To solve this problem, the goal is to determine the value of $Q_3$ (the third quartile) when $Q_1$ (the first quartile) is given as $51.75$. Let's proceed step by step:

1. Extracting the Cumulative Frequency

We begin by calculating the cumulative frequency (CF) for the given data:

2. Formula for Quartile

The formula for finding the quartile position is: $Q_k = L + \left( \frac{kN}{4} - CF_{\text{prev}} \right) \frac{h}{f}$ Where:

• $L$: Lower boundary of the class containing $Q_k$
• $N$: Total frequency ($N = 45$ here)
• $k$: Quartile number (1 for $Q_1$, 3 for $Q_3$)
• $CF_{\text{prev}}$: Cumulative frequency before the quartile class
• $h$: Class width
• $f$: Frequency of the quartile class

3. Determine the Quartile Class for $Q_3$

• Total frequency $N = 45$
• Position of $Q_3$: $\frac{3N}{4} = \frac{3 \times 45}{4} = 33.75$

The cumulative frequency just greater than $33.75$ is $41$, corresponding to the class $52-56$. Thus:

• $L = 52$
• $CF_{\text{prev}} = 32$
• $f = 9$
• $h = 4$

4. Calculate $Q_3$

Substitute the values into the formula: $Q_3 = 52 + \left( \frac{33.75 - 32}{9} \right) \times 4$ $Q_3 = 52 + \left( \frac{1.75}{9} \right) \times 4$ $Q_3 = 52 + 0.7778 \times 4$ $Q_3 = 52 + 3.1112$ $Q_3 \approx 55.11$

Final Answer:

The value of $Q_3$ is approximately 55.11.

Let me know if you want details or have any questions.
Here are five related questions for further exploration:

1. How do we determine the class width in a frequency distribution?
2. Can the same approach be used to find the median?
3. What happens if $Q_1$ or $Q_3$ falls exactly at a class boundary?
4. How does the total frequency affect the positions of quartiles?
5. Why is cumulative frequency essential for locating quartiles?

Tip: Always verify the total cumulative frequency matches the sum of all frequencies before proceeding with quartile calculations.