IIT JEE Main Maths -Unit 13- Standard deviation, variance, mean deviation- Study Notes-New Syllabus
IIT JEE Main Maths -Unit 13- Standard deviation, variance, mean deviation – Study Notes – New syllabus
IIT JEE Main Maths -Unit 13- Standard deviation, variance, mean deviation – Study Notes -IIT JEE Main Maths – per latest Syllabus.
Key Concepts:
- Range
- Measures of Dispersion
- Combined Variance and Standard Deviation
- Coefficient of Variation
Range
Range is the simplest measure of dispersion. It gives the spread of a dataset by considering the difference between the largest and the smallest values.
Definition of Range
\( \text{Range} = \text{Maximum value} – \text{Minimum value} \)
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It shows how widely the data is spread but ignores all intermediate values, so it is not a very reliable measure for irregular data.
Coefficient of Range
Useful for comparing variability of two different datasets.
\( \text{Coefficient of Range} = \dfrac{L – S}{L + S} \)
Where:
- \( L \) = largest value
- \( S \) = smallest value
Range for Grouped Data
For grouped data (class intervals):
\( \text{Range} = \text{Upper limit of last class} – \text{Lower limit of first class} \)
This works only for continuous classes.
Advantages
- Very easy to compute
- Gives quick estimate of variability
Disadvantages
- Depends only on two extreme values
- Highly affected by outliers
- Not reliable for skewed distributions
Example
Find the range of the data: 4, 10, 6, 8, 12.
▶️ Answer / Explanation
Maximum = 12
Minimum = 4
Range = \( 12 – 4 = 8 \)
Range = 8
Example
Find the coefficient of range for the data: 15, 8, 20, 5, 12.
▶️ Answer / Explanation
Largest value \( L = 20 \)
Smallest value \( S = 5 \)
Coefficient of Range \( = \dfrac{L – S}{L + S} = \dfrac{20 – 5}{20 + 5} = \dfrac{15}{25} = 0.6 \)
Coefficient of Range = 0.6
Example
For the grouped data:
Classes: 0–10, 10–20, 20–30, 30–40
Frequencies: 5, 7, 12, 6
Find the range.
▶️ Answer / Explanation
Upper limit of last class = 40
Lower limit of first class = 0
Range = \( 40 – 0 = 40 \)
Range = 40
Measures of Dispersion
Measures of dispersion tell us how spread out the values of a dataset are around the average. The three most important measures in JEE are:
- Mean Deviation (M.D.)

- Variance
- Standard Deviation (S.D.)
Mean Deviation (M.D.)
Mean deviation is the average of absolute deviations from mean or median.
(a) Ungrouped Data
\( \text{MD about mean} = \dfrac{\sum |x_i – \bar{x}|}{n} \)
\( \text{MD about median} = \dfrac{\sum |x_i – M|}{n} \)
(b) Grouped Data
\( \text{MD about mean} = \dfrac{\sum f_i |x_i – \bar{x}|}{\sum f_i} \)
\( \text{MD about median} = \dfrac{\sum f_i |x_i – M|}{\sum f_i} \)
Mean deviation about the mean is smaller than about the median for symmetrical distributions.
Variance
Variance measures the average squared deviation from the mean.
(a) Ungrouped Data
\( \sigma^2 = \dfrac{\sum (x_i – \bar{x})^2}{n} \)
Shortcut formula:
\( \sigma^2 = \dfrac{\sum x_i^2}{n} – \bar{x}^2 \)
(b) Grouped Data
\( \sigma^2 = \dfrac{\sum f_i (x_i – \bar{x})^2}{\sum f_i} \)
Shortcut:
\( \sigma^2 = \dfrac{\sum f_i x_i^2}{\sum f_i} – \bar{x}^2 \)
Standard Deviation (S.D.)
S.D. is the positive square root of variance.
\( \sigma = \sqrt{\sigma^2} \)
S.D. is the most important measure of dispersion for JEE.
Coding Method (Shortcut Method) Used for large values in grouped data.
If
\( u_i = \dfrac{x_i – a}{h} \)
Then
\( \bar{x} = a + h\dfrac{\sum f_iu_i}{\sum f_i} \)
\( \sigma = h\sqrt{\dfrac{\sum f_i u_i^2}{\sum f_i} – \left(\dfrac{\sum f_i u_i}{\sum f_i}\right)^2} \)
Comparison of Dispersions
- \( \text{S.D.} \ge \text{Mean Deviation} \ge \text{Quartile Deviation} \)
- S.D. is widely used because of algebraic convenience.
Example
For numbers 2, 4, 6, 8, find the standard deviation.
▶️ Answer / Explanation
Step 1: Mean
\( \bar{x} = \dfrac{2 + 4 + 6 + 8}{4} = 5 \)
Step 2: Variance
\( (x_i – \bar{x})^2 = 9, 1, 1, 9 \)
Variance \( \sigma^2 = \dfrac{20}{4} = 5 \)
Step 3: Standard Deviation
\( \sigma = \sqrt{5} \)
Answer: \( \sigma = \sqrt{5} \)
Example
The frequency distribution is:
x: 10, 20, 30
f: 3, 4, 3
Find the variance.
▶️ Answer / Explanation
Total frequency: \( N = 3 + 4 + 3 = 10 \)
Step 1: Mean
\( \bar{x} = \dfrac{3(10) + 4(20) + 3(30)}{10} = \dfrac{30 + 80 + 90}{10} = 20 \)
Step 2: Compute \( \sum f x^2 \)
\( 3(100) + 4(400) + 3(900) = 300 + 1600 + 2700 = 4600 \)
Step 3: Apply shortcut formula
\( \sigma^2 = \dfrac{\sum f x^2}{N} – \bar{x}^2 = \dfrac{4600}{10} – 20^2 = 460 – 400 = 60 \)
Answer: Variance = 60
Example
Find the mean deviation about the mean for the grouped data:
Class intervals: 0–10, 10–20, 20–30
Frequencies: 5, 3, 2
▶️ Answer / Explanation
Step 1: Midpoints
\( x_i = 5, 15, 25 \)
Step 2: Mean
\( \bar{x} = \dfrac{5(5) + 3(15) + 2(25)}{10} = \dfrac{25 + 45 + 50}{10} = 12 \)
Step 3: Compute \( |x_i – \bar{x}| \)
\( |5 – 12| = 7 \)
\( |15 – 12| = 3 \)
\( |25 – 12| = 13 \)
Step 4: Find M.D.
\( \text{MD} = \dfrac{5(7) + 3(3) + 2(13)}{10} \)
\( = \dfrac{35 + 9 + 26}{10} = \dfrac{70}{10} = 7 \)
Answer: Mean Deviation = 7
Combined Variance and Standard Deviation
When two or more groups are merged, their variance and standard deviation are not simply averaged. You must use weighted formulas based on the size of each group.
Combined Variance Formula (Two Groups)
If two groups have:
- Sizes: \( n_1,\ n_2 \)
- Means: \( \bar{x}_1,\ \bar{x}_2 \)
- Variances: \( \sigma_1^2,\ \sigma_2^2 \)
Then combined variance:
\( \sigma^2 = \dfrac{n_1(\sigma_1^2 + \bar{x}_1^2) + n_2(\sigma_2^2 + \bar{x}_2^2)}{n_1 + n_2} – \bar{x}^2 \)
Where \( \bar{x} \) is combined mean:
\( \bar{x} = \dfrac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} \)
Combined Standard Deviation
\( \sigma = \sqrt{\sigma^2} \)
Shortcut Form
Let difference in means be \( d = \bar{x}_1 – \bar{x}_2 \)
Then:
\( \sigma^2 = \dfrac{n_1\sigma_1^2 + n_2\sigma_2^2}{n_1 + n_2} + \dfrac{n_1n_2}{(n_1 + n_2)^2}d^2 \)
This form is heavily used in JEE.
Example
Two groups: Group 1: \( n_1 = 20, \sigma_1 = 3 \) Group 2: \( n_2 = 30, \sigma_2 = 4 \) Means are same. Find combined SD.
▶️ Answer / Explanation
Since means are same, \( d = 0 \)
\( \sigma^2 = \dfrac{20(9) + 30(16)}{50} = \dfrac{180 + 480}{50} = \dfrac{660}{50} = 13.2 \)
\( \sigma = \sqrt{13.2} \)
Answer: \( \sigma = \sqrt{13.2} \)
Example
Group 1: \( n_1 = 10,\ \bar{x}_1 = 40,\ \sigma_1 = 4 \) Group 2: \( n_2 = 20,\ \bar{x}_2 = 50,\ \sigma_2 = 5 \) Find combined variance.
▶️ Answer / Explanation
Combined mean:
\( \bar{x} = \dfrac{10(40) + 20(50)}{30} = \dfrac{400 + 1000}{30} = 46.\overline{6} \)
Using formula:
\( \sigma^2 = \dfrac{10(16 + 1600) + 20(25 + 2500)}{30} – (46.\overline{6})^2 \)
Calculate: Group1 → \( 10(1616) = 16160 \) Group2 → \( 20(2525) = 50500 \)
Total = 66660 Divide by 30 → 2222 Now subtract mean²: \( (46.\overline{6})^2 = 2177.\overline{7} \)
\( \sigma^2 = 2222 – 2177.\overline{7} = 44.\overline{3} \)
Answer: Combined variance = \( 44.\overline{3} \)
Example
Two groups have same variance 9. Their means differ by 6. If \( n_1 = n_2 = 20 \), find the combined variance.
▶️ Answer / Explanation
Using shortcut:
\( \sigma^2 = \dfrac{20(9) + 20(9)}{40} + \dfrac{20\cdot20}{40^2}6^2 \)
First term: \( \dfrac{360}{40} = 9 \)
Second term: \( \dfrac{400}{1600} \times 36 = \dfrac{1}{4} \times 36 = 9 \)
Combined variance = \( 9 + 9 = 18 \)
Answer: \( \sigma^2 = 18 \)
Coefficient of Variation
The coefficient of variation measures relative dispersion and is used to compare consistency between two datasets.
Formula
\( \text{C.V.} = \dfrac{\sigma}{\bar{x}} \times 100 \)
Where:
- \( \sigma \) = standard deviation
- \( \bar{x} \) = mean
Interpretation
- Smaller C.V. → more consistent (less variation)
- Larger C.V. → less consistent
Important JEE Conclusion
A dataset with lower coefficient of variation is more consistent, regardless of mean or SD alone.
Example
A dataset has \( \bar{x} = 50 \) and \( \sigma = 5 \). Find C.V.
▶️ Answer / Explanation
C.V. = \( \dfrac{5}{50} \times 100 = 10\% \)
Answer: 10 percent
Example
Dataset A: \( \bar{x} = 40,\ \sigma = 8 \) Dataset B: \( \bar{x} = 50,\ \sigma = 12 \) Which is more consistent?
▶️ Answer / Explanation
C.V.(A) = \( \dfrac{8}{40}\times100 = 20\% \)
C.V.(B) = \( \dfrac{12}{50}\times100 = 24\% \)
Lower C.V. = more consistent Dataset A is more consistent.
Answer: Dataset A
Example
Two machines A and B produce bulbs. Their performance is: Machine A: mean = 200, SD = 10 Machine B: mean = 150, SD = 6 Which machine is more consistent?
▶️ Answer / Explanation
C.V.(A) = \( \dfrac{10}{200}\times100 = 5\% \)
C.V.(B) = \( \dfrac{6}{150}\times100 = 4\% \)
Lower C.V. → more consistent Machine B is better.
Answer: Machine B
