Mean

Concept Explanation: Mean is the average value of a set of numbers. It is calculated by summing all observations and dividing by the total number of observations.

Formula Derivation

For ungrouped data with observations \( x_1, x_2, \ldots, x_n \), mean \( \bar{x} \) is given by:

\[ \bar{x} = \frac{\sum_{i=1}^n x_i}{n} \]

For grouped data, where \( f_i \) is the frequency of the \( i^{th} \) class and \( x_i \) is the class mark (midpoint), mean is:

\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \]

Class mark \( x_i \) is calculated as:

\[ x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \]

Assumed Mean Method (shortcut):

\[ \bar{x} = A + \frac{\sum f_i d_i}{\sum f_i} \]

where \( A \) is the assumed mean and \( d_i = x_i - A \) are deviations.

Worked Illustrations

Given class intervals and frequencies, calculate mean using both direct and assumed mean methods.

Solved Example

Marks obtained by 30 students are grouped as follows:

Class Interval	Frequency (\( f_i \))	Class Mark (\( x_i \))	Deviation (\( d_i = x_i - 47.5 \))	\( f_i x_i \)	\( f_i d_i \)
10-25	1	17.5	-30	17.5	-30
25-40	3	32.5	-15	97.5	-45
40-55	5	47.5	0	237.5	0
55-70	4	62.5	15	250	60
70-85	7	77.5	30	542.5	210
85-100	10	92.5	45	925	450
Total	30			2070	645

Calculate mean by:

• Direct Method: \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2070}{30} = 69 \)
• Assumed Mean Method: \( \bar{x} = 47.5 + \frac{645}{30} = 47.5 + 21.5 = 69 \)

Practice Set

• Level 1: Find mean of ungrouped data: 12, 15, 18, 20, 25.
• Level 2: Calculate mean using assumed mean method for grouped data: Class intervals 0-10, 10-20, 20-30 with frequencies 5, 8, 7.
• Level 3: Given grouped data with class intervals and frequencies, calculate mean using both methods and compare.

Answer Key

• Level 1: Mean = \( \frac{12+15+18+20+25}{5} = 18 \)
• Level 2: Assume mean \( A=15 \), calculate deviations and mean accordingly.
• Level 3: Detailed steps required for both methods.

Quick Reference

• Mean (ungrouped): \( \bar{x} = \frac{\sum x_i}{n} \)
• Mean (grouped direct): \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \)
• Mean (assumed): \( \bar{x} = A + \frac{\sum f_i d_i}{\sum f_i} \)

Glossary

• Class Mark: Midpoint of class interval.
• Frequency: Number of observations in a class.
• Deviation: Difference between class mark and assumed mean.

Median

Concept Explanation: Median is the middle value of an ordered data set dividing it into two equal halves.

Formula Derivation

For ungrouped data:

• If number of observations \( n \) is odd, median is the \( \left(\frac{n+1}{2}\right)^{th} \) value.
• If \( n \) is even, median is average of \( \frac{n}{2}^{th} \) and \( \left(\frac{n}{2}+1\right)^{th} \) values.

For grouped data, median is calculated using:

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

where:

• \( l \) = lower limit of median class
• \( N \) = total frequency
• \( c.f. \) = cumulative frequency before median class
• \( f \) = frequency of median class
• \( h \) = class width

Worked Illustrations

Identify median class by locating \( \frac{N}{2} \) in cumulative frequency, then apply formula.

Solved Example

Given cumulative frequency data for heights of 51 girls:

Class Interval	Frequency (\( f \))	Cumulative Frequency (\( c.f. \))
Below 140	4	4
140-145	7	11
145-150	18	29
150-155	15	44
155-160	7	51

Total \( N = 51 \), so \( \frac{N}{2} = 25.5 \). Median class is 145-150.

Applying formula:

\[ \text{Median} = 145 + \left( \frac{25.5 - 11}{18} \right) \times 5 = 145 + \frac{14.5}{18} \times 5 = 145 + 4.03 = 149.03 \]

Practice Set

• Level 1: Find median of ungrouped data: 12, 15, 18, 20, 25.
• Level 2: Calculate median for grouped data with given frequencies and class intervals.
• Level 3: Given cumulative frequency table, find median and interpret results.

Answer Key

• Level 1: Median = 18 (middle value)
• Level 2: Identify median class and apply formula.
• Level 3: Detailed calculation as shown in example.

Quick Reference

• Median (ungrouped odd): middle value
• Median (ungrouped even): average of two middle values
• Median (grouped): \( l + \left( \frac{\frac{N}{2} - c.f.}{f} \right) \times h \)

Glossary

• Median Class: Class interval containing the median.
• Cumulative Frequency: Sum of frequencies up to a class.
• Class Width: Difference between upper and lower class limits.

Mode

Concept Explanation: Mode is the value that occurs most frequently in a data set.

For ungrouped data, mode is the observation with highest frequency.

For grouped data, mode is estimated using the modal class (class with highest frequency) and the formula:

\[ \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \]

where:

• \( l \) = lower limit of modal class
• \( h \) = class width
• \( f_1 \) = frequency of modal class
• \( f_0 \) = frequency of class before modal class
• \( f_2 \) = frequency of class after modal class

Worked Illustrations

Identify modal class by highest frequency, then apply formula.

Solved Example

Family size data:

Class Interval	Frequency
1-3	7
3-5	8
5-7	2
7-9	3
9-11	0

Modal class is 3-5 with frequency \( f_1 = 8 \).

Applying formula:

\[ \text{Mode} = 3 + \left( \frac{8 - 7}{2 \times 8 - 7 - 2} \right) \times 2 = 3 + \frac{1}{7} \times 2 = 3 + 0.286 = 3.286 \]

Practice Set

• Level 1: Find mode of ungrouped data: 2, 3, 3, 5, 7.
• Level 2: Calculate mode for grouped data with given frequencies.
• Level 3: Given frequency distribution, find modal class and mode using formula.

Answer Key

• Level 1: Mode = 3
• Level 2: Identify modal class and apply formula.
• Level 3: Detailed calculation as shown in example.

Quick Reference

• Mode (ungrouped): most frequent value
• Mode (grouped): \( l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)

Glossary

• Modal Class: Class with highest frequency.
• Frequency: Number of observations in a class.
• Class Width: Difference between class limits.

Empirical Relationship Between Mean, Median and Mode

Concept Explanation: The mean, median and mode of a moderately skewed distribution are related by empirical formulas.

Formula Derivation

The empirical relations are:

\[ \text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean} \] \[ \text{Median} = \frac{1}{3} \times \text{Mode} + \frac{2}{3} \times \text{Mean} \] \[ \text{Mean} = \frac{3}{2} \times \text{Median} - \frac{1}{2} \times \text{Mode} \]

Worked Illustrations

Given any two of mean, median, mode, calculate the third using these formulas.

Solved Example

If mean = 50 and median = 52, find mode:

\[ \text{Mode} = 3 \times 52 - 2 \times 50 = 156 - 100 = 56 \]

Practice Set

• Level 1: Given mean and median, find mode.
• Level 2: Given mode and mean, find median.
• Level 3: Given median and mode, find mean.

Answer Key

• Level 1: Use \( \text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean} \)
• Level 2: Use \( \text{Median} = \frac{1}{3} \times \text{Mode} + \frac{2}{3} \times \text{Mean} \)
• Level 3: Use \( \text{Mean} = \frac{3}{2} \times \text{Median} - \frac{1}{2} \times \text{Mode} \)

Quick Reference

• Mode = 3 Median − 2 Mean
• Median = (1/3) Mode + (2/3) Mean
• Mean = (3/2) Median − (1/2) Mode

Glossary

• Empirical Relation: Approximate relationship between mean, median and mode.