JEE Main 2022 (26 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Statistics: The mean and standard deviation of $50$ observations are $15$ and $2$ respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is $70$ . If the correct mean is $16$ , then the correct variance is equal to :

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Visualized Solution

Identify Given Parameters

Number of observations: $n = 50$
Old mean: $\overset{x}{ˉ}_{o l d} = 15$
Old standard deviation: $σ_{o l d} = 2$
Let $x$ be the incorrect observation and $x^{'}$ be the correct one.
Given relation: $x + x^{'} = 70$

Calculate Old Sum of Observations

Sum of old observations: $\sum x_{o l d} = n \times \overset{x}{ˉ}_{o l d}$
$\sum x_{o l d} = 50 \times 15 = 750$

Calculate New Sum of Observations

Correct mean: $\overset{x}{ˉ}_{n e w} = 16$
Sum of new observations: $\sum x_{n e w} = n \times \overset{x}{ˉ}_{n e w}$
$\sum x_{n e w} = 50 \times 16 = 800$

Establish the System of Equations

Difference in sums: $\sum x_{n e w} - \sum x_{o l d} = x^{'} - x$
$x^{'} - x = 800 - 750 = 50$
Equation 1: $x^{'} + x = 70$
Equation 2: $x^{'} - x = 50$

Solve for Correct and Incorrect Observations

Adding the equations: $(x^{'} + x) + (x^{'} - x) = 70 + 50$
$2 x^{'} = 120 ⟹ x^{'} = 60$
Substitute $x^{'} = 60$ into $x^{'} + x = 70$ :
$60 + x = 70 ⟹ x = 10$

Old Variance Formula Setup

Old variance: $σ_{o l d}^{2} = 2^{2} = 4$
Formula: $σ_{o l d}^{2} = \frac{\sum x _{o l d}^{2}}{n} - \overset{x}{ˉ}_{o l d}^{2}$

Calculate Old Sum of Squares

$4 = \frac{\sum x _{o l d}^{2}}{50} - 1 5^{2}$
$4 = \frac{\sum x _{o l d}^{2}}{50} - 225$
$\frac{\sum x _{o l d}^{2}}{50} = 229$
$\sum x_{o l d}^{2} = 50 \times 229 = 11450$

Update Sum of Squares

Removing incorrect $x^{2}$ : $\sum x_{o l d}^{2} - x^{2} = 11450 - 1 0^{2} = 11350$
Adding correct $(x^{'})^{2}$ : $\sum x_{n e w}^{2} = 11350 + (60)^{2}$
$\sum x_{n e w}^{2} = 11350 + 3600 = 14950$

Calculate New Variance

New variance: $σ_{n e w}^{2} = \frac{\sum x _{n e w}^{2}}{n} - \overset{x}{ˉ}_{n e w}^{2}$
First term: $\frac{14950}{50} = 299$
Second term: $\overset{x}{ˉ}_{n e w}^{2} = 1 6^{2} = 256$
Correct variance: $σ_{n e w}^{2} = 299 - 256 = 43$

The Way Forward

Key Takeaway:
Update $\sum x$ to find individual values of $x$ and $x^{'}$ .
Update $\sum x^{2}$ to find the new variance using the corrected values.

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The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

We begin with a dataset of

n = 50

observations. The initial mean is given as

\overset{x}{ˉ}_{o l d} = 15

The total sum of these observations is calculated using the definition

\overset{x}{ˉ} = \frac{\sum x}{n}

. Therefore, the old sum is:

\sum x_{o l d} = 50 \times 15 = 750

Unmasking the Culprit

We are informed that one observation

x

was recorded incorrectly and should have been

x^{'}

. We are given the constraint

x^{'} + x = 70

After replacing the incorrect value, the new mean becomes

\overset{x}{ˉ}_{n e w} = 16

. The new sum is:

\sum x_{n e w} = 50 \times 16 = 800

The difference between the new sum and the old sum is

800 - 750 = 50

. This difference represents the net change:

x^{'} - x = 50

We now solve the system of linear equations: 1)

x^{'} + x = 70

x^{'} - x = 50

Adding these equations yields

2 x^{'} = 120

, which gives

x^{'} = 60

. Substituting this back, we find the incorrect value

x = 10

The Energy of the Data

To find the variance, we utilize the formula:

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

Given the old variance

σ_{o l d}^{2} = 2^{2} = 4

, we substitute the known values:

4 = \frac{\sum x _{o l d}^{2}}{50} - 1 5^{2}

This simplifies to

4 = \frac{\sum x _{o l d}^{2}}{50} - 225

. Solving for the sum of squares, we obtain:

\sum x_{o l d}^{2} = 50 \times 229 = 11450

The Final Restoration

We now perform the data correction by removing the contribution of the incorrect value and adding the correct one:

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

Substituting our values:

\sum x_{n e w}^{2} = 11450 - 1 0^{2} + 6 0^{2} = 11450 - 100 + 3600 = 14950

Finally, we calculate the new variance using the corrected sum of squares and the new mean:

σ_{n e w}^{2} = \frac{14950}{50} - 1 6^{2}

The first term simplifies to

299

, and the second term is

256

. Subtracting these gives:

σ_{n e w}^{2} = 299 - 256 = 43

The correct variance is 43.

Common Traps & FAQs

Why do we need to find the sum of squares? Can't we just adjust the variance directly?

Variance is a measure of spread, and it is inherently tied to the sum of squares of the data points. Because the variance formula involves the square of the mean, it is not a linear operator. You cannot simply add or subtract a value from the variance. You must reconstruct the 'sum of squares'—the raw energy of the data—to perform the correction accurately.

Is there a risk of getting the signs wrong when updating the sum of squares?

Absolutely! This is the most common trap. Always think of it as a 'ledger'. You are removing the 'bad' data point (subtract its square) and adding the 'good' data point (add its square). If you treat it like a bank account, you will never lose your way.

Why does the mean change so much?

The mean is sensitive to outliers. By replacing a small value (10) with a much larger one (60), you are significantly shifting the 'balance point' of your data set. This is a perfect example of how one single data point can influence the statistical properties of an entire collection.

Select Answer:

Visualized Solution

Identify Given Parameters

Calculate Old Sum of Observations

Calculate New Sum of Observations

Establish the System of Equations

Solve for Correct and Incorrect Observations

Old Variance Formula Setup

Calculate Old Sum of Squares

Update Sum of Squares

Calculate New Variance

The Way Forward

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

Unmasking the Culprit

The Energy of the Data

The Final Restoration

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