Question

Calculate the mean from the following data:

Marks	$0 - 10$	$10 - 30$	$30 - 60$	$60 - 80$	$80 - 90$
No. of students	$5$	$15$	$30$	$8$	$2$

Answer 1

Hint: The mean of the following data is the average of the data set. The mean is the ratio of summation of product of the midpoint and the frequency to the summation of frequency. The midpoint is the average taken between the marks given. The fraction this will give the mean value. The summation is nothing but the addition of all terms.
Formula:
Mean $M = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{x_i}} }}$, where ${f_i}$Is the midpoint and ${x_i}$is the students of the given data set.
Midpoint ${x_i} = \dfrac{{{m_i} + {m_{i + 1}}}}{2}$, where ${m_i}$is the first mark in the given data set and ${m_{i + 1}}$is the next first mark in the given data set.

Complete answer:
Given,
The given data set are

Marks	$0 - 10$	$10 - 30$	$30 - 60$	$60 - 80$	$80 - 90$
No. of students	$5$	$15$	$30$	$8$	$2$

The marks given are $0 - 10$, $10 - 30$ , $30 - 60$, $60 - 80$ and $80 - 90$.
The number of students is $5$ , $15$, $30$, $8$ and $2$.
We need to find the midpoint of marks,
The midpoint of the value is nothing but the average of the given data.
${m_1} = 0$
${m_2} = 10$
${m_3} = 30$
${m_4} = 60$
${m_5} = 80$and
${m_6} = 90$
Let us find the midpoint
${x_1} = \dfrac{{{m_1} + {m_2}}}{2}$
Substitute ${m_1} = 0$ and ${m_2} = 10$
${x_1} = \dfrac{{0 + 10}}{2}$
Add the above terms,
${x_1} = \dfrac{{10}}{2}$
Divide the terms in the numerator and the denominator,
${x_1} = 5$
Substitute ${m_3} = 30$ and ${m_2} = 10$
${x_2} = \dfrac{{30 + 10}}{2}$
Add the above terms,
${x_2} = \dfrac{{40}}{2}$
Divide the terms in the numerator and the denominator,
${x_2} = 8$
Substitute ${m_3} = 30$ and ${m_4} = 60$
${x_3} = \dfrac{{30 + 60}}{2}$
Add the above terms,
${x_3} = \dfrac{{90}}{2}$
Divide the terms in the numerator and the denominator,
${x_3} = 45$
Substitute ${m_4} = 60$ and ${m_5} = 80$
${x_4} = \dfrac{{60 + 80}}{2}$
Add the above terms,
${x_4} = \dfrac{{140}}{2}$
Divide the terms in the numerator and the denominator,
${x_4} = 70$
Substitute ${m_5} = 80$ and ${m_6} = 90$
${x_5} = \dfrac{{80 + 90}}{2}$
Add the above terms,
${x_5} = \dfrac{{170}}{2}$
Divide the terms in the numerator and the denominator,
${x_5} = 85$
The midpoint and no. of students are multiplied
${f_1} = 5$
${f_2} = 15$
${f_3} = 30$
${f_4} = 8$ and
${f_5} = 2$
The midpoints are
${x_1} = 5$
${x_2} = 8$
${x_3} = 45$
${x_4} = 70$ and ${x_5} = 85$
$\sum {{f_i}{x_i} = } {f_1}{x_1} + {f_2}{x_2} + {f_3}{x_3} + {f_4}{x_4} + {f_5}{x_5}$
Substitute ${f_1} = 5$, ${f_2} = 15$,${f_3} = 30$,${f_4} = 8$ and ${f_5} = 2$
${x_1} = 5$,${x_2} = 8$,${x_3} = 45$,${x_4} = 70$ and ${x_5} = 85$
$\sum {{f_i}{x_i} = } 5 \times 5 + 15 \times 8 + 30 \times 45 + 8 \times 70 + 8 \times 85$
Multiply the above terms
$\sum {{f_i}{x_i} = } 25 + 120 + 1350 + 560 + 680$
Add the terms in the above equation,
$\sum {{f_i}{x_i} = } 2735$
Now we should calculate $\sum {{x_i}} $is the sum of the ${x_i}$
$\sum {{x_i} = } {x_1} + {x_2} + {x_3} + {x_4} + {x_5}$
Substitute ${x_1} = 5$,${x_2} = 8$,${x_3} = 45$,${x_4} = 70$ and ${x_5} = 85$
$\sum {{x_i} = } 5 + 8 + 45 + 70 + 85$
Add the terms in the above equation,
$\sum {{x_i} = } 213$
Calculate $M = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{x_i}} }}$
Substitute $\sum {{x_i} = } 213$ and $\sum {{f_i}{x_i} = } 2735$,
$M = \dfrac{{2735}}{{213}}$
Divide the numerator and the denominator
$M = 12.84$

Note:
The mean of the following data is the average of the data set. The mean is the ratio of summation of product of the midpoint and the frequency to the summation of frequency. The midpoint is the average taken between the marks given. The fraction this will give the mean value. The summation is nothing but the addition of all terms. The formula must be added.