Find missing frequency if the arithmetic mean is 21.14 and class Interval is 0-10 10-20 20-30 30-40 40-50 and frequency is 8, 22, 31, x and 2

Given the class intervals and frequencies, let's denote the missing frequency by \(x\). The arithmetic mean (A.M.) is given as 21.14. We need to find the missing frequency \(x\).

First, we need to find the class marks (midpoints) of each class interval:

- For 0-10, the class mark is \(\frac{0+10}{2} = 5\)
- For 10-20, the class mark is \(\frac{10+20}{2} = 15\)
- For 20-30, the class mark is \(\frac{20+30}{2} = 25\)
- For 30-40, the class mark is \(\frac{30+40}{2} = 35\)
- For 40-50, the class mark is \(\frac{40+50}{2} = 45\)

Next, we write down the frequencies and their corresponding class marks:

\[
\begin{array}{|c|c|}
\hline
\text{Class Interval} & \text{Frequency} (f_i) & \text{Class Mark} (x_i) & f_i \cdot x_i \\
\hline
0-10 & 8 & 5 & 8 \cdot 5 = 40 \\
10-20 & 22 & 15 & 22 \cdot 15 = 330 \\
20-30 & 31 & 25 & 31 \cdot 25 = 775 \\
30-40 & x & 35 & 35x \\
40-50 & 2 & 45 & 2 \cdot 45 = 90 \\
\hline
\end{array}
\]

The formula for the arithmetic mean is given by:

\[
\text{A.M.} = \frac{\sum f_i x_i}{\sum f_i}
\]

We know the arithmetic mean (A.M.) is 21.14, so:

\[
21.14 = \frac{40 + 330 + 775 + 35x + 90}{8 + 22 + 31 + x + 2}
\]

Simplifying the sums in the numerator and the denominator, we get:

\[
21.14 = \frac{1235 + 35x}{63 + x}
\]

Now we solve for \(x\):

\[
21.14(63 + x) = 1235 + 35x
\]

Expanding and simplifying:

\[
1331.82 + 21.14x = 1235 + 35x
\]

Rearranging terms to isolate \(x\):

\[
1331.82 - 1235 = 35x - 21.14x
\]

\[
96.82 = 13.86x
\]

Solving for \(x\):

\[
x = \frac{96.82}{13.86} \approx 6.99 \approx 7
\]

Therefore, the missing frequency \(x\) is approximately 7.

Do you want details or have any questions?

Here are some questions you might find interesting:

1. How do you find the class marks for a given class interval?
2. What is the formula for the arithmetic mean of grouped data?
3. How do you solve equations involving arithmetic mean and frequencies?
4. What are the steps to solve for a missing frequency in a frequency distribution?
5. How does changing one frequency affect the overall arithmetic mean?

**Tip:** When solving for a missing value in a frequency distribution, carefully handle the algebraic manipulation and ensure the correct use of formulas to maintain accuracy.

Learn how to find the missing frequency in a frequency distribution problem with an arithmetic mean of 21.14. This problem involves calculating class marks, using the arithmetic mean formula, and solving equations with frequency distributions. Suitable for advanced high school students studying statistics and data analysis.

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