Does the order of subtraction in $(\bar{x}_1 - \bar{x}_2)^2$ matter?

Not at all! Because the term is squared, the sign of the difference disappears. Whether you do $(12 - 17)^2$ or $(17 - 12)^2$, you get $(-5)^2$ or $(5)^2$, both of which equal $25$. The math is robust against your choice of order, so don't worry about which mean comes first.

JEE Main 2021 (27 Aug Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Statistics: An online exam is attempted by 50 candidates out of which 20 are boys. The average marks obtained by boys is 12 with a variance 2. The variance of marks obtained by 30 girls is also 2. The average marks of all 50 candidates is 15. If $μ$ is the average marks of girls and $σ^{2}$ is the variance of marks of 50 candidates, then $μ + σ^{2}$ is equal to .

Q: Is there a shortcut to avoid the long variance formula?

While you could calculate the sum of squares $\sum x^2$ for both groups separately, it is prone to calculation errors. The combined variance formula is the most elegant and efficient path. Trust the formula; it is derived from the definition of variance and is your best friend in JEE Advanced. It saves you from calculating the raw sum of squares for all 50 candidates.

Enter Numerical Value:

Progressive Hints

Hint 1

Think of the total population as a union of two distinct sets. The combined mean is simply the weighted average of the individual means. Visualize the boys and girls as two separate clusters of data points, and remember that the total sum of marks is the sum of the marks of the boys plus the sum of the marks of the girls.

Hint 2

When dealing with combined variance, remember that it is not just the average of the variances. You must account for the 'spread' created by the difference between the two group means. The formula is

σ^{2} = \frac{n _{1} σ _{1}^{2} + n _{2} σ _{2}^{2}}{n _{1} + n _{2}} + \frac{n _{1} n _{2}}{( n _{1} + n _{2} ) ^{2}} (\overset{x}{ˉ}_{1} - \overset{x}{ˉ}_{2})^{2}

. This formula is your best friend for merging datasets.

Hint 3

Start by isolating

μ

using the weighted mean equation:

15 = \frac{20 ( 12 ) + 30 ( μ )}{50}

. Once you have

μ

, substitute it directly into the second term of the variance formula. The rest is just arithmetic, so keep your calculations precise!

Visualized Solution

Problem Overview

Total candidates: $n = 50$
Boys: $n_{1} = 20$
Girls: $n_{2} = 30$

Analyzing Boys' Data

For Boys ( $n_{1} = 20$ ):
Average marks $\overset{x}{ˉ}_{b} = 12$
Variance $σ_{b}^{2} = 2$

Analyzing Girls' Data

For Girls ( $n_{2} = 30$ ):
Average marks $\overset{x}{ˉ}_{g} = μ$
Variance $σ_{g}^{2} = 2$

Combined Data Parameters

Total candidates $n = 50$
Combined average $\overset{ˉ}{X} = 15$
Combined variance $= σ^{2}$

The Combined Mean Formula

$\overset{ˉ}{X} = \frac{n _{1} x ˉ _{b} + n _{2} x ˉ _{g}}{n _{1} + n _{2}}$

Substituting Values for Mean

$15 = \frac{20 ( 12 ) + 30 ( μ )}{50}$

Solving for $μ$

$15 \times 50 = 240 + 30 μ$
$750 = 240 + 30 μ$
$30 μ = 510$
$μ = 17$

The Combined Variance Formula

$σ^{2} = \frac{n _{1} σ _{1}^{2} + n _{2} σ _{2}^{2}}{n _{1} + n _{2}} + \frac{n _{1} n _{2}}{( n _{1} + n _{2} ) ^{2}} (\overset{x}{ˉ}_{1} - \overset{x}{ˉ}_{2})^{2}$

Calculating Within-Group Variance

Term 1: $\frac{n _{1} σ _{1}^{2} + n _{2} σ _{2}^{2}}{n _{1} + n _{2}}$
$= \frac{20 ( 2 ) + 30 ( 2 )}{50}$
$= \frac{100}{50} = 2$

Calculating Between-Group Variance

Term 2: $\frac{n _{1} n _{2}}{( n _{1} + n _{2} ) ^{2}} (\overset{x}{ˉ}_{1} - \overset{x}{ˉ}_{2})^{2}$
$= \frac{20 \times 30}{( 50 ) ^{2}} (12 - 17)^{2}$

Simplifying the Second Term

$= \frac{600}{2500} \times (- 5)^{2}$
$= \frac{6}{25} \times 25 = 6$

Total Combined Variance $σ^{2}$

$σ^{2} = Term 1 + Term 2$
$σ^{2} = 2 + 6 = 8$

Final Calculation $μ + σ^{2}$

We need to find: $μ + σ^{2}$
$μ = 17$
$σ^{2} = 8$
$μ + σ^{2} = 17 + 8 = 25$

00:00 / 00:00

The Sigma Insight: Variance and Standard Deviation

Solution Diagram

Analyzing the Setup

We are given two groups of students: boys and girls. Let

n_{1} = 20

be the number of boys and

n_{2} = 30

be the number of girls.

For the boys, the mean is

\overset{x}{ˉ}_{b} = 12

and the variance is

σ_{1}^{2} = 2

. For the girls, the variance is

σ_{2}^{2} = 2

, while their mean

μ

is unknown. The combined mean of the entire group of

50

students is

\overset{ˉ}{X} = 15

Finding the Mean of the Girls

The combined mean is the weighted average of the two groups. We use the formula:

\overset{ˉ}{X} = \frac{n _{1} x ˉ _{b} + n _{2} x ˉ _{g}}{n _{1} + n _{2}}

Substituting the known values into the equation:

15 = \frac{20 ( 12 ) + 30 ( μ )}{50}

Multiplying both sides by

50

yields

750 = 240 + 30 μ

. Solving for

μ

30 μ = 750 - 240 = 510

μ = \frac{510}{30} = 17

Calculating the Combined Variance

The combined variance

σ^{2}

is calculated using the formula that accounts for both the internal variance of the groups and the variance between their means:

σ^{2} = \frac{n _{1} σ _{1}^{2} + n _{2} σ _{2}^{2}}{n _{1} + n _{2}} + \frac{n _{1} n _{2}}{( n _{1} + n _{2} ) ^{2}} (\overset{x}{ˉ}_{1} - \overset{x}{ˉ}_{2})^{2}

The first part (weighted average of variances) is:

\frac{20 ( 2 ) + 30 ( 2 )}{50} = \frac{40 + 60}{50} = 2

The second part (between-group variance) is:

\frac{20 \times 30}{( 50 ) ^{2}} (12 - 17)^{2} = \frac{600}{2500} \times (- 5)^{2} = \frac{6}{25} \times 25 = 6

Adding these two components together, we find the combined variance:

σ^{2} = 2 + 6 = 8

Final Calculation

The problem asks for the sum of the girls' mean

μ

and the combined variance

σ^{2}

μ + σ^{2} = 17 + 8 = 25

The final answer is 25.

Common Traps & FAQs

Why can't we just take the average of the two variances? Why is there a second term?

Great question! The variance of a combined set is not just the average of the individual variances because the groups might be centered at different values. The second term,

\frac{n _{1} n _{2}}{( n _{1} + n _{2} ) ^{2}} (\overset{x}{ˉ}_{1} - \overset{x}{ˉ}_{2})^{2}

, accounts for the 'distance' between the means of the two groups. If the means were identical, this term would vanish, and the variance would indeed be the weighted average. But here, the shift in means adds extra spread to the total population.

Does the order of subtraction in

(\overset{x}{ˉ}_{1} - \overset{x}{ˉ}_{2})^{2}

matter?

Not at all! Because the term is squared, the sign of the difference disappears. Whether you do

(12 - 17)^{2}

(17 - 12)^{2}

, you get

(- 5)^{2}

(5)^{2}

, both of which equal

25

. The math is robust against your choice of order, so don't worry about which mean comes first.

Is there a shortcut to avoid the long variance formula?

While you could calculate the sum of squares

\sum x^{2}

for both groups separately, it is prone to calculation errors. The combined variance formula is the most elegant and efficient path. Trust the formula; it is derived from the definition of variance and is your best friend in JEE Advanced. It saves you from calculating the raw sum of squares for all 50 candidates.

Enter Numerical Value:

Visualized Solution

Problem Overview

Analyzing Boys' Data

Analyzing Girls' Data

Combined Data Parameters

The Combined Mean Formula

Substituting Values for Mean

Solving for $μ$

The Combined Variance Formula

Calculating Within-Group Variance

Calculating Between-Group Variance

Simplifying the Second Term

Total Combined Variance $σ^{2}$

Final Calculation $μ + σ^{2}$

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

Finding the Mean of the Girls

Calculating the Combined Variance

Final Calculation

Similar Questions

The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and $σ^{2}$ respectively. If the variance of all the 30 numbers in the two sets is 13, then $σ^{2}$ is equal to

If the mean and the variance of the data \begin{array}{|c|c|c|c|c|} \hline Class & 4-8 & 8-12 & 12-16 & 16-20 \\ \hline Frequency & 3 & $λ$ & 4 & 7 \\ \hline \end{array} are $μ$ and 19 respectively, then the value of $λ + μ$ is

If the mean and variance of the data 65, 68, 58, 44, 48, 45, 60, $α, β, 60$ where $α > β$ are 56 and 66.2 respectively, then $α^{2} + β^{2}$ is equal to

Suppose a class has 7 students. The average marks of these students in the mathematics examination is 62, and their variance is 20. A student fails in the examination if he/she gets less than 50 marks, then in worst case, the number of students can fail is

5 students of a class have an average height 150 cm and variance 18 cm². A new student, whose height is 156 cm, joined them. The variance (in cm²) of the height of these six students is:

Let the mean and variance of 8 numbers $x, y, 10, 12, 6, 12, 4, 8$ be 9 and 9.25 respectively. If $x > y$ , then $3 x - 2 y$ is equal to ______.

A student scores the following marks in five tests : 45, 54, 41, 57, 43. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is

If the mean and variance of the frequency distribution $x_{i} f_{i} 2444 6 α 815108 12 β 144165$ are 9 and 15.08, respectively, then the value of $α^{2} + β^{2} - α β$ is _____.

Enter Numerical Value:

Visualized Solution

Problem Overview

Analyzing Boys' Data

Analyzing Girls' Data

Combined Data Parameters

The Combined Mean Formula

Substituting Values for Mean

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } μ

The Combined Variance Formula

Calculating Within-Group Variance

Calculating Between-Group Variance

Simplifying the Second Term

Total Combined Variance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } σ2

Final Calculation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } μ+σ2

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

Finding the Mean of the Girls

Calculating the Combined Variance

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the mean and variance of the data 65, 68, 58, 44, 48, 45, 60, α,β,60 where α>β are 56 and 66.2 respectively, then α2+β2 is equal to

Suppose a class has 7 students. The average marks of these students in the mathematics examination is 62, and their variance is 20. A student fails in the examination if he/she gets less than 50 marks, then in worst case, the number of students can fail is

5 students of a class have an average height 150 cm and variance 18 cm². A new student, whose height is 156 cm, joined them. The variance (in cm²) of the height of these six students is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the mean and variance of 8 numbers x,y,10,12,6,12,4,8 be 9 and 9.25 respectively. If x>y, then 3x−2y is equal to ______.

A student scores the following marks in five tests : 45, 54, 41, 57, 43. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is

Solving for $μ$

Total Combined Variance $σ^{2}$

Final Calculation $μ + σ^{2}$

If the mean and variance of the data 65, 68, 58, 44, 48, 45, 60, $α, β, 60$ where $α > β$ are 56 and 66.2 respectively, then $α^{2} + β^{2}$ is equal to

Let the mean and variance of 8 numbers $x, y, 10, 12, 6, 12, 4, 8$ be 9 and 9.25 respectively. If $x > y$ , then $3 x - 2 y$ is equal to ______.