From the image, you have a frequency distribution table with class intervals and their corresponding frequencies. To find the variance and standard deviation, let's break it down.

Frequency Distribution:

Steps to calculate variance and standard deviation:

1. Find the midpoint of each class: The midpoint for each class is calculated as the average of the lower and upper bounds of the class interval.

   • $M_1 = \frac{130 + 138}{2} = 134$
   • $M_2 = \frac{139 + 147}{2} = 143$
   • $M_3 = \frac{148 + 156}{2} = 152$
   • $M_4 = \frac{157 + 165}{2} = 161$
   • $M_5 = \frac{166 + 174}{2} = 170$

2. Calculate the mean: The mean is the weighted average of the midpoints. $\text{Mean} = \frac{\sum (M_i \cdot f_i)}{\sum f_i}$ where $M_i$ is the midpoint of each class and $f_i$ is the frequency.

3. Variance formula: $\text{Variance} = \frac{\sum f_i (M_i - \text{Mean})^2}{\sum f_i}$

4. Standard deviation: This is the square root of the variance.

Let me calculate these values for you.The calculations are as follows:

• Mean: 147.5
• Variance: 122.85
• Standard Deviation: 11.08 (rounded to two decimal places)

Would you like further details or explanations on these calculations?

Here are 5 related questions to expand your understanding:

1. How does the choice of class intervals affect the variance and standard deviation?
2. What is the importance of midpoints in grouped data calculations?
3. How would the results change if the frequencies were modified slightly?
4. Why is standard deviation more commonly used than variance in data interpretation?
5. How would you interpret a higher versus a lower standard deviation?

Tip: In grouped data, always use the class midpoints to calculate measures of central tendency and dispersion (like mean and variance).