Question

For grouped data, Arithmetic Mean by Assumed Mean method: -
(a) \[A+\dfrac{Sd}{N}\]
(b) \[A+\dfrac{Sfd}{Sf}\]
(c) \[\dfrac{SfX}{Sf}\]
(d) None of these

Answer 1

Hint: Consider an example of table containing grouped data of marks of some students. Assume ‘\[{{X}_{i}}\]’ as the mid – point of the interval of marks. Here, i = 1, 2, 3, ……n. Assume ‘A’ as the assumed mean that can be considered as any \[{{X}_{i}}\] value. Find \[{{d}_{i}}={{X}_{i}}-A\], where ‘\[{{d}_{i}}\]’ is the deviation for each \[{{X}_{i}}\]. Find out the summation of product of f and d, denoted by \[\sum{fd}\] then take its ratio with summation of f, denoted by \[\sum{f}\]. Finally add this ratio with assumed mean ‘A’ to get the answer.

Complete step by step answer:
We have to find, Arithmetic Mean by Assumed Mean method for grouped data. In general, this method is recommended rather than a direct method if the given data is large. Assumed Mean method helps in reducing the calculations.
Let us consider an example to find the formula for grouped data. In general, this method is recommended rather than a direct method if the given data is large. Assumed Mean method helps in reducing the calculations.
Let us consider an example to find the formula for mean by assumed mean method. Let us consider the following observations in a grouped form: -

Marks obtained	\[{{X}_{i}}\]	\[{{f}_{i}}\]
10 – 20	\[{{X}_{1}}=15\]	\[{{f}_{1}}=4\]
20 – 30	\[{{X}_{2}}=25\]	\[{{f}_{2}}=3\]
30 – 40	\[{{X}_{3}}=35\]	\[{{f}_{3}}=2\]
40 – 50	\[{{X}_{4}}=45\]	\[{{f}_{4}}=6\]
50 – 60	\[{{X}_{5}}=55\]	\[{{f}_{5}}=1\]

The following table indicates the marks obtained by 16 students. Here, marks are classified in interval. This is an example of grouped data.
Now, \[{{X}_{i}}\] is the mid – point of marks intervals and \[{{f}_{i}}\] is the frequency or number of students in that interval of marks. So, we have to assume a mean ‘A’, called assumed mean, which can be any one of \[{{X}_{i}}\] value. Let us assume A = 35, that is \[{{X}_{3}}\].
Now, we have to calculate deviation (\[{{d}_{i}}\]) for each \[{{X}_{i}}\], given by the relation: - \[{{d}_{i}}={{X}_{i}}-A\]. Therefore,
\[\begin{align}
  & \Rightarrow {{d}_{1}}={{X}_{1}}-A=15-35=-20 \\
 & \Rightarrow {{d}_{2}}={{X}_{2}}-A=25-35=-10 \\
 & \Rightarrow {{d}_{3}}={{X}_{3}}-A=35-35=0 \\
 & \Rightarrow {{d}_{4}}={{X}_{4}}-A=45-35=10 \\
 & \Rightarrow {{d}_{5}}={{X}_{5}}-A=55-35=20 \\
\end{align}\]
So, here the mean is calculated by taking the sum of ‘A’ with the ratio of \[\sum{{{f}_{i}}{{d}_{i}}}\] to \[\sum{{{f}_{i}}}\].
\[\Rightarrow \] Mean = \[A+\dfrac{\sum{{{f}_{i}}{{d}_{i}}}}{\sum{{{f}_{i}}}}\]
 \[\Rightarrow \] Mean = \[35+\dfrac{{{f}_{1}}{{d}_{1}}+{{f}_{2}}{{d}_{2}}+{{f}_{3}}{{d}_{3}}+{{f}_{4}}{{d}_{4}}+{{f}_{5}}{{d}_{5}}}{{{f}_{1}}+{{f}_{2}}+{{f}_{3}}+{{f}_{4}}+{{f}_{5}}}\]
\[\Rightarrow \] Mean = \[35+\dfrac{4\times \left( -20 \right)+3\times \left( -10 \right)+2\times 0+6\times 10+1\times 20}{4+3+2+6+1}\]
\[\Rightarrow \] Mean = \[35+\dfrac{-80-30+60+20}{16}\]
\[\Rightarrow \] Mean = \[35-\dfrac{30}{16}\approx 33.125\]
So, the formula for mean using assumed mean method, for grouped data is, \[A+\dfrac{\sum{{{f}_{i}}{{d}_{i}}}}{\sum{{{f}_{i}}}}=A+\dfrac{\sum{fd}}{\sum{f}}\] (without using indices).

So, the correct answer is “Option A”.

Note: One may note that, in the option we have been provided with \[A+\dfrac{Sfd}{Sf}\]. Actually, here ‘S’ denotes the summation, that is ‘\[\sum{{}}\]’. So, we must not think that ‘S’ is any particular term or variable. Also, note that \[\sum{f}\] can be also written as N, which denotes the total number of observations (here it means students). So, the required formula can also be written as: - Mean = \[A+\dfrac{\sum{fd}}{N}\].