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Comprehensive Statistical Methods: Mean, Median, Mode, Variance, and Standard Deviation Explained

Explore key statistical concepts including the relationships between mean, median, and mode, plus detailed methods to calculate variance and standard deviation with practical examples.

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STATISTICAL METHODS Dr. Abhishek Kumar Lecture-06
Relation between Mean, Median, and Mode  Basic Definitions • Mean (औसत) Sum of all values ÷ total values → • Median (मध्यिका) Middle value after sorting → • Mode (बहुलक) Most frequent value →  Important Relation (Formula) • For a moderately skewed distribution, the empirical relation is:  Rearranged forms: Mean = (Mode + 2×Median) / 3 • Median = (Mode + 2×Mean) / 3
📈 Case-wise Relation 1. ✅ Symmetrical Distribution •Mean = Median = Mode Example: 10, 20, 30, 40, 50 👉 ✔ All are 30 2. Positively Skewed (Right ➡️ Skewed) •Mean > Median > Mode 👉 Example: 2, 3, 4, 5, 100 •Mean = 22.8 •Median = 4 •Mode = 2–5 (approx lower values) ✔ High value (100) pulls mean to right
3. Negatively Skewed (Left Skewed) ⬅️ • Mean < Median < Mode 👉 Example: 1, 50, 60, 70, 80 •Mean = 52.2 •Median = 60 •Mode ≈ 60–80 ✔ Low value (1) pulls mean to left
Example 1: The median and mode for a given set of data points is 20 and 30 respectively. Find out the mean. Given, Mode = 30, Median = 20 To find the Mean: Mode = 3×Median 2×Mean − 30=3×20 2×Mean − 30=60 2×Mean − 2×Mean=30 Mean=15 Solution:
Example 2: The mean and mode for a given set of data points is 20 and 30 respectively. Find out the mean. Given: Mode = 25, Mean = 12 To find the Median: Mode=3×Median 2×Mean − 25=3×Median 2×12 − 25=3×Median 24 − 3×Median=49 Median 16.33 ≈ Solution:
Example 3: The median and mean for a given set of data points is 15 and 10 respectively. Find out the mean. Given: Median = 15, Mean = 10 To find the Mode: Mode=3×Median 2×Mean − Mode=3×15 2×10 − Mode=45 20 − Mode=25 Solution:
Q1. Given: Mean = 24, Mode = 28. Calculate the Median and verify if the distribution is positively or negatively skewed. Q2. Given: Mean = 40, Median = 35. Calculate the Mode and comment on the data distribution's symmetry. Q3. Given: Mode = 48, Mean = 45. Calculate the Median and determine the implications for the data's central tendency. Practice Questions on Relation between Mean, Median and Mode
Variance  Variance is the average squared deviation from the mean of a set of data.  It is used to find the standard deviation.
Find the Mean of the data.  Mean is the average so add up the values and divide by the number of items.  Subtract the mean from each value – the result is called the deviation from the mean.  Square each deviation of the mean.  Find the sum of the squares.  Divide the total by the number of items. {Processes To Find Variance}
Variance Formula The variance formula includes the Summation Notation, which represents the sum of all the items to the right of Sigma.  For population variance Mean is represented by & x ̅ and n & N is the number of items.    2 2 x X N       2 2 1 x X s n     For sample variance
Standard Deviation  Standard Deviation shows the variation in data.  If the data is close together, the standard deviation will be small.  If the data is spread out, the standard deviation will be large.  Standard Deviation is often denoted by the lowercase Greek letter sigma, . 
Find the variance. a) Find the Mean of the data. b) Subtract the mean from each value. c) Square each deviation of the mean. d) Find the sum of the squares. e) Divide the total by the number of items.  Take the square root of the variance. {Processes To Find Variance & Standard Deviation}
Standard Deviation Formula The standard deviation formula can be represented using Sigma Notation: Note:- The standard deviation formula is the square root of the variance.   2 1 x X s n       2 x N      sample standard deviation population standard deviation
Find The Variance And Standard Deviation Example : 1 The math test scores of five students are: 92,88,80,68 and 52.
2) Find the deviation from the mean: (x - ) x ̅ 92-76=16 88-76=12 80-76=4 68-76= -8 52-76= -24 1) Find the Mean ( ): x ̅ (92+88+80+68+52)/5 = 76.
3) Square the deviation from the mean: (x - ) x ̅ 2 (16)² = 256 (12)² = 144 (4)² = 16 (-8)² = 64 (-24)² = 576
4) Find the sum of the squares of the deviation from the mean (x - ) x ̅ 2 : 256+144+16+64+576= 1056 5) Divide by the number of data items to find the variance: 2 1056/5 = 211.2 ( ) x X N  
6) Find the square root of the variance: 211.2 14.53  Thus, the standard deviation of the test scores is 14.53.
Standard Deviation Example Example : 2 Measure the height of all the persons in your room.  If your room contains four members and their heights are : 165 145 153 150
1) Now you have to find the mean of the height’s measured : Mean ( ) x ̅ = 165+145+153+150 =153.25 4 2) Next step is to find the variance.  It is the average of the squares of the differences from mean.  Subtract individual heights from the mean and square each value.
3) Find the deviation from the mean: (x -x ) ̅ 165-153.25= 11.75 145-153.25= -8.25 153-153.25= -0.25 150-153.25= -3.25 4) Square the deviation from the mean: (x - )² x ̅ (11.75)²=138.0625 (-8.25)²=68.0625 (-0.25)²=0.0625 (-3.25)²=10.5625
5) Find the sum of the squares of the deviation from the mean(x - )² x ̅ 138.0625+68.0625+0.0625+10.5625=216.75  Sum of the square of deviation is: 216.75  For population standard deviation, we would calculate variance without subtracting “1” from the denominator.  But here we subtracting “1” from the denominator.  This Process is called degree of freedom.
6) Divide by the number of data items and subtracting 1 from the denominator to find the variance: 216.75 = 72.25 4-1 7) Find the square root of the variance: = √72.25 =8.5 Thus, the standard deviation of the heights of the members is 8.5.   2 2 1 x X s n       2 1 x X s n    