Question

Class interval
0-20
20-40
40-60
60-80
80-100
100-120
20
35
Frequency
52
44
38
31
please explain me in details by using Shortcut Method (Assumed Mean Method)​

Answer 1

Question : -

$\begin{tabular}{|c|c|c|c|c|c|c|}\cline{1-7} \sf Class interval (CI) & 0-20 & 20-40 & 40-60 & 60-80 & 80-100 & 100-120 \\ \cline{1-7} \sf Frequency (f)</p><p>& 20 & 35 & 52 & 44 & 38 & 31 \\ \cline{1-7}\end{tabular}$

ANSWER

Given : -

$\begin{tabular}{|c|c|c|c|c|c|c|}\cline{1-7} \sf Class interval (CI) & 0-20 & 20-40 & 40-60 & 60-80 & 80-100 & 100-120 \\ \cline{1-7} \sf Frequency (f)</p><p>& 20 & 35 & 52 & 44 & 38 & 31 \\ \cline{1-7}\end{tabular}$

Required to find : -

• Mean of the whole data ?

Condition mentioned : -

Solve this using the assumed mean method !

Formula used : -

$\sf{\bf{ Mean (\bar{x}) = a + \dfrac{ \sum fi di }{\sum fi}}}$

Solution : -

$\begin{tabular}{|c|c|c|c|c|c|c|}\cline{1-7} \sf Class interval (CI) & 0-20 & 20-40 & 40-60 & 60-80 & 80-100 & 100-120 \\ \cline{1-7} \sf Frequency (f)</p><p>& 20 & 35 & 52 & 44 & 38 & 31 \\ \cline{1-7}\end{tabular}$

We need to find the mean ( using the assumed mean method ) !

In order to find the mean we need to prepare some more columns in the given stratical data .

The extra columns which we needed are ; Mid values (xi) & The product of fi & xi values of the respective class intervals.

So,

In order to find the mid values (xi) we need to take the average of the corresponding upper bound & lower bound of the class interval.

Similarly,

The xi ( mid value ) corresponding to the highest frequency is considered as the Assumed mean (a) [ In general]

Let's prepare the table ...

This data will be represented as;

$\begin{tabular}{|c|c|c|c|c|} \cline{1-5} \sf Class Interval (CI) & \sf Frequency (f) & xi & di(di = xi - a)& fidi \\ \cline{1-5} 0-20 & 20 & 10 & - 40& - 800 \\ \cline{1-5} 20-40 & 35 & 30 & - 20& - 700 \\ \cline{1-5} 40-60 & \qquad 52 \longrightarrow & \qquad \: 50 \longrightarrow \sf{( a)}&0&0 \\ \cline{1-5} 60-80 & 44 &70 &20&880 \\ \cline{1-5} 80-100 & 38 &90 &40&1520 \\ \cline{1-5} 100-120 & 31 &110 &60&1860 \\ \cline{1-5} & \sum \sf fi =220 &&& \sum \sf fidi =2760 & \cline{1 - 5} \end{tabular}$

Using the formula;

$\sf{\bf{ Mean (\bar{x}) = a + \dfrac{ \sum fi di }{\sum fi}}}$

Here,

fidi = product of the frequency & deviation

fi = frequency

a = assumed mean

The values of these are;

• fidi = 2760
• fi = 220
• a = 50

Substituting the values;

$\: \sf \bf Mean(\bar{x}) = 50 + \dfrac{2260}{220} \left\lgroup \pink{ \because \: \rm Mean = a + \frac{ \sum fidi}{ \sum fi} }\right\rgroup \\ \\ \sf \bf Mean (\bar{x}) = 50 + \dfrac{226}{22} \\ \\ \sf \bf Mean (\bar{x} ) = 50 + 10.2727 \\ \\ \sf \bf Mean ( \bar x ) = 60.2727$

Therefore,

Mean of the above data = 60.2727