Question

A survey conducted on 20 households in a locality by a group of statements resulted in the following frequency table for the number of family members in a household.

Family size	1-3	3-5	5-7	7-9	9-11
Number of families	7	8	2	2	1

Find the mode of the given data.

Answer 1

Hint: We start solving the problem by checking whether the given data in the frequency table is continuous or not. If the data is continuous, we proceed further to find the mode otherwise, we add the correction factor to lower and upper boundaries of family size intervals to make it continuous. We then find the family size with the highest number of families to get the modal group. We then use the formula ${{\operatorname{m}}_{e}}=L+\dfrac{{{f}_{m}}-{{f}_{m-1}}}{\left( {{f}_{m}}-{{f}_{m-1}} \right)+\left( {{f}_{m}}-{{f}_{m+1}} \right)}\times I$ and substitute the required values to get the required value of mode.

Complete step by step answer:
According to the problem, we are given a frequency table representing the survey conducted on 20 households in a locality by a group of statements. We need to find the mode of that given data.
Let us rewrite the given table

Family size	1-3	3-5	5-7	7-9	9-11
Number of families	7	8	2	2	1

We can see that the upper limit of the previous interval of family size is equal to the lower limit of the next interval of family size, which makes the given data in the frequency table continuous.
We know that the mode is defined as the value which appears most often in a data set. Let us find the interval of family size in which the number of families is higher. We can see that the interval $3-5$ has a higher number of families 8.
We know that the mode of a grouped distribution is defined as ${{\operatorname{m}}_{e}}=L+\dfrac{{{f}_{m}}-{{f}_{m-1}}}{\left( {{f}_{m}}-{{f}_{m-1}} \right)+\left( {{f}_{m}}-{{f}_{m+1}} \right)}\times I$.
Where, L = lower boundary of the interval of family size in which we get the mode = 3.
${{f}_{m-1}}$ = no. of families in the family size which is just before the interval we get the mode = 7.
${{f}_{m}}$ = no. of families in the family size interval in which we get the mode = 8.
${{f}_{m+1}}$ = no. of families in the family size which is just after the interval we get the mode = 2.
I = size of the interval in family size = 2.
Now, let us these values in the formula ${{\operatorname{m}}_{e}}=L+\dfrac{{{f}_{m}}-{{f}_{m-1}}}{\left( {{f}_{m}}-{{f}_{m-1}} \right)+\left( {{f}_{m}}-{{f}_{m+1}} \right)}\times I$.
\[\Rightarrow {{\operatorname{m}}_{e}}=3+\dfrac{8-7}{\left( 8-7 \right)+\left( 8-2 \right)}\times 2\].
\[\Rightarrow {{\operatorname{m}}_{e}}=3+\dfrac{1}{1+6}\times 2\].
\[\Rightarrow {{\operatorname{m}}_{e}}=3+\dfrac{2}{7}\].
\[\Rightarrow {{\operatorname{m}}_{e}}=3+0.29\].
\[\therefore {{\operatorname{m}}_{e}}=3.29\].
So, we have found the mean of the given frequency distribution is 3.29.

Note:
We should know that the obtained mode is an estimate not an absolute value as the raw data is exactly not known to us. We should know that the obtained mode is near to the absolute mode value as we can see that the number of families is higher in family sizes $1-3$ and $3-5$. We should not confuse the notations and make calculation mistakes while solving this problem. Similarly, we expect problems to find the mean, median for the given data.