Calculate the mean from the following data:
Marks $0 - 10$ $10 - 30$ $30 - 60$ $60 - 80$ $80 - 90$ No. of students $5$ $15$ $30$ $8$ $2$
| Marks | $0 - 10$ | $10 - 30$ | $30 - 60$ | $60 - 80$ | $80 - 90$ |
| No. of students | $5$ | $15$ | $30$ | $8$ | $2$ |
Answer
537.9k+ views
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
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.
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.
Recently Updated Pages
A Paragraph on Pollution in about 100-150 Words

What is BLO What is the full form of BLO class 8 social science CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE

1 GB equals how many MB?

10 examples of evaporation in daily life with explanations

What is the full form of POSCO class 10 social science CBSE

Which is the hottest planet in the Solar system A Earth class 10 social science CBSE

Name any four life processes in living things class 10 biology CBSE

