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Animated Solution for Mathematics - Statistics: The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is

Q: Is the number of observations ($n$) changing?

No, the number of observations remains $n=10$. We are simply replacing one incorrect value with a correct one, so the total count of data points is constant.

Select Answer:

Visualized Solution

Identify Given Parameters

Number of observations $n = 10$
Incorrect Mean $\overset{x}{ˉ}_{o l d} = 20$
Incorrect Standard Deviation $σ_{o l d} = 8$
Incorrect value $= 50$ , Correct value $= 40$

Calculate Incorrect Sum of Observations

Mean formula: $\overset{x}{ˉ} = \frac{\sum x _{i}}{n}$
$\frac{\sum x _{in cor r ec t}}{10} = 20$
$\sum x_{in cor r ec t} = 20 \times 10 = 200$

Calculate Correct Sum

Correct Sum = Incorrect Sum - Incorrect Value + Correct Value
$\sum x_{cor r ec t} = 200 - 50 + 40 = 190$

Calculate Correct Mean

Correct Mean $\overset{x}{ˉ}_{n e w} = \frac{190}{10} = 19$

Define the Variance Formula

Variance formula: $σ^{2} = \frac{\sum x _{i}^{2}}{n} - (\overset{x}{ˉ})^{2}$
Incorrect Variance $σ_{o l d}^{2} = 8^{2} = 64$

Setup Equation for Incorrect Sum of Squares

$64 = \frac{\sum x _{i, in cor r ec t}^{2}}{10} - (20)^{2}$
$64 = \frac{\sum x _{i, in cor r ec t}^{2}}{10} - 400$

Calculate Incorrect Sum of Squares

$\frac{\sum x _{i, in cor r ec t}^{2}}{10} = 64 + 400 = 464$
$\sum x_{i, in cor r ec t}^{2} = 464 \times 10 = 4640$

Logic for Correcting Sum of Squares

Correct $\sum x^{2} = Incorrect \sum x^{2} - (Incorrect Value)^{2} + (Correct Value)^{2}$

Calculate Correct Sum of Squares

$\sum x_{i, cor r ec t}^{2} = 4640 - 5 0^{2} + 4 0^{2}$
$\sum x_{i, cor r ec t}^{2} = 4640 - 2500 + 1600 = 3740$

Setup Final Variance Calculation

Correct Variance $σ_{n e w}^{2} = \frac{\sum x _{i, cor r ec t}^{2}}{n} - (\overset{x}{ˉ}_{n e w})^{2}$
$σ_{n e w}^{2} = \frac{3740}{10} - (19)^{2}$

Calculate Final Variance

$σ_{n e w}^{2} = 374 - 361$
$σ_{n e w}^{2} = 13$
The correct variance is 13.

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The Sigma Insight: Measures of Dispersion

Analyzing the Setup

To solve this problem, we must recognize that the mean and variance are derived from two fundamental pillars: the sum of observations,

\sum x_{i}

, and the sum of squares of observations,

\sum x_{i}^{2}

We are given

n = 10

, an initial mean

\overset{x}{ˉ} = 20

, and an initial standard deviation

σ = 8

. One observation was incorrectly recorded as

50

instead of

40

Phase 1

Correcting the Mean

First, we calculate the incorrect sum of observations using the formula

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

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

Now, we perform the correction by subtracting the incorrect value and adding the correct one:

\sum x_{n e w} = 200 - 50 + 40 = 190

With the new sum, the correct mean is:

\overset{x}{ˉ}_{n e w} = \frac{190}{10} = 19

Phase 2

The Trap of the Sum of Squares

We must determine the sum of squares,

\sum x_{i}^{2}

, using the variance formula

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

. For the incorrect data:

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

Solving for the sum of squares:

\frac{\sum x _{o l d}^{2}}{10} = 64 + 400 = 464 \Rightarrow \sum x_{o l d}^{2} = 4640

Next, we apply the correction to the sum of squares by subtracting the square of the incorrect value and adding the square of the correct one:

\sum x_{n e w}^{2} = 4640 - 5 0^{2} + 4 0^{2}

\sum x_{n e w}^{2} = 4640 - 2500 + 1600 = 3740

Phase 3

The Final Victory

We now have the correct sum of squares,

3740

, and the correct mean,

19

. We plug these into the variance formula one last time:

σ_{n e w}^{2} = \frac{3740}{10} - (19)^{2}

This simplifies to:

σ_{n e w}^{2} = 374 - 361 = 13

By systematically correcting the raw sums, we have navigated the trap and arrived at the correct variance of $13$ . Remember, in JEE, it is about respecting the mathematical structure of the data.

Common Traps & FAQs

Wait, why can't I just adjust the variance directly by the difference between 50 and 40?

That is the classic trap! Variance is a non-linear statistic. It depends on the sum of squares of the observations, not just the values themselves. When you change an observation, the sum of squares changes by the difference of the squares of the values, not the difference of the values. Always go back to the raw sums!

Do I need to calculate the standard deviation or just the variance?

The question specifically asks for the correct variance. You do not need to take the square root at the end. Keep your answer as

σ^{2}

Is the number of observations (

n

) changing?

No, the number of observations remains

n = 10

. We are simply replacing one incorrect value with a correct one, so the total count of data points is constant.

Animated Solution for Mathematics - Statistics: The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is

Select Answer:

Visualized Solution

Identify Given Parameters

Calculate Incorrect Sum of Observations

Calculate Correct Sum

Calculate Correct Mean

Define the Variance Formula

Setup Equation for Incorrect Sum of Squares

Calculate Incorrect Sum of Squares

Logic for Correcting Sum of Squares

Calculate Correct Sum of Squares

Setup Final Variance Calculation

Calculate Final Variance

The Sigma Insight: Measures of Dispersion

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

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Animated Solution for Mathematics - Statistics: The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is

Select Answer:

Visualized Solution

Identify Given Parameters

Calculate Incorrect Sum of Observations

Calculate Correct Sum

Calculate Correct Mean

Define the Variance Formula

Setup Equation for Incorrect Sum of Squares

Calculate Incorrect Sum of Squares

Logic for Correcting Sum of Squares

Calculate Correct Sum of Squares

Setup Final Variance Calculation

Calculate Final Variance

The Sigma Insight: Measures of Dispersion

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

The mean and variance of 20 observations are found to be 10 and 4, respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was 11. Then the correct variance is:

The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is

The mean and variance of 20 observations are found to be 10 and 4, respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was 11. Then the correct variance is :

In calculating the mean and variance of 10 readings, a student wrongly used the figure 52 for the correct figure of 25. He obtained the mean and variance as 45.0 and 16.0 respectively. Determine the correct mean and variance.

The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12. If the new mean of the marks is 10.2, then their new variance is equal to :

The mean and variance of 8 observations are 10 and 13.5, respectively. If 6 of these observations are 5, 7, 10, 12, 14, 15, then the absolute difference of the remaining two observations is :

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