If the median of the following frequency. distribution is 26.11. Find frequency the missing frequency
class interval 0-10 10-20 20-30 30-40 40-50
Frequency 8 F 36 24 12

To find the missing frequency \( F \), we will use the median formula for a grouped frequency distribution. Let's solve it step by step.

### Given Data:
- Class intervals: \( 0-10, 10-20, 20-30, 30-40, 40-50 \)
- Frequencies: \( 8, F, 36, 24, 12 \)
- Median: 26.11

### Step 1: Identify the median class
The median class is the class interval where the cumulative frequency just exceeds or equals half of the total frequency.

#### Let's first calculate the total frequency (including \( F \)):

\[
\text{Total frequency } = 8 + F + 36 + 24 + 12 = 80 + F
\]

Half of the total frequency is:

\[
\frac{80 + F}{2}
\]

### Step 2: Cumulative frequencies
We'll now compute the cumulative frequencies for the given data.

| Class Interval | Frequency | Cumulative Frequency |
|----------------|------------|----------------------|
| 0-10           | 8          | 8                    |
| 10-20          | F          | \( 8 + F \)          |
| 20-30          | 36         | \( 8 + F + 36 = 44 + F \) |
| 30-40          | 24         | \( 44 + F + 24 = 68 + F \) |
| 40-50          | 12         | \( 68 + F + 12 = 80 + F \) |

We need to find the median class, where the cumulative frequency just exceeds \( \frac{80 + F}{2} \).

### Step 3: Determine the median class
The median value is 26.11, which lies in the class interval \( 20-30 \), as \( 26.11 \) is between 20 and 30.

Thus, the median class is \( 20-30 \), with:
- \( l = 20 \) (lower boundary of the median class)
- \( cf = 8 + F \) (cumulative frequency before the median class)
- \( f = 36 \) (frequency of the median class)
- \( h = 10 \) (class width)

### Step 4: Apply the median formula
The formula for the median in a grouped frequency distribution is:

\[
\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h
\]

Substituting the values:

\[
26.11 = 20 + \left( \frac{\frac{80 + F}{2} - (8 + F)}{36} \right) \times 10
\]

Simplify the equation:

\[
26.11 = 20 + \left( \frac{40 + \frac{F}{2} - 8 - F}{36} \right) \times 10
\]

\[
26.11 = 20 + \left( \frac{32 - \frac{F}{2}}{36} \right) \times 10
\]

\[
26.11 = 20 + \left( \frac{320 - 5F}{36} \right)
\]

### Step 5: Solve for \( F \)
Now, solve the equation:

\[
26.11 = 20 + \frac{320 - 5F}{36}
\]

\[
6.11 = \frac{320 - 5F}{36}
\]

Multiply both sides by 36:

\[
6.11 \times 36 = 320 - 5F
\]

\[
219.96 = 320 - 5F
\]

Solve for \( F \):

\[
5F = 320 - 219.96
\]

\[
5F = 100.04
\]

\[
F = \frac{100.04}{5} = 20.008
\]

Rounding to the nearest integer, \( F \approx 20 \).

Thus, the missing frequency \( F \) is approximately **20**.

### Would you like more details or have any questions?

Here are 5 related questions:
1. What is the formula for calculating the median of grouped data?
2. How do you identify the median class in a frequency distribution?
3. How does changing the class width affect the calculation of the median?
4. What are the steps to calculate cumulative frequency in a distribution table?
5. How can you check if your calculated median is accurate?

**Tip:** Always double-check the class boundaries and cumulative frequencies to ensure the correct median class is selected.

If the median of the following frequency distribution is 26.11, find the missing frequency. Class interval: 0-10, 10-20, 20-30, 30-40, 40-50. Frequency: 8, F, 36, 24, 12.

Learn how to find the missing frequency in a grouped frequency distribution with a given median of 26.11. This problem involves class intervals, cumulative frequencies, and the median formula. Suitable for students in Grades 10-12.

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