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

We are given the following data with some missing frequencies, \( x \) and \( y \), and the median of the data is 868. The total frequency is 100. To find \( x \) and \( y \), we need to apply the concept of the median in a frequency distribution and use the cumulative frequency to determine the unknowns.

### Step 1: Organize the Data

| Class Interval | Frequency | Cumulative Frequency |
|----------------|-----------|----------------------|
| 800 – 820      | 7         | 7                    |
| 820 – 840      | 14        | 7 + 14 = 21          |
| 840 – 860      | \( x \)   | 21 + \( x \)         |
| 860 – 880      | 25        | 21 + \( x \) + 25    |
| 880 – 900      | \( y \)   | 21 + \( x \) + 25 + \( y \) |
| 900 – 920      | 10        | 21 + \( x \) + 25 + \( y \) + 10 |
| 920 – 940      | 5         | 100                  |

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

### Step 2: Median Class
The median is given as 868, which falls in the class interval 860 – 880. Therefore, the median class is 860 – 880.

### Step 3: Apply the Median Formula
The formula for the median in a grouped frequency distribution is:
\[
\text{Median} = L + \left( \frac{\frac{N}{2} - C_f}{f} \right) \times h
\]
Where:
- \( L \) is the lower boundary of the median class (860),
- \( N \) is the total frequency (100),
- \( C_f \) is the cumulative frequency before the median class,
- \( f \) is the frequency of the median class (25),
- \( h \) is the class width (20 for each class).

From the formula:
\[
\text{Median} = 868
\]
Also, \( \frac{N}{2} = 50 \), and the cumulative frequency before the median class is \( C_f = 21 + x \).

Substitute into the median formula:
\[
868 = 860 + \left( \frac{50 - (21 + x)}{25} \right) \times 20
\]

### Step 4: Solve for \( x \)
Simplify the equation:
\[
868 = 860 + \left( \frac{29 - x}{25} \right) \times 20
\]
\[
868 - 860 = \left( \frac{29 - x}{25} \right) \times 20
\]
\[
8 = \frac{29 - x}{25} \times 20
\]
\[
8 = \frac{20(29 - x)}{25}
\]
Multiply both sides by 25:
\[
8 \times 25 = 20(29 - x)
\]
\[
200 = 580 - 20x
\]
\[
20x = 580 - 200
\]
\[
20x = 380
\]
\[
x = \frac{380}{20} = 19
\]

### Step 5: Solve for \( y \)
From Equation 1, we have:
\[
x + y = 39
\]
Substitute \( x = 19 \) into the equation:
\[
19 + y = 39 \quad \Rightarrow \quad y = 39 - 19 = 20
\]

### Final Answer:
- \( x = 19 \)
- \( y = 20 \)

Would you like more details or have any questions? Here are some related questions you might explore:

1. How is the median class determined in a frequency distribution?
2. What is the significance of the cumulative frequency when calculating the median?
3. Can the median formula be used for open-ended class intervals?
4. How does the class width affect the calculation of the median?
5. What would change if the class intervals were unequal?

**Tip**: Always verify your calculations by checking the total frequency and ensuring that the median class is chosen correctly!

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

Learn how to find the missing frequencies x and y in a grouped frequency distribution where the median is 868 and the total frequency is 100. This step-by-step guide uses the median formula and cumulative frequency calculations, suitable for students in Grades 9-10.

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