Consider the following frequency distribution table. If the total frequency is 685 and the median is 42.6, find the values of \(f_1\) and \(f_2\).
\[\begin{array}{c|ccccccc}\text{Class Interval} & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 & 60-70 & 70-80 \\ \hline \text{Frequency} & 180 & f_1 & 34 & 180 & 136 & f_2 & 50\end{array}\]
Concept:
The median of a grouped frequency distribution is calculated using:
\[\text{Median}=l+\left(\frac{\frac{N}{2}-cf}{f}\right)h\]
where \(l\) is the lower boundary of the median class, \(N\) is the total frequency, \(cf\) is the cumulative frequency before the median class, \(f\) is the frequency of the median class, and \(h\) is the class width.
Step 1: Use the total frequency.
\[180+f_1+34+180+136+f_2+50=685\]
\[f_1+f_2+580=685\]
\[f_1+f_2=105\quad ...(1)\]
Step 2: Identify the median class.
\[\frac{N}{2}=\frac{685}{2}=342.5\]
The median class is \(40-50\). Therefore, \(l=40\), \(cf=180+f_1+34=214+f_1\), \(f=180\), and \(h=10\).
Step 3: Apply the median formula.
\[42.6=40+\left(\frac{342.5-(214+f_1)}{180}\right)10\]
\[2.6=\frac{128.5-f_1}{18}\]
\[46.8=128.5-f_1\]
\[f_1=81.7\]
Reinforce your conceptual grasp with related questions from NIMCET 2023:
Take timed full-length mock tests with authentic question distributions, negative marking (+12/-3 in Math), and instant All India Rank calculation.
Start Free Mock Test Series →