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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

IIT JEE Main Maths -Study Notes – All Topics

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

\( \text{Range} = \text{Maximum value} – \text{Minimum value} \)

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

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