Math Problem Statement

		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

Solution

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:

N2=2592=129.5\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:

Median=L+(N2Ff)×h\text{Median} = L + \left( \frac{\frac{N}{2} - F}{f} \right) \times h

Where:

  • LL = lower boundary of the median class = 55.5
  • NN = total frequency = 259
  • FF = cumulative frequency before the median class = 0 (since the median class is the first one)
  • ff = frequency of the median class = 243
  • hh = class width = 5135.5 - 55.5 = 5080

Substitute the values into the formula:

Median=55.5+(129.50243)×5080\text{Median} = 55.5 + \left( \frac{129.5 - 0}{243} \right) \times 5080

Median=55.5+(129.5243)×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.

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Math Problem Analysis

Mathematical Concepts

Statistics
Median
Class Intervals
Cumulative Frequency

Formulas

Median = L + [(N/2 - F) / f] * h
Where L = lower boundary of median class, N = total frequency, F = cumulative frequency before the median class, f = frequency of the median class, h = class width

Theorems

Median calculation for grouped data

Suitable Grade Level

Grades 10-12