Standard deviation for the following distribution is:</p> Here is the table formatted for better readability: \[ \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Size of Item} & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline \text{Frequency} & 3 & 6 & 9 & 13 & 8 & 5 & 4 \\ \hline \end{array} \]
Step-by-step Solution:
The formula for standard deviation is: \[ \sigma = \sqrt{\frac{\sum f x^2}{n} - (\bar{x})^2} \] Where \( \bar{x} \) is the mean of the data, and it is given by: \[ \bar{x} = \frac{\sum f x}{n} \] Now, from the given data: \[ \bar{x} = \frac{3 \cdot 6 + 6 \cdot 7 + 9 \cdot 8 + \dots + 4 \cdot 12}{3 + 6 + 9 + \dots + 4} = 9. \] Also, we are given that: \[ \sum f x^2 = 4012. \] Now, we can calculate the standard deviation as follows: \[ \sigma = \sqrt{\frac{4012}{48} - 9^2} = \sqrt{83.67 - 81} = \sqrt{2.67} \approx 1.607. \] Thus, the standard deviation is approximately \( \sigma = 1.607 \).