32.71
First, calculate the mean of the data: \[ \text{Mean} = \frac{1 + 2 + 3 + 5 + 8 + 13 + 17}{7} = \frac{49}{7} = 7 \] Next, compute the sum of the squared differences from the mean: \[ \text{Sum of squares} = (1-7)^2 + (2-7)^2 + (3-7)^2 + (5-7)^2 + (8-7)^2 + (13-7)^2 + (17-7)^2 \] \[ = 36 + 25 + 16 + 4 + 1 + 36 + 100 = 218 \] Finally, the variance (\(\sigma^2\)) is calculated as: \[ \sigma^2 = \frac{\text{Sum of squares}}{n} = \frac{218}{7} \approx 31.14 \]
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:
\[\begin{array}{|c|c|c|c|c|c|c|c|} \hline \textbf{Class-interval} & 11-13 & 13-15 & 15-17 & 17-19 & 19-21 & 21-23 & 23-25 \\ \hline \text{Frequency} & \text{7} & \text{6} & \text{9} & \text{13} & \text{20} & \text{5} & \text{4} \\ \hline \end{array}\]