Let's go through each part of the question:

1. Class Width (a): The class width is the difference between the lower boundaries of consecutive classes. In this case:

   • $53 - 50 = 3$ Therefore, the class width is 3.

2. Class Midpoints of the First Class (b): The midpoint of a class is calculated by taking the average of the lower and upper limits of that class.

   • For the first class (50 - 52), the midpoint is $\frac{50 + 52}{2} = 51$. So, the class midpoint of the first class is 51.

3. Class Boundaries of the First Class (c): Class boundaries are calculated by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit.

   • For the first class (50 - 52), the boundaries are:
     • Lower boundary: $50 - 0.5 = 49.5$
     • Upper boundary: $52 + 0.5 = 52.5$ Therefore, the class boundaries for the first class are 49.5 - 52.5.

Now let's match these with the options provided:

• The correct answer is Option B: $\text{(a) 3 | (b) 51 | (c) 49.5-52.5}$.

Would you like more details or have any questions?

Here are some related questions for practice:

1. How do you calculate the frequency density of a class in a frequency distribution?
2. What are class intervals, and why are they important in grouped data?
3. How can you determine the total number of data points from a frequency distribution?
4. What would happen to the class width if we added another class from 65-67?
5. How does calculating class boundaries help when creating histograms?

Tip: Remember, class boundaries help avoid gaps between classes in continuous data, which is useful in graphical representations like histograms.