The image contains mathematical problems related to statistics, particularly frequency distributions, measures of central tendency, and quartiles. Below is the breakdown of the visible tasks, and I'll address one problem as an example.

Problem 8:

Frequency Table:

Tasks:

(a) Find the modal class:
The modal class is the class interval with the highest frequency.

• From the table, the frequency values are: 3, 4, 5, 4, 6.
• The highest frequency is 6, corresponding to the class interval 40 - 50.
  Answer: Modal class is 40 - 50.

(b) Find the median using the formula:
The formula for the median is:
$\text{Median} = L + \left(\frac{\frac{N}{2} - cf}{f}\right) \cdot h$ Where:

• $L$ = lower boundary of the median class
• $N$ = total frequency
• $cf$ = cumulative frequency before the median class
• $f$ = frequency of the median class
• $h$ = class width

Solution Steps:

1. $N = \sum \text{Frequency} = 3 + 4 + 5 + 4 + 6 = 22$.
2. $\frac{N}{2} = 11$.
3. Locate the cumulative frequency that contains 11:
   • Cumulative frequencies: $3, 7, 12, 16, 22$.
   • Median class is 20 - 30 (cumulative frequency just above 11).
4. Values:
   • $L = 20$
   • $cf = 7$ (cumulative frequency before 20 - 30)
   • $f = 5$
   • $h = 10$

Substitute into the formula:
$\text{Median} = 20 + \left(\frac{11 - 7}{5}\right) \cdot 10 = 20 + \left(\frac{4}{5}\right) \cdot 10 = 20 + 8 = 28$
Answer: Median is 28.

(c) Find the average (mean):
The formula for the mean is:
$\text{Mean} = \frac{\sum f \cdot x}{\sum f}$
Where $x$ is the midpoint of each class interval.

• Midpoints: $5, 15, 25, 35, 45$
• Multiply midpoints by frequency:
  $(5 \cdot 3) + (15 \cdot 4) + (25 \cdot 5) + (35 \cdot 4) + (45 \cdot 6)$
  $= 15 + 60 + 125 + 140 + 270 = 610$
• Total frequency ($\sum f$) = 22.

$\text{Mean} = \frac{610}{22} \approx 27.73$
Answer: Mean is 27.73.

(d) Convert marks obtained to a scale of 100:
If marks obtained are converted to marks out of 100, multiply each value by $2$. For example, if the mean was 27.73 out of 50:
$\text{Converted Mean} = 27.73 \times 2 = 55.46$
Answer: Converted mean is 55.46.

Would you like to go over Problem 9 or Problem 10 in detail, or clarify any specific part of the solution?

Relative Questions:

1. How is the modal class determined in grouped frequency data?
2. What are the steps for finding the median in a frequency distribution?
3. Why do we use midpoints when calculating the mean for grouped data?
4. How does cumulative frequency help in identifying the median class?
5. How can we scale marks to a different range (e.g., from 50 to 100)?

Tip:

To avoid calculation errors, always verify cumulative frequency sums and check the correct interval for the median class.