| Class : | 4 – 6 | 7 – 9 | 10 – 12 | 13 – 15 |
| Frequency : | 5 | 4 | 9 | 10 |
Step 1: Find the class midpoints
The midpoints ($x_i$) are calculated as the average of the lower and upper limits of each class:
\[x_1 = \frac{4 + 6}{2} = 5, \quad x_2 = \frac{7 + 9}{2} = 8, \quad x_3 = \frac{10 + 12}{2} = 11, \quad x_4 = \frac{13 + 15}{2} = 14.\]
Step 2: Multiply midpoints by frequencies
Construct the table as follows:
\[\begin{array}{|c|c|c|c|}\hline\text{Class} & \text{Frequency } (f_i) & \text{Midpoint } (x_i) & f_i x_i \\\hline4 - 6 & 5 & 5 & 25 \\7 - 9 & 4 & 8 & 32 \\10 - 12 & 9 & 11 & 99 \\13 - 15 & 10 & 14 & 140 \\\hline\end{array}\]
Step 3: Calculate the total frequency and total $f_i x_i$
\[\sum f_i = 5 + 4 + 9 + 10 = 28.\]
\[\sum f_i x_i = 25 + 32 + 99 + 140 = 296.\]
Step 4: Calculate the mean
The formula for the mean is:
\[\text{Mean} = \frac{\sum f_i x_i}{\sum f_i}.\]
Substitute the values:
\[\text{Mean} = \frac{296}{28} = 10.57.\]
Correct Answer: Mean = 10.57
Let the Mean and Variance of five observations $ x_i $, $ i = 1, 2, 3, 4, 5 $ be 5 and 10 respectively. If three observations are $ x_1 = 1, x_2 = 3, x_3 = a $ and $ x_4 = 7, x_5 = b $ with $ a>b $, then the Variance of the observations $ n + x_n $ for $ n = 1, 2, 3, 4, 5 $ is
Find the variance of the following frequency distribution:
| Class Interval | ||||
| 0--4 | 4--8 | 8--12 | 12--16 | |
| Frequency | 1 | 2 | 2 | 1 |
आप अदिति / आदित्य हैं। आपकी दादीजी को खेलों में अत्यधिक रुचि है। ओलंपिक खेल-2024 में भारत के प्रदर्शन के बारे में जानकारी देते हुए लगभग 100 शब्दों में पत्र लिखिए।