Problem Breakdown:

We are given a frequency distribution table with the following class intervals and frequencies:

The arithmetic mean is 67.45 inches, and we need to find the value of $f_3$.

Step-by-Step Solution:

Step 1: Formula for Arithmetic Mean

The arithmetic mean for a frequency distribution is calculated as:

$\bar{x} = \frac{\sum f \cdot x}{\sum f}$

Where:

• $x$ is the mid-point of each class interval.
• $f$ is the frequency of the class.
• $\bar{x}$ is the mean (given as 67.45).

Step 2: Calculate Midpoints ($x$) for Each Class

The midpoints are calculated as:

$x = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$

• For 60-62: $x = \frac{60 + 62}{2} = 61$
• For 63-65: $x = \frac{63 + 65}{2} = 64$
• For 66-68: $x = \frac{66 + 68}{2} = 67$
• For 69-71: $x = \frac{69 + 71}{2} = 70$
• For 72-74: $x = \frac{72 + 74}{2} = 73$

Now, the table becomes:

Step 3: Expand the Formula

Substitute into the mean formula:

$67.45 = \frac{\sum f \cdot x}{\sum f}$

First, expand $\sum f \cdot x$ (frequency × midpoint):

$\sum f \cdot x = (15 \cdot 61) + (54 \cdot 64) + (f_3 \cdot 67) + (81 \cdot 70) + (24 \cdot 73)$

$\sum f \cdot x = 915 + 3456 + 67f_3 + 5670 + 1752$

$\sum f \cdot x = 11793 + 67f_3$

Next, expand $\sum f$ (total frequency):

$\sum f = 15 + 54 + f_3 + 81 + 24$

$\sum f = 174 + f_3$

Step 4: Solve for $f_3$

Substitute back into the mean formula:

$67.45 = \frac{11793 + 67f_3}{174 + f_3}$

Multiply through by $174 + f_3$:

$67.45 (174 + f_3) = 11793 + 67f_3$

Distribute $67.45$:

$11735.3 + 67.45f_3 = 11793 + 67f_3$

Simplify:

$11735.3 - 11793 = 67f_3 - 67.45f_3$

$-57.7 = -0.45f_3$

Solve for $f_3$:

$f_3 = \frac{-57.7}{-0.45}$

$f_3 = 128.44 \approx 128$

Final Answer:

The value of $f_3$ is 128.

Would you like me to explain any step in further detail or solve a similar example?

Here are 5 related questions for practice:

1. How do you calculate the arithmetic mean for grouped data if the total frequency is not given?
2. What happens if there is a missing midpoint in the frequency table?
3. How would you solve if two frequencies were unknown?
4. How is the weighted mean formula related to this problem?
5. Can you calculate the median and mode for the same frequency table?

Tip: Always double-check your calculations for midpoint and product terms when working with frequency distributions.