To find the missing frequency $f_3$, given that the mean of the distribution is 30, we can use the formula for the mean of a frequency distribution. Here's a step-by-step solution:

Step 1: Set up the Midpoints for Each Class Interval

We need the midpoint ($x$) of each class interval to use in calculating the mean.

• For $0 - 10$: Midpoint, $x_1 = \frac{0 + 10}{2} = 5$
• For $10 - 20$: Midpoint, $x_2 = \frac{10 + 20}{2} = 15$
• For $20 - 30$: Midpoint, $x_3 = \frac{20 + 30}{2} = 25$
• For $30 - 40$: Midpoint, $x_4 = \frac{30 + 40}{2} = 35$
• For $40 - 50$: Midpoint, $x_5 = \frac{40 + 50}{2} = 45$

Step 2: Set Up the Mean Formula

The formula for the mean $\bar{x}$ of a frequency distribution is:

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

where:

• $f$ is the frequency,
• $x$ is the midpoint of each interval.

Step 3: Substitute Known Values

We know:

• $\bar{x} = 30$
• Frequencies and midpoints for each class.

Let’s set up the equation with $f_3$ as the unknown.

$30 = \frac{(5 \cdot 5) + (10 \cdot 15) + (15 \cdot 25) + (f_3 \cdot 35) + (8 \cdot 45)}{5 + 10 + 15 + f_3 + 8}$

Step 4: Simplify the Equation

1. Calculate each term in the numerator:

   • $5 \cdot 5 = 25$
   • $10 \cdot 15 = 150$
   • $15 \cdot 25 = 375$
   • $f_3 \cdot 35 = 35f_3$
   • $8 \cdot 45 = 360$

   So, the numerator is $25 + 150 + 375 + 35f_3 + 360$.

2. Simplify the denominator:

   • $5 + 10 + 15 + f_3 + 8 = 38 + f_3$

Thus, we have:

$30 = \frac{910 + 35f_3}{38 + f_3}$

Step 5: Solve for $f_3$

Multiply both sides by $38 + f_3$:

$30(38 + f_3) = 910 + 35f_3$

Expanding and simplifying:

$1140 + 30f_3 = 910 + 35f_3$

Rearrange to isolate $f_3$:

$1140 - 910 = 35f_3 - 30f_3$

$230 = 5f_3$

$f_3 = \frac{230}{5} = 46$

Solution

The missing frequency $f_3$ is $46$.

Would you like further clarification on this process or details on related topics? Here are some related questions you might find helpful:

1. How do you calculate the mean for a grouped frequency distribution?
2. What are midpoints, and why are they used in frequency distributions?
3. How would this calculation change if the mean were different?
4. What is the impact of changing one frequency on the overall mean?
5. How can this approach be applied to distributions with more intervals?

Tip: Always double-check calculations by substituting the final answer back into the original equation to confirm accuracy.