JEE Main 2021 (26 Aug Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Statistics: The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively. It was found that by mistake one data value was taken as 25 instead of 35. If $α$ and $β$ are the mean and standard deviation respectively for correct data, then $(α, β)$ is:

Q: The question asks for $\sqrt{\beta}$ as the standard deviation, but the options look like integers. Did I miss a square root?

Great catch! The question defines $\sqrt{\beta}$ as the standard deviation. This means $\beta$ is the variance. If you calculate $\beta = 26$, then the standard deviation is $\sqrt{26}$. The question asks for the pair $(\alpha, \beta)$, not $(\alpha, \sqrt{\beta})$. Always read the target variable carefully!

Select Answer:

Progressive Hints

Hint 1

Think of the mean as the 'center of gravity' of your data. When one value changes, the entire center shifts. Your first step isn't to calculate the new mean directly, but to recover the 'total sum' of all observations before the error occurred. How does the mean formula

\overset{x}{ˉ} = \frac{\sum x _{i}}{n}

help you find that total?

Hint 2

Now, the variance is a different beast. It measures the spread, which depends on the squares of the values. You cannot simply adjust the variance by adding or subtracting a constant. You must isolate the 'sum of squares' term,

\sum x_{i}^{2}

, using the variance formula

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

. This is your secret weapon.

Hint 3

To execute, first find the old sum of squares using the incorrect variance. Then, apply the correction:

\sum x_{n e w}^{2} = \sum x_{o l d}^{2} - (25)^{2} + (35)^{2}

. Once you have this corrected sum of squares, plug it back into the variance formula along with your new mean to find

β

Visualized Solution

Analyzing the Given Data

Number of observations: $n = 20$
Incorrect Mean: $\overset{x}{ˉ} = 10$
Incorrect Standard Deviation: $σ = 2.5$
Mistake: $25$ was recorded instead of $35$ .

Formula for the Sum of Observations

Mean formula: $\overset{x}{ˉ} = \frac{\sum x _{i}}{n}$
Rearranging gives the sum: $\sum x_{i} = n \cdot \overset{x}{ˉ}$

Calculating the Incorrect Sum

Incorrect Sum: $\sum x_{o l d} = 20 \cdot 10$
$\sum x_{o l d} = 200$

Correcting the Sum

Correct Sum: $\sum x_{n e w} = \sum x_{o l d} - Wrong Value + Right Value$
$\sum x_{n e w} = 200 - 25 + 35$
$\sum x_{n e w} = 210$

Finding the Correct Mean ( $α$ )

Correct Mean: $α = \frac{\sum x _{n e w}}{n}$
$α = \frac{210}{20}$
$α = 10.5$

Formula for Variance

Variance is the square of Standard Deviation: $σ^{2}$
Variance formula: $σ^{2} = \frac{\sum x _{i}^{2}}{n} - (\overset{x}{ˉ})^{2}$

Setting up the Old Sum of Squares

Old Variance: $(2.5)^{2} = 6.25$
$6.25 = \frac{\sum x _{o l d}^{2}}{20} - (10)^{2}$

Calculating $\sum x_{o l d}^{2}$

$6.25 = \frac{\sum x _{o l d}^{2}}{20} - 100$
$\frac{\sum x _{o l d}^{2}}{20} = 106.25$
$\sum x_{o l d}^{2} = 106.25 \cdot 20 = 2125$

Correcting the Sum of Squares

Correct Sum of Squares: $\sum x_{n e w}^{2} = \sum x_{o l d}^{2} - (Wrong)^{2} + (Right)^{2}$
$\sum x_{n e w}^{2} = 2125 - 2 5^{2} + 3 5^{2}$

Calculating $\sum x_{n e w}^{2}$

$3 5^{2} - 2 5^{2} = 1225 - 625 = 600$
$\sum x_{n e w}^{2} = 2125 + 600$
$\sum x_{n e w}^{2} = 2725$

Finding the Correct Variance ( $β$ )

New Variance ( $β$ ) = $\frac{\sum x _{n e w}^{2}}{n} - (α)^{2}$
$β = \frac{2725}{20} - (10.5)^{2}$

Calculating $β$

$\frac{2725}{20} = 136.25$
$(10.5)^{2} = 110.25$
$β = 136.25 - 110.25 = 26$

Final Answer

Correct Mean ( $α$ ) = $10.5$
Correct Variance ( $β$ ) = $26$
Final Answer: $(α, β) = (10.5, 26)$

00:00 / 00:00

The Sigma Insight: Variance and Standard Deviation

Solution Diagram

Analyzing the Setup

We are given a dataset of

n = 20

observations with an initial mean of

10

and a standard deviation of

2.5

. A data entry error occurred where a value of

25

was recorded instead of the correct value of

35

Our objective is to determine the true mean,

α

, and the true variance,

β

Phase 1

The Linear Correction

The mean

\overset{x}{ˉ}

is defined as the sum of all values divided by the count:

\overset{x}{ˉ} = \frac{\sum x _{i}}{n}

Given the initial parameters, the incorrect sum is:

\sum x_{o l d} = 20 \times 10 = 200

To correct this, we remove the incorrect value and add the correct one:

\sum x_{n e w} = 200 - 25 + 35 = 210

The new mean

α

is calculated as:

α = \frac{210}{20} = 10.5

Phase 2

The Quadratic Trap

Variance is defined by the relationship between the sum of squares and the square of the mean:

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

Using the incorrect variance

σ^{2} = (2.5)^{2} = 6.25

, we solve for the incorrect sum of squares:

6.25 = \frac{\sum x _{o l d}^{2}}{20} - (10)^{2}

Rearranging the terms:

\frac{\sum x _{o l d}^{2}}{20} = 6.25 + 100 = 106.25

\sum x_{o l d}^{2} = 2125

Phase 3

The Surgical Repair

We must now update the sum of squares by removing the square of the incorrect value and adding the square of the correct value:

\sum x_{n e w}^{2} = \sum x_{o l d}^{2} - (25)^{2} + (35)^{2}

Substituting the values:

\sum x_{n e w}^{2} = 2125 - 625 + 1225 = 2725

Finally, we calculate the corrected variance

β

using the new sum of squares and the new mean

α = 10.5

β = \frac{2725}{20} - (10.5)^{2}

β = 136.25 - 110.25 = 26

The true mean is

α = 10.5

and the true variance is

β = 26

. The final result is the ordered pair $(10.5, 26)$ .

Common Traps & FAQs

Why can't I just add the difference (35-25) to the standard deviation? It seems like a linear shift!

Ah, the classic trap! Standard deviation is not a linear operator. It measures the 'spread' or 'distance' from the mean. When you change a data point, you aren't just shifting the data; you are changing the 'squared distances' of every single point from the mean. That's why we must go back to the sum of squares,

\sum x_{i}^{2}

, to find the truth.

Do I really need to calculate the sum of squares? It feels like so much extra work.

I know it feels tedious, but in statistics, the sum of squares is the 'DNA' of the variance. Without it, you are blind to how the data is distributed. Trust the process—once you master this, you'll never fear a variance problem again.

The question asks for

β

as the standard deviation, but the options look like integers. Did I miss a square root?

Great catch! The question defines

β

as the standard deviation. This means

β

is the variance. If you calculate

β = 26

, then the standard deviation is

26

. The question asks for the pair

(α, β)

, not

(α, β)

. Always read the target variable carefully!

Select Answer:

Visualized Solution

Analyzing the Given Data

Formula for the Sum of Observations

Calculating the Incorrect Sum

Correcting the Sum

Finding the Correct Mean ( $α$ )

Formula for Variance

Setting up the Old Sum of Squares

Calculating $\sum x_{o l d}^{2}$

Correcting the Sum of Squares

Calculating $\sum x_{n e w}^{2}$

Finding the Correct Variance ( $β$ )

Calculating $β$

Final Answer

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

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Select Answer:

Visualized Solution

Analyzing the Given Data

Formula for the Sum of Observations

Calculating the Incorrect Sum

Correcting the Sum

Finding the Correct Mean ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α)

Formula for Variance

Setting up the Old Sum of Squares

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑xold2​

Correcting the Sum of Squares

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑xnew2​

Finding the Correct Variance ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β)

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

Final Answer

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is

The mean and variance of 10 observations were calculated as 15 and 15 respectively by a student who took by mistake 25 instead of 15 for one observation. Then, the correct standard deviation is

Finding the Correct Mean ( $α$ )

Calculating $\sum x_{o l d}^{2}$

Calculating $\sum x_{n e w}^{2}$

Finding the Correct Variance ( $β$ )

Calculating $β$