			Boundaries		Limit Test		Cummulative	
x	f	Xm	LB	UB	LL	HL	<CF	>CF
56 - 5135	243	2595.5	55.5	5135.5	56	5135	243	259
5136 - 10215	11	7675.5	5135.5	10215.5	5136	10215	254	16
10216 - 15295	1	12755.5	10215.5	15295.5	10216	15295	255	5
15296 - 20375	1	17835.5	15295.5	20375.5	15296	20375	256	4
20376 - 25455	2	22915.5	20375.5	25455.5	20376	25455	258	3
25456 - 30535	0	27995.5	25455.5	30535.5	25456	30535	258	1
30536 - 35615	0	33075.5	30535.5	35615.5	30536	35615	258	1
35616 - 40695	0	38155.5	35615.5	40695.5	35616	40695	258	1
40696 - 45772	1	43234	40695.5	45772.5	40696	45772	259	1
	259							                                                                                                                             USING THESE DATA, COMPUTE FOR MEDIAN CLASS AND MEDIAN

To compute the median class and the median from the given data, we will follow the steps below:

### Step 1: Identify the total cumulative frequency (CF)
From the data, the total cumulative frequency is already provided as 259.

### Step 2: Determine the position of the median
The formula for determining the median position is:

\[
\frac{N}{2} = \frac{259}{2} = 129.5
\]

This means the median lies in the class where the cumulative frequency first exceeds 129.5.

### Step 3: Locate the median class
Looking at the cumulative frequency column, we need to find the class where the cumulative frequency first surpasses 129.5:

- The first cumulative frequency greater than 129.5 is 243, which corresponds to the class **56 - 5135**.

Thus, the **median class** is **56 - 5135**.

### Step 4: Apply the formula for the median
The formula for calculating the median is:

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

Where:
- \( L \) = lower boundary of the median class = **55.5**
- \( N \) = total frequency = **259**
- \( F \) = cumulative frequency before the median class = **0** (since the median class is the first one)
- \( f \) = frequency of the median class = **243**
- \( h \) = class width = **5135.5 - 55.5 = 5080**

Substitute the values into the formula:

\[
\text{Median} = 55.5 + \left( \frac{129.5 - 0}{243} \right) \times 5080
\]

\[
\text{Median} = 55.5 + \left( \frac{129.5}{243} \right) \times 5080
\]

Now let's compute that.

The median is approximately **2762.74**.

### Summary:
- **Median class**: 56 - 5135
- **Median**: 2762.74

Would you like more details or have any questions about this process?

Here are some follow-up questions to deepen understanding:
1. How does changing the class width affect the median?
2. What happens if the frequencies were more evenly distributed across classes?
3. How would we calculate the mode from this data?
4. Can the median be computed if the data is presented in a different form, like a list of individual values?
5. How can we verify the accuracy of this median calculation?

**Tip:** The median is less affected by extreme values (outliers) compared to the mean, making it a robust measure of central tendency in skewed distributions.

USING THESE DATA, COMPUTE FOR MEDIAN CLASS AND MEDIAN

This guide explains how to calculate the median class and median for grouped data using cumulative frequencies. It uses real data to show step-by-step calculations for students at Grades 10-12. The median class and the final median value are determined using statistical formulas.

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