JEE Main 2019 (9 January)

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Animated Solution for Mathematics - Statistics: 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:

Q: Why do we use $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ instead of the definition $\sigma^2 = \frac{\sum(x_i - \bar{x})^2}{n}$?

Excellent question. The definition $\sigma^2 = \frac{\sum(x_i - \bar{x})^2}{n}$ requires you to know every individual value of $x_i$. In this problem, we do not know the individual heights of the five students; we only know the aggregate statistics (mean and variance). The computational formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ allows us to extract the 'sum of squares' ($\sum x_i^2$) from the given data, which is the only information we have. It is a powerful algebraic tool that bypasses the need for individual data points.

Select Answer:

Progressive Hints

Hint 1

Think of variance not just as a number, but as a measure of the 'spread' or 'uncertainty' in your data. When you add a new student, you are not just changing the mean; you are fundamentally altering the distribution. The key is to realize that you do not need to know the individual heights of the first five students—you only need their aggregate properties, specifically the sum of their heights and the sum of the squares of their heights.

Hint 2

To solve this without knowing individual values, you must rely on the computational formula for variance:

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

. This formula is your best friend in JEE statistics because it allows you to work with the 'sum of squares' (

\sum x_{i}^{2}

) directly, which is an invariant property of the initial group. Treat this formula as a bridge between the known variance and the unknown sum of squares.

Hint 3

Start by isolating

\sum x_{i}^{2}

for the initial 5 students using the given variance and mean. Once you have that, update your total sum of heights and your total sum of squares by including the new student's height (

156

cm). Finally, apply the variance formula again with

n = 6

to find the new variance. The calculation is straightforward arithmetic, but keep a close eye on your squaring operations!

Visualized Solution

Initial Data of 5 Students

Number of students $n_{1} = 5$
Initial Mean $\overset{x}{ˉ}_{1} = 150$ cm
Initial Variance $σ_{1}^{2} = 18$ cm $^{2}$

Sum of Initial Heights

Mean formula: $\overset{x}{ˉ} = \frac{\sum x _{i}}{n}$
$\sum x_{i} = n_{1} \times \overset{x}{ˉ}_{1}$
$\sum x_{i} = 5 \times 150 = 750$ cm

The Variance Formula

Variance formula: $σ^{2} = \frac{\sum x _{i}^{2}}{n} - (\overset{x}{ˉ})^{2}$
We know $σ_{1}^{2} = 18$ , $n = 5$ , and $\overset{x}{ˉ}_{1} = 150$

Finding $\sum x_{i}^{2}$ for 5 Students

Substitute values: $18 = \frac{\sum x _{i}^{2}}{5} - (150)^{2}$
$18 = \frac{\sum x _{i}^{2}}{5} - 22500$

Calculating Initial Sum of Squares

$\frac{\sum x _{i}^{2}}{5} = 18 + 22500 = 22518$
$\sum x_{i}^{2} = 22518 \times 5 = 112590$

Adding the New Student

New student height $x_{6} = 156$ cm
Total students $n_{2} = 6$

Calculating the New Mean

New Sum $\sum x_{n e w} = 750 + 156 = 906$
New Mean $\overset{x}{ˉ}_{2} = \frac{906}{6} = 151$ cm

Calculating New Sum of Squares

New $\sum x_{i}^{2} = 112590 + (156)^{2}$
$(156)^{2} = 24336$
New $\sum x_{i}^{2} = 112590 + 24336 = 136926$

Setting up the New Variance

New Variance $σ_{2}^{2} = \frac{136926}{6} - (151)^{2}$

Final Variance Calculation

$\frac{136926}{6} = 22821$
$(151)^{2} = 22801$
$σ_{2}^{2} = 22821 - 22801 = 20$ cm $^{2}$

Conclusion

Final Variance = 20
Adding a value far from the mean increases the variance.

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The Sigma Insight: Variance and Standard Deviation

Solution Diagram

Analyzing the Initial State

We begin with a set of

n_{1} = 5

students. We are given the initial mean

\overset{x}{ˉ}_{1} = 150

cm and the initial variance

σ_{1}^{2} = 18

^{2}

To track the "energy" of this system, we utilize the computational formula for variance:

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

By rearranging this to solve for the sum of squares (

\sum x_{i}^{2}

), we establish our baseline:

18 = \frac{\sum x _{i}^{2}}{5} - (150)^{2}

18 = \frac{\sum x _{i}^{2}}{5} - 22500

\sum x_{i}^{2} = 5 \times (22500 + 18) = 112590

Incorporating the New Data

A sixth student joins the group with a height of

x_{6} = 156

cm. The total number of students is now

n_{2} = 6

First, we calculate the new sum of heights. Since the original mean was

150

, the original sum was

5 \times 150 = 750

. The new sum is:

\sum x_{n e w} = 750 + 156 = 906

The new mean

\overset{x}{ˉ}_{2}

is therefore:

\overset{x}{ˉ}_{2} = \frac{906}{6} = 151 cm

Next, we update the total sum of squares by adding the square of the new student's height:

\sum x_{n e w}^{2} = 112590 + (156)^{2} = 112590 + 24336 = 136926

Final Calculation of Variance

With the updated parameters, we apply the variance formula once more to find the new spread of the distribution:

σ_{2}^{2} = \frac{136926}{6} - (151)^{2}

σ_{2}^{2} = 22821 - 22801

σ_{2}^{2} = 20

The new variance of the group is $20$ cm $^{2}$ . This increase demonstrates how the addition of a value further from the original mean expands the overall dispersion of the dataset.

Common Traps & FAQs

Why can't I just take the average of the old variance and the new student's deviation from the mean?

That is a very intuitive trap! Variance is not a linear property. It is based on the sum of squares, which is a quadratic function. When you add a new data point, you are changing the mean of the entire set, which shifts the reference point for every single student. Because the mean changes, the 'deviation from the mean' for the original five students also changes. Therefore, you cannot simply average the variances; you must recalculate the sum of squares for the entire new population.

Why do we use

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

instead of the definition

σ^{2} = \frac{\sum ( x _{i} - x ˉ ) ^{2}}{n}

Excellent question. The definition

σ^{2} = \frac{\sum ( x _{i} - x ˉ ) ^{2}}{n}

requires you to know every individual value of

x_{i}

. In this problem, we do not know the individual heights of the five students; we only know the aggregate statistics (mean and variance). The computational formula

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

allows us to extract the 'sum of squares' (

\sum x_{i}^{2}

) from the given data, which is the only information we have. It is a powerful algebraic tool that bypasses the need for individual data points.

Does the variance always increase when we add a new data point?

Not necessarily! The variance increases if the new data point is 'far' from the mean, effectively increasing the spread of the data. If you were to add a student whose height was exactly equal to the current mean, the variance would actually decrease because you are adding a value that contributes zero to the squared deviation, while increasing the denominator (

n

). In this problem, the new student's height (

156

cm) is further from the mean (

150

cm) than the average, which is why the variance increased from

18

20

Animated Solution for Mathematics - Statistics: 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:

Select Answer:

Visualized Solution

Initial Data of 5 Students

Sum of Initial Heights

The Variance Formula

Finding $\sum x_{i}^{2}$ for 5 Students

Calculating Initial Sum of Squares

Adding the New Student

Calculating the New Mean

Calculating New Sum of Squares

Setting up the New Variance

Final Variance Calculation

Conclusion

The Sigma Insight: Variance and Standard Deviation

Analyzing the Initial State

Incorporating the New Data

Final Calculation of Variance

Similar Questions

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Animated Solution for Mathematics - Statistics: 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:

Select Answer:

Visualized Solution

Initial Data of 5 Students

Sum of Initial Heights

The Variance Formula

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑xi2​ for 5 Students

Calculating Initial Sum of Squares

Adding the New Student

Calculating the New Mean

Calculating New Sum of Squares

Setting up the New Variance

Final Variance Calculation

Conclusion

The Sigma Insight: Variance and Standard Deviation

Analyzing the Initial State

Incorporating the New Data

Final Calculation of Variance

Similar Questions

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Finding $\sum x_{i}^{2}$ for 5 Students

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