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.