Question 107

Mathematics Statistics Easy

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 \(f_2\).

\[\begin{array}{c|ccccccc}\text{Class Interval} & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 & 60-70 & 70-80 \\ \hline \text{Frequency} & 180 & f_1 & 34 & 180 & 136 & f_2 & 50\end{array}\]

(A) 80, 25
(B) 83, 22
(C) 79, 26
(D) 82, 23
✓ Correct Answer: Option D Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

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.
\[42.6=40+\left(\frac{342.5-(214+f_1)}{180}\right)10\]
\[2.6=\frac{128.5-f_1}{18}\]
\[46.8=128.5-f_1\]
\[f_1=81.7\]

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