Why can't we just use the definition $\sigma = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n}}$?

You absolutely can! However, using the definition requires you to calculate $(x_i - \bar{x})^2$ for each term, which involves the mean $\bar{x} = k/4$. This leads to more complex algebraic expressions. The computational formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ is a standard JEE technique to save time and reduce the chance of calculation errors. It's all about choosing the most efficient path to the solution.

JEE Main 2019 (09 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Statistics: If the standard deviation of the numbers $- 1, 0, 1, k$ is $5$ where $k > 0$ , then $k$ is equal to

Q: Wait, why do we reject the negative root?

Great catch! In the final step, we get $k^2 = 24$, which means $k = \pm\sqrt{24}$. However, the problem explicitly states $k > 0$. In JEE problems, always re-read the constraints at the end. They are often the key to selecting the correct answer among the options.

Q: Is there a shortcut for the mean?

Yes! Notice the symmetry of the numbers $-1, 0, 1$. Their sum is $0$. When you add $k$ to this set, the sum becomes $0 + k = k$. Since there are $4$ observations, the mean is simply $k/4$. Recognizing symmetry is a hallmark of an elite problem solver.

Select Answer:

Visualized Solution

Identify the Data Set

Given data set: ${- 1, 0, 1, k}$
Number of observations $n = 4$
Standard Deviation $σ = 5$
Constraint: $k > 0$

The Variance Formula

Standard Deviation $σ = 5 ⟹$ Variance $σ^{2} = 5$
Formula: $σ^{2} = \frac{\sum x _{i}^{2}}{n} - (\overset{x}{ˉ})^{2}$

Calculate the Mean $\overset{x}{ˉ}$

Mean $\overset{x}{ˉ} = \frac{\sum x _{i}}{n}$
$\overset{x}{ˉ} = \frac{- 1 + 0 + 1 + k}{4}$
$\overset{x}{ˉ} = \frac{k}{4}$

Sum of Squares $\sum x_{i}^{2}$

$\sum x_{i}^{2} = (- 1)^{2} + 0^{2} + 1^{2} + k^{2}$
$\sum x_{i}^{2} = 1 + 0 + 1 + k^{2}$
$\sum x_{i}^{2} = 2 + k^{2}$

Set up the Equation

Substitute the values into the variance equation:
$σ^{2} = \frac{\sum x _{i}^{2}}{n} - (\overset{x}{ˉ})^{2}$
$5 = \frac{2 + k ^{2}}{4} - (\frac{k}{4})^{2}$

Expand the Mean Square

$5 = \frac{2 + k ^{2}}{4} - \frac{k ^{2}}{16}$

Clear the Denominators

Multiply the entire equation by $16$ :
$16 \times 5 = 16 \times (\frac{2 + k ^{2}}{4}) - 16 \times (\frac{k ^{2}}{16})$
$80 = 4 (2 + k^{2}) - k^{2}$

Expand and Simplify

$80 = 8 + 4 k^{2} - k^{2}$
$80 = 8 + 3 k^{2}$

Isolate $k^{2}$

$3 k^{2} = 80 - 8$
$3 k^{2} = 72$
$k^{2} = \frac{72}{3} = 24$

Find the value of $k$

$k = \pm 24$
Since $k > 0$ , $k = 24$
$k = 4 \times 6 = 26$
Matches Option (2)

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The Sigma Insight: Variance and Standard Deviation

Solution Diagram

Analyzing the Setup

We are given a data set

{- 1, 0, 1, k}

with

n = 4

observations. The standard deviation is given as

σ = 5

To simplify our calculations, we work with the variance,

σ^{2}

σ^{2} = 5

The Master Equation

We utilize the computational formula for variance, which is a staple in the JEE syllabus:

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

First, we calculate the mean

\overset{x}{ˉ}

\overset{x}{ˉ} = \frac{- 1 + 0 + 1 + k}{4} = \frac{k}{4}

Next, we calculate the sum of the squares of the observations,

\sum x_{i}^{2}

\sum x_{i}^{2} = (- 1)^{2} + 0^{2} + 1^{2} + k^{2} = 2 + k^{2}

Solving for k

Substituting these values into the variance formula, we obtain:

5 = \frac{2 + k ^{2}}{4} - (\frac{k}{4})^{2}

Expanding the squared term gives:

5 = \frac{2 + k ^{2}}{4} - \frac{k ^{2}}{16}

To eliminate the denominators, multiply the entire equation by

16

16 \times 5 = 16 \times (\frac{2 + k ^{2}}{4}) - 16 \times (\frac{k ^{2}}{16})

80 = 4 (2 + k^{2}) - k^{2}

Expanding and simplifying the expression:

80 = 8 + 4 k^{2} - k^{2}

80 = 8 + 3 k^{2}

72 = 3 k^{2}

Solving for

k^{2}

k^{2} = 24

k = \pm 24

Final Result

Given the constraint

k > 0

, we discard the negative root. Simplifying the radical:

k = 24 = 4 \times 6 = * * 26 * *

Common Traps & FAQs

Why can't we just use the definition

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

You absolutely can! However, using the definition requires you to calculate

(x_{i} - \overset{x}{ˉ})^{2}

for each term, which involves the mean

\overset{x}{ˉ} = k /4

. This leads to more complex algebraic expressions. The computational formula

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

is a standard JEE technique to save time and reduce the chance of calculation errors. It's all about choosing the most efficient path to the solution.

Wait, why do we reject the negative root?

Great catch! In the final step, we get

k^{2} = 24

, which means

k = \pm 24

. However, the problem explicitly states

k > 0

. In JEE problems, always re-read the constraints at the end. They are often the key to selecting the correct answer among the options.

Is there a shortcut for the mean?

Yes! Notice the symmetry of the numbers

- 1, 0, 1

. Their sum is

0

. When you add

k

to this set, the sum becomes

0 + k = k

. Since there are

4

observations, the mean is simply

k /4

. Recognizing symmetry is a hallmark of an elite problem solver.

Animated Solution for Mathematics - Statistics: If the standard deviation of the numbers $- 1, 0, 1, k$ is $5$ where $k > 0$ , then $k$ is equal to

Select Answer:

Visualized Solution

Identify the Data Set

The Variance Formula

Calculate the Mean $\overset{x}{ˉ}$

Sum of Squares $\sum x_{i}^{2}$

Set up the Equation

Expand the Mean Square

Clear the Denominators

Expand and Simplify

Isolate $k^{2}$

Find the value of $k$

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Master Equation

Solving for k

Final Result

Similar Questions

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Let $a_{1}, a_{2}, \dots, a_{10}$ be 10 observations such that $\sum_{k = 1}^{10} a_{k} = 50$ and $\sum_{\forall k < j} a_{k} \cdot a_{j} = 1100$ . Then the standard deviation of $a_{1}, a_{2}, \dots, a_{10}$ is equal to :

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Animated Solution for Mathematics - Statistics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the standard deviation of the numbers −1,0,1,k is 5​ where k>0, then k is equal to

Select Answer:

Visualized Solution

Identify the Data Set

The Variance Formula

Calculate the Mean .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xˉ

Sum of Squares .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑xi2​

Set up the Equation

Expand the Mean Square

Clear the Denominators

Expand and Simplify

Isolate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k2

Find the value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Master Equation

Solving for k

Final Result

Similar Questions

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Animated Solution for Mathematics - Statistics: If the standard deviation of the numbers $- 1, 0, 1, k$ is $5$ where $k > 0$ , then $k$ is equal to

Calculate the Mean $\overset{x}{ˉ}$

Sum of Squares $\sum x_{i}^{2}$

Isolate $k^{2}$

Find the value of $k$

If $\sum_{i = 1}^{n} (x_{i} - a) = n$ and $\sum_{i = 1}^{n} (x_{i} - a)^{2} = na$ , $(n, a > 1)$ then the standard deviation of $n$ observations $x_{1}, x_{2}, \dots, x_{n}$ is:

A data consists of $n$ observations: $x_{1}, x_{2}, ....., x_{n}$ . If $\sum_{i = 1}^{n} (x_{i} + 1)^{2} = 9 n$ and $\sum_{i = 1}^{n} (x_{i} - 1)^{2} = 5 n$ , then the standard deviation of this data is :

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