30. The median of the following data is 868. Find the values of x and y, if the total frequency is 100. (2019)

Class

Frequency

800-820

7

820-840

14

840-860

X

860-880 25

880-900 900-920 10

y

920-940

5

We are tasked with finding the values of \( x \) and \( y \) in the given frequency distribution table, where the total frequency is 100 and the median class is 860-880 with a median value of 868.

### Step 1: Organizing the Data
Let's first rewrite the data properly:

| Class Interval | Frequency |
|----------------|------------|
| 800-820        | 7          |
| 820-840        | 14         |
| 840-860        | \( x \)    |
| 860-880        | 25         |
| 880-900        | \( y \)    |
| 900-920        | 10         |
| 920-940        | 5          |

The total frequency is 100, meaning:
\[
7 + 14 + x + 25 + y + 10 + 5 = 100
\]
Simplifying this:
\[
61 + x + y = 100
\]
\[
x + y = 39 \quad \text{(Equation 1)}
\]

### Step 2: Identifying the Median Class
We are told the median is 868. From the class intervals, the median class is 860-880.

Now, let's find the cumulative frequencies:

| Class Interval | Frequency | Cumulative Frequency |
|----------------|------------|----------------------|
| 800-820        | 7          | 7                    |
| 820-840        | 14         | 21                   |
| 840-860        | \( x \)    | \( 21 + x \)         |
| 860-880        | 25         | \( 46 + x \)         |
| 880-900        | \( y \)    | \( 46 + x + y \)     |
| 900-920        | 10         | \( 56 + x + y \)     |
| 920-940        | 5          | \( 61 + x + y \)     |

The total frequency is 100, so the median is at the \( \frac{100}{2} = 50 \)-th position. The class containing the 50th value is the median class (860-880).

### Step 3: Using the Median Formula
The formula for the median is:

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

Where:
- \( L \) is the lower boundary of the median class = 860,
- \( N \) is the total frequency = 100,
- \( F \) is the cumulative frequency before the median class = \( 21 + x \),
- \( f \) is the frequency of the median class = 25,
- \( h \) is the class width = 20 (since 880 - 860 = 20).

Substitute the values:

\[
868 = 860 + \left( \frac{50 - (21 + x)}{25} \right) \times 20
\]

Simplify:

\[
868 = 860 + \left( \frac{29 - x}{25} \right) \times 20
\]
\[
868 = 860 + \frac{580 - 20x}{25}
\]
\[
868 = 860 + 23.2 - 0.8x
\]
\[
8 = 23.2 - 0.8x
\]
\[
0.8x = 15.2
\]
\[
x = 19
\]

### Step 4: Finding \( y \)
From Equation 1:
\[
x + y = 39
\]
Substitute \( x = 19 \):
\[
19 + y = 39
\]
\[
y = 20
\]

### Final Answer:
The values of \( x \) and \( y \) are:
\[
x = 19, \quad y = 20
\]

Would you like further details or have any questions?

Here are five related questions for further practice:
1. How would you find the mode of a frequency distribution?
2. If the median class changes, how would it affect the value of \( x \)?
3. How would you calculate the mean from the given data?
4. What happens if \( N \) (total frequency) changes in a median calculation?
5. How is the class width relevant in calculating the median?

**Tip:** In a frequency distribution, the cumulative frequency helps identify the median class and is key to solving median-related problems.

30. The median of the following data is 868. Find the values of x and y, if the total frequency is 100. (2019) Class Frequency 800-820 7 820-840 14 840-860 X 860-880 25 880-900 900-920 10 y 920-940 5

Solve for x and y in a frequency distribution where the median is 868, using the median formula and cumulative frequency. This step-by-step solution explores key statistical concepts and is suitable for students in Grades 10-12.

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