Find the mean of the following frequency distribution:
Show Hint
To calculate the mean of a frequency distribution, multiply each midpoint by its corresponding frequency, sum the results, and then divide by the total frequency.
The given frequency distribution is:
Step 1: Find the midpoints of each class-interval.
The midpoint of a class-interval \( [l, u] \) is given by:
\[
\text{Midpoint} = \frac{l + u}{2}.
\]
Thus, the midpoints for each class-interval are:
- \( 20-60: \frac{20 + 60}{2} = 40 \)
- \( 60-100: \frac{60 + 100}{2} = 80 \)
- \( 100-150: \frac{100 + 150}{2} = 125 \)
- \( 150-250: \frac{150 + 250}{2} = 200 \)
- \( 250-350: \frac{250 + 350}{2} = 300 \)
- \( 350-450: \frac{350 + 450}{2} = 400 \)
Step 2: Multiply the frequency by the corresponding midpoint.
Now, multiply each midpoint by the corresponding frequency:
- \( 40 \times 7 = 280 \)
- \( 80 \times 5 = 400 \)
- \( 125 \times 16 = 2000 \)
- \( 200 \times 12 = 2400 \)
- \( 300 \times 2 = 600 \)
- \( 400 \times 3 = 1200 \)
Step 3: Calculate the sum of the frequencies and the sum of the products.
The sum of the frequencies is:
\[
7 + 5 + 16 + 12 + 2 + 3 = 45.
\]
The sum of the products of midpoints and frequencies is:
\[
280 + 400 + 2000 + 2400 + 600 + 1200 = 5880.
\]
Step 4: Find the mean.
The mean is given by:
\[
\text{Mean} = \frac{\sum f \times x}{\sum f},
\]
where \( \sum f \times x \) is the sum of the products and \( \sum f \) is the sum of the frequencies.
Substitute the values:
\[
\text{Mean} = \frac{5880}{45} = 130.67.
\]
Conclusion:
The mean of the given frequency distribution is \( 130.67 \).
