in an interval of $10-15$ .
To find the value of ${{D}_{4}}$the formula used will be:
$\Rightarrow {{D}_{4}}=L+\dfrac{h}{f}\left( \dfrac{4N}{10}-c.f. \right)$
As per the class interval the values of \[L,\text{ }h,\text{ }f,\text{ }c.f.\] are $10,5,25,25$ respectively. On putting these value in the formula we get:
$\Rightarrow {{D}_{4}}=10+\dfrac{5}{25}\left( \dfrac{4\times 100}{10}-25 \right)$
$\Rightarrow {{D}_{4}}=10+\dfrac{5}{25}\left( 40-25 \right)$
$\Rightarrow {{D}_{4}}=10+\dfrac{5}{25}\left( 15 \right)$
$\Rightarrow {{D}_{4}}=10+3$
$\Rightarrow {{D}_{4}}=13$
Similarly we will find the value of ${{P}_{48}}$.
${{P}_{48}}$class = class containing ${{\left( \dfrac{48N}{100} \right)}^{th}}$ observation
On putting the values in the formula we get:
$\Rightarrow {{\left( \dfrac{48N}{100} \right)}^{th}}={{\left( \dfrac{48\times 100}{100} \right)}^{th}}$
$\Rightarrow {{48}^{th}}$ observation
For knowing the class in which ${{48}^{th}}$ observation lies we will see the cumulative frequency after $48$. Now the Cumulative frequency above $48$ is $50$. This shows ${{D}_{4}}$lies in an interval of $10-15$ .
To find the value of ${{P}_{48}}$the formula used will be:
$\Rightarrow {{P}_{48}}=L+\dfrac{h}{f}\left( \dfrac{48N}{10}-c.f. \right)$
As per the class interval the values of \[L,\text{ }h,\text{ }f,\text{ }c.f.\] are $10,5,25,25$ respectively. On putting these value in the formula we get:
$\Rightarrow {{P}_{48}}=10+\dfrac{5}{25}\left( \dfrac{48\times 100}{100}-25 \right)$
$\Rightarrow {{P}_{48}}=10+\dfrac{1}{5}\left( 23 \right)$
$\Rightarrow {{P}_{48}}=10+4.6$
$\Rightarrow {{P}_{48}}=14.6$
$\therefore $ The value ${{D}_{4}}$ and ${{P}_{48}}$ from the following data given are $13$ and $14.6$ respectively.

Note:To solve this question the first step should be to find the cumulative frequency as per the data given. To find the length of the class interval we subtract any of the two consecutive mid- values given in the question.