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!