Let's solve the problems shown in the image.

1. Weights of 11 Students (in kgs):

Weights: 48, 53, 42, 59, 47, 45, 51, 68, 48, 51

(a) Arithmetic Mean (AM):

The arithmetic mean is calculated by summing all the values and dividing by the total number of values.

$\text{AM} = \frac{\sum \text{Weights}}{\text{Number of students}} = \frac{48 + 53 + 42 + 59 + 47 + 45 + 51 + 68 + 48 + 51}{11}$

(b) Mode:

The mode is the number that appears most frequently in the data set.

2. Frequency Distribution (Marks and Number of Students):

We need to find the:

• Arithmetic Mean (AM)
• Median
• Mode

Let me compute the solutions for each part.Here are the results for both problems:

1. Weights of 11 Students:

• Arithmetic Mean (AM): 51.2 kg
• Mode: 48 kg (appears most frequently)

2. Frequency Distribution (Marks and Number of Students):

• Arithmetic Mean (AM): 58.0 marks
• Median: 60.5 marks
• Mode: The mode class is the interval $60-80$, which contains the most students (40 students).

Would you like more details on any of the steps, or do you have any questions?

Here are 5 related questions to deepen your understanding:

1. How do you calculate the median when there are an even number of data points?
2. What is the difference between the mean, median, and mode in terms of data interpretation?
3. Why is it important to use midpoints when calculating the mean of a frequency distribution?
4. How would the mode change if two values occurred with the same highest frequency?
5. How can outliers affect the mean but not the median?

Tip: For frequency distributions, the mode is the class interval with the highest frequency, but the exact mode can sometimes be calculated using a formula for grouped data.