To solve this problem, let’s denote the midpoints of each class interval as $x$ and use the frequency given in each class to calculate the mean. Given that the mean of the distribution is 18, we can set up the equation for the mean and solve for the unknown frequency $f$ of the class interval 19–21.

Step 1: Calculate Midpoints for Each Class Interval

The midpoint $x_i$ of each class interval is calculated as follows:

• For 11–13: $x = \frac{11 + 13}{2} = 12$
• For 13–15: $x = \frac{13 + 15}{2} = 14$
• For 15–17: $x = \frac{15 + 17}{2} = 16$
• For 17–19: $x = \frac{17 + 19}{2} = 18$
• For 19–21: $x = \frac{19 + 21}{2} = 20$
• For 21–23: $x = \frac{21 + 23}{2} = 22$
• For 23–25: $x = \frac{23 + 25}{2} = 24$

Step 2: Set Up the Mean Formula

The mean $\bar{x}$ is calculated using: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$ where $f_i$ is the frequency of each class.

Step 3: Create the Equation

Let’s substitute the values we know:

• Frequencies $f_i$: 3, 6, 9, 13, $f$, 5, 4
• Midpoints $x_i$: 12, 14, 16, 18, 20, 22, 24
• Given mean $\bar{x} = 18$

So, $18 = \frac{(3 \times 12) + (6 \times 14) + (9 \times 16) + (13 \times 18) + (f \times 20) + (5 \times 22) + (4 \times 24)}{3 + 6 + 9 + 13 + f + 5 + 4}$

Step 4: Simplify and Solve for $f$

1. Calculate each term in the numerator and denominator separately.

   • Numerator: $(3 \times 12) + (6 \times 14) + (9 \times 16) + (13 \times 18) + (f \times 20) + (5 \times 22) + (4 \times 24)$
   • Denominator: $3 + 6 + 9 + 13 + f + 5 + 4$

2. Solve for $f$ by equating this expression to 18.

Let me calculate these values for you.The frequency $f$ for the class interval 19–21 is 8.

Would you like further details on the calculations or have any questions?

Here are some related questions you might find interesting:

1. How is the mean calculated for grouped data?
2. What is the effect of changing a frequency on the mean?
3. How would the calculation change if the class intervals were different?
4. What other measures of central tendency could we calculate with this data?
5. How can we determine if this distribution is skewed based on these intervals?

Tip: In frequency distributions, using midpoints for each class interval simplifies the calculation of the mean, especially when dealing with large datasets.