Explanation
Step 1: Create a Cumulative Frequency Table
Let's compute the cumulative frequency (cf) for each class interval in terms of f1:
Class Interval | Frequency (f) | Cumulative Frequency (cf) |
10−20 | 180 | 180 |
20−30 | f1 | 180+f1 |
30−40 | 34 | 214+f1 |
40−50 | 180 | 394+f1 |
50−60 | 136 | 530+f1 |
60−70 | f2 | 530+f1+f2 |
70−80 | 50 | 580+f1+f2 |
Step 2: Use Total Frequency to form Equation 1
We are given that the total frequency (N) is 685:
580+f1+f2=685
f1+f2=685−580
f1+f2=105
--- (Equation 1)
Step 3: Identify the Median Class and Apply Formula
The given median is 42.6. Since 42.6 lies in the interval 40−50, the median class is 40−50.
From this median class, we extract the required values for the median formula:
Lower limit of median class (l) = 40
Frequency of median class (f) = 180
Cumulative frequency of the preceding class (cf) = 214+f1
Class width (h) = 10
Total frequency (N) = 685⟹2N=2685=342.5
The formula for the median of a grouped frequency distribution is:
Median=l+(f2N−cf)×h
Substitute the values into the formula:
42.6=40+(180342.5−(214+f1))×10
Step 4: Solve for f1 and f2
Subtract 40 from both sides:
42.6−40=(180342.5−214−f1)×10
2.6=18128.5−f1
Cross-multiply to clear the fraction:
2.6×18=128.5−f1
46.8=128.5−f1
f1=128.5−46.8
f1=81.7≈82
Now substitute f1=82 into Equation 1 to find f2:
82+f2=105
f2=105−82
f2=23
Thus, the values are f1=82 and f2=23.
Correct Answer:
Correct Option: D
