Question

Find the mode and median of the following frequency distribution:

\[x\]	10	11	12	13	14	15
\[f\]	1	4	7	5	9	8

Answer 1

Hint: The mean of a frequency distribution is the ratio of the sum of the products of the observation and the frequency to the total numbers of frequencies in the distribution, i.e.,
\[\text{Mean} = \dfrac{\sum_{i=1} (f_{i}\times x_{i})}{\sum f}\]
Mode of a frequency distribution corresponds to that class interval which has the highest frequency.

Complete step-by-step answer:
According to the given question, the frequency distribution is,

\[x\]	\[10\]	\[11\]	\[12\]	\[13\]	\[14\]	\[15\]
\[f\]	\[1\]	\[4\]	\[7\]	\[5\]	\[9\]	\[8\]

Since the class interval has equal distribution, the mean of the above frequency distribution is given as,
Mean \[=\dfrac{\sum\limits_{i=1}{\left( {{f}_{i}}\times {{x}_{i}} \right)}}{\sum{f}}\]
Expanding the values of the observation and frequency uptil six places of \[i\], we have the formula of mean re-written as:
Mean \[=\dfrac{{{x}_{1}}{{f}_{1}}+{{x}_{2}}{{f}_{2}}+{{x}_{3}}{{f}_{3}}+{{x}_{4}}{{f}_{4}}+{{x}_{4}}{{f}_{4}}+{{x}_{4}}{{f}_{4}}}{{{f}_{1}}+{{f}_{2}}+{{f}_{3}}+{{f}_{4}}+{{f}_{5}}+{{f}_{6}}}\]
Now placing the values in the expanded valued formula of mean, we get the mean as:
\[=\dfrac{1\times 10+4\times 11+7\times 12+5\times 13+9\times 14+8\times 15}{1+4+7+5+9+8}\]
\[=\dfrac{10+44+84+65+126+120}{34}\]
\[=\dfrac{449}{34}\]
\[=13.2\]
The mode of any frequency distribution is that observation which has the maximum frequency.
Here, the maximum frequency amongst the other given frequency is \[9\] corresponding to the observation \[14\].
So, mode of the above frequency distribution is \[14\].

Note: If the data are in class intervals, then the mean can be calculated using the same formula. Here the middle point of the class interval is taken as the observation and is multiplied with the frequency.