Question

Let M denote the median of the following frequency distribution.

\(x_i\)	\(f_i\)
0 - 4	2
4 - 8	4
8 - 12	7
12 - 16	8
16 - 20	6

Then 20M is equal to:

• A. 416
• B. 104
• C. 52
• D. 208

Answer 1

The Correct Option is D

First, calculate the cumulative frequency.

Class	Frequency	Cumulative Frequency
0-4	3	3
4-8	9	12
8-12	10	22
12-16	8	30
16-20	6	36

The total frequency \( N = 36 \), so \( \frac{N}{2} = 18 \).

The median class is 8-12, as it is the class where the cumulative frequency first exceeds 18.

Lower limit \( l = 8 \) Frequency \( f = 10 \) Cumulative frequency of the class before the median class \( C = 12 \) Class width \( h = 4 \)

Using the median formula:

\[ M = l + \left( \frac{\frac{N}{2} - C}{f} \right) \times h \]

Substitute the values:

\[ M = 8 + \left( \frac{18 - 12}{10} \right) \times 4 \] \[ = 8 + \left( \frac{6}{10} \right) \times 4 \] \[ = 8 + 0.6 \times 4 \] \[ = 8 + 2.4 = 10.4 \]

Then,

\[ 20M = 20 \times 10.4 = 208 \].

Answer 2

The problem requires us to calculate the value of \(20M\), where \(M\) is the median of the given frequency distribution.

Concept Used:

The median for a grouped frequency distribution is calculated using the formula:

\[ M = l + \left( \frac{\frac{N}{2} - c.f.}{f} \right) \times h \]

where:

• \(l\) is the lower limit of the median class.
• \(N\) is the sum of all frequencies (\(\sum f_i\)).
• \(c.f.\) is the cumulative frequency of the class preceding the median class.
• \(f\) is the frequency of the median class.
• \(h\) is the class size (width) of the median class.

Step-by-Step Solution:

Step 1: Construct the cumulative frequency distribution table.

First, we find the total frequency \(N\) and then create a cumulative frequency (\(c.f.\)) column.

Class	Frequency (\(f_i\))	Cumulative Frequency (\(c.f.\))
0 - 4	3	3
4 - 8	9	3 + 9 = 12
8 - 12	10	12 + 10 = 22
12 - 26	8	22 + 8 = 30
16 - 20	6	30 + 6 = 36

The total frequency is \(N = 36\).

Step 2: Identify the median class.

To find the median class, we first calculate the value of \(\frac{N}{2}\):

\[ \frac{N}{2} = \frac{36}{2} = 18 \]

The median class is the class interval whose cumulative frequency is just greater than or equal to 18. Looking at the \(c.f.\) column, the value just greater than 18 is 22. The class corresponding to this cumulative frequency is 8 - 12.

Therefore, the median class is 8 - 12.

Step 3: Extract the values needed for the median formula.

• The lower limit of the median class, \(l = 8\).
• The total frequency, \(N = 36\).
• The frequency of the median class, \(f = 10\).
• The cumulative frequency of the class preceding the median class (4 - 8), \(c.f. = 12\).
• The class size of the median class, \(h = 12 - 8 = 4\).

Note: Even though the class intervals are not uniform, the formula uses the width \(h\) of the median class only.

Step 4: Substitute the values into the formula to calculate the median \(M\).

\[ M = l + \left( \frac{\frac{N}{2} - c.f.}{f} \right) \times h \] \[ M = 8 + \left( \frac{18 - 12}{10} \right) \times 4 \] \[ M = 8 + \left( \frac{6}{10} \right) \times 4 \] \[ M = 8 + 0.6 \times 4 \] \[ M = 8 + 2.4 = 10.4 \]

Final Computation & Result:

Step 5: Calculate the value of \(20M\).

\[ 20M = 20 \times 10.4 \] \[ 20M = 208 \]

The value of 20M is 208. This corresponds to option (4).