The problem asks to find the missing frequency in the frequency distribution table, given that the mean of the distribution is 27.

The classes and frequencies are:

• Classes: 0-10, 10-20, 20-30, 30-40, 40-50
• Frequencies: 5, $f_1$, 15, 16, 6

We are also given that the mean of the distribution is 27.

Step 1: Find midpoints of the class intervals

The midpoint ($x_i$) for each class interval is calculated as: $x_i = \frac{\text{Lower limit} + \text{Upper limit}}{2}$

• Midpoint for 0-10: $\frac{0 + 10}{2} = 5$
• Midpoint for 10-20: $\frac{10 + 20}{2} = 15$
• Midpoint for 20-30: $\frac{20 + 30}{2} = 25$
• Midpoint for 30-40: $\frac{30 + 40}{2} = 35$
• Midpoint for 40-50: $\frac{40 + 50}{2} = 45$

So, the midpoints are: $5, 15, 25, 35, 45$

Step 2: Apply the formula for the mean

The formula for the mean in a frequency distribution is: $\text{Mean} = \frac{\sum f_i x_i}{\sum f_i}$ Where $f_i$ are the frequencies and $x_i$ are the midpoints. The given mean is 27, so we will use that.

Let $f_1$ be the missing frequency for the class 10-20. Now, we will calculate the sum of $f_i x_i$ and $\sum f_i$ including the unknown frequency $f_1$.

Step 3: Calculate $\sum f_i x_i$ and $\sum f_i$

$\sum f_i = 5 + f_1 + 15 + 16 + 6 = f_1 + 42$ $\sum f_i x_i = (5 \times 5) + (f_1 \times 15) + (15 \times 25) + (16 \times 35) + (6 \times 45)$ $\sum f_i x_i = 25 + 15f_1 + 375 + 560 + 270 = 15f_1 + 1230$

Step 4: Use the mean formula

The mean is 27, so: $27 = \frac{15f_1 + 1230}{f_1 + 42}$ Now, cross-multiply to solve for $f_1$: $27(f_1 + 42) = 15f_1 + 1230$ $27f_1 + 1134 = 15f_1 + 1230$ $27f_1 - 15f_1 = 1230 - 1134$ $12f_1 = 96$ $f_1 = \frac{96}{12} = 8$

Conclusion:

The missing frequency $f_1$ is 8.

Would you like more details or have any questions?

Here are some related questions to deepen your understanding:

1. How would the solution change if the mean were different?
2. Can you verify the calculated mean using the frequencies found?
3. What is the total frequency for the given distribution?
4. How do you calculate the standard deviation for this distribution?
5. What is the impact of outliers on the mean in a frequency distribution?

Tip: Always check your calculations by substituting back the values into the original formula to ensure correctness.