The table below shows the daily expenditure on the food of 25 households in a locality.

Daily expenses (in Rs)	100 - 150	150 - 200	200 - 250	250 - 300	300 - 350
No of households	4	5	12	2	2

Find the mean daily expenses on food by a suitable method.

Hint: Apply the assumed mean method to find the mean daily expenses on food. Draw a frequency distribution table containing three columns. In column 1 consider the daily expenses, in column 2 consider the frequency, that is the number of households and in column 3 consider ${{X}_{i}}$, which is the midpoint of the interval of expenses. 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 ${{X}_{i}}$. Now, apply the formula : Mean = $A+\dfrac{\sum{{{f}_{i}}{{d}_{i}}}}{\sum{{{f}_{i}}}}$ to get the mean.

Complete step-by-step solution:
Here we have been provided with the following frequency distribution table.

We have to find the mean daily expenses for food.
Here, we will use the assumed mean method to calculate the mean because the given data is large. So, let us draw a table containing 3 columns. In column 1, we will consider the daily expenses, in column 2 frequency, that is the number of households will be considered and in column 3 we will consider ${{X}_{i}}$, which is the midpoint of the interval of expenses.

In the above frequency table, ${{f}_{i}}$ denotes the frequency or the number of households in a particular interval of expenses. Now, we have to assume a mean A, called as assumed mean which can be any one of ${{X}_{i}}$ values. Let us assume A = 225, 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,
$\begin{align}
  & {{d}_{1}}={{X}_{1}}-A=125-225=-100 \\
 & {{d}_{2}}={{X}_{2}}-A=175-225=-50 \\
 & {{d}_{3}}={{X}_{3}}-A=225-225=0 \\
 & {{d}_{4}}={{X}_{4}}-A=275-225=50 \\
 & {{d}_{5}}={{X}_{5}}-A=325-225=100 \\
\end{align}$
So, applying the formula to find mean by assumed mean method, we get,
Mean = $A+\dfrac{\sum{{{f}_{i}}{{d}_{i}}}}{\sum{{{f}_{i}}}}$
$=A+\left( \dfrac{{{f}_{1}}{{d}_{1}}+{{f}_{2}}{{d}_{2}}+{{f}_{3}}{{d}_{3}}+{{f}_{4}}{{d}_{4}}+{{f}_{5}}{{d}_{5}}}{{{f}_{1}}+{{f}_{2}}+{{f}_{3}}+{{f}_{4}}+{{f}_{5}}} \right)$
Therefore, substituting the required values, we get,
$\begin{align}
  & \Rightarrow \text{Mean}=225+\left[ \dfrac{4\times \left( -100 \right)+5\times \left( -50 \right)+12\times 0+2\times 50+2\times 100}{4+5+12+2+2} \right] \\
 & \Rightarrow \text{Mean}=225+\left[ \dfrac{-400-250+100+200}{25} \right] \\
 & \Rightarrow \text{Mean}=225+\left[ \dfrac{-350}{25} \right] \\
 & \Rightarrow \text{Mean}=225-14 \\
 & \Rightarrow \text{Mean}=211 \\
\end{align}$
Hence, mean daily expenses on food is 211.

Note: One may note that there are two ways to find the mean of the given grouped data. They are the Direct Method and the Assumed mean method, but here, we have preferred to use the method of the assumed mean. This is because the numbers in the daily expenses are large and therefore if the direct mean method is applied, then it will make our calculations much harder. At last, remember that we must know the formula for assumed mean such that we can proceed step-by-step accordingly.