The data provided is grouped into class intervals along with the number of students in each group. To find the variance and standard deviation, we follow these steps:

Step 1: Organize the data

Step 2: Calculate midpoints ( $x_i$ )

For each class interval, the midpoint is calculated as:

$\text{Midpoint} = \frac{\text{Lower Bound} + \text{Upper Bound}}{2}$

Midpoints for the intervals:

$x_i = [5, 15, 25, 35, 45, 55, 65]$

Step 3: Calculate the mean ( $\bar{x}$ )

The mean is calculated as:

$\bar{x} = \frac{\sum (f_i x_i)}{\sum f_i}$

Where $f_i$ is the frequency and $x_i$ is the midpoint. I'll calculate this next.

Step 4: Calculate variance

Variance is given by:

$\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$

Step 5: Calculate standard deviation

The standard deviation is the square root of the variance:

$\sigma = \sqrt{\sigma^2}$

Let's calculate the mean, variance, and standard deviation now.### Results:

• Mean ( $\bar{x}$ ): 30.77
• Variance ( $\sigma^2$ ): 277.92
• Standard Deviation ( $\sigma$ ): 16.67

Let me know if you would like more details or have further questions.

Here are some related questions for practice:

1. How would the variance change if the class intervals were narrower (e.g., 5 marks)?
2. What is the range of the dataset, and how does it relate to standard deviation?
3. How does increasing the frequency in the highest class interval (60-70) affect the standard deviation?
4. Can you calculate the mean if the data were ungrouped?
5. What other statistical measures (like median or mode) could be derived from this grouped data?

Tip: Variance gives a sense of how spread out the data is, but the standard deviation is often more interpretable because it's in the same units as the data (marks in this case).