We are given that \( \bar{x} = 7 \) and \( \sum_{i=1}^{8} f_i x_i = 315 \), and we are asked to find \( \sum_{i=1}^{8} f_i \).
Recall that the mean \( \bar{x} \) is given by:
\[ \bar{x} = \frac{\sum_{i=1}^{8} f_i x_i}{\sum_{i=1}^{8} f_i} \] Substitute the given values into the equation: \[ 7 = \frac{315}{\sum_{i=1}^{8} f_i} \] Solving for \( \sum_{i=1}^{8} f_i \): \[ \sum_{i=1}^{8} f_i = \frac{315}{7} = 45 \]
Answer: 45
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 mean of the following distribution:
\[\begin{array}{|c|c|c|c|c|c|c|c|} \hline \textbf{Class-interval} & 11-13 & 13-15 & 15-17 & 17-19 & 19-21 & 21-23 & 23-25 \\ \hline \text{Frequency} & \text{7} & \text{6} & \text{9} & \text{13} & \text{20} & \text{5} & \text{4} \\ \hline \end{array}\]