JEE Main 2020 (8 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Statistics: 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:

Select Answer:

Visualized Solution

Initial Parameters: $n, \overset{x}{ˉ}, σ^{2}$

Number of observations ( $n$ ) = $20$
Given Mean ( $\overset{x}{ˉ}_{o l d}$ ) = $10$
Given Variance ( $σ_{o l d}^{2}$ ) = $4$

Calculate Old Sum: $\sum x_{i, o l d}$

Formula: $\overset{x}{ˉ} = \frac{\sum x _{i}}{n}$
$\sum x_{i, o l d} = n \times \overset{x}{ˉ}_{o l d}$
$\sum x_{i, o l d} = 20 \times 10 = 200$

Calculate Old Sum of Squares: $\sum x_{i, o l d}^{2}$

Formula: $σ^{2} = \frac{\sum x _{i}^{2}}{n} - (\overset{x}{ˉ})^{2}$
Substitute values: $4 = \frac{\sum x _{i, o l d}^{2}}{20} - (10)^{2}$
$\frac{\sum x _{i, o l d}^{2}}{20} = 4 + 100 = 104$
$\sum x_{i, o l d}^{2} = 104 \times 20 = 2080$

Identify the Error: $x_{w r o n g} \to x_{r i g h t}$

Incorrect observation ( $x_{w r o n g}$ ) = $9$
Correct observation ( $x_{r i g h t}$ ) = $11$

Calculate New Sum: $\sum x_{i, n e w}$

New Sum ( $\sum x_{i, n e w}$ ) = $\sum x_{i, o l d} - 9 + 11$
$\sum x_{i, n e w} = 200 - 9 + 11 = 202$

Calculate New Mean: $\overset{x}{ˉ}_{n e w}$

New Mean ( $\overset{x}{ˉ}_{n e w}$ ) = $\frac{202}{20} = 10.1$

Calculate New Sum of Squares: $\sum x_{i, n e w}^{2}$

New Sum of Squares ( $\sum x_{i, n e w}^{2}$ ) = $\sum x_{i, o l d}^{2} - (9)^{2} + (11)^{2}$
$\sum x_{i, n e w}^{2} = 2080 - 81 + 121$
$\sum x_{i, n e w}^{2} = 2080 + 40 = 2120$

Setup Correct Variance: $σ_{n e w}^{2}$

New Variance ( $σ_{n e w}^{2}$ ) = $\frac{\sum x _{i, n e w}^{2}}{n} - (\overset{x}{ˉ}_{n e w})^{2}$
$σ_{n e w}^{2} = \frac{2120}{20} - (10.1)^{2}$

Final Computation: $σ_{n e w}^{2} = 3.99$

$σ_{n e w}^{2} = 106 - 102.01$
$σ_{n e w}^{2} = 3.99$
Correct Variance = $3.99$

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The Sigma Insight: Measures of Dispersion

Solution Diagram

Analyzing the Setup

To correct our statistics, we must first understand the fundamental definitions. The mean,

\overset{x}{ˉ}

, is the center of gravity of our data, defined as:

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

The variance,

σ^{2}

, is the measure of spread, defined as:

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

We are given

n = 20

\overset{x}{ˉ}_{o l d} = 10

, and

σ_{o l d}^{2} = 4

. Before we can fix the error, we need the raw totals.

Deconstructing the Data

The old sum is calculated as:

\sum x_{i, o l d} = n \cdot \overset{x}{ˉ}_{o l d} = 20 \cdot 10 = 200

To find the sum of squares, we rearrange the variance formula:

\sum x_{i, o l d}^{2} = n (σ^{2} + \overset{x}{ˉ}^{2}) = 20 (4 + 1 0^{2}) = 20 (104) = 2080

We now have our raw materials: a sum of

200

and a sum of squares of

2080

Surgical Precision

We perform the correction by removing the incorrect value

9

and inserting the correct value

11

. Our new sum becomes:

\sum x_{i, n e w} = 200 - 9 + 11 = 202

The new mean is:

\overset{x}{ˉ}_{n e w} = \frac{202}{20} = 10.1

The variance is more sensitive, so we must update the sum of squares by subtracting the square of the wrong value and adding the square of the right value:

\sum x_{i, n e w}^{2} = 2080 - 9^{2} + 1 1^{2} = 2080 - 81 + 121 = 2120

The Final Reveal

With our new sum of squares,

\sum x_{i, n e w}^{2} = 2120

, and our new mean,

\overset{x}{ˉ}_{n e w} = 10.1

, we calculate the correct variance using the formula:

σ_{n e w}^{2} = \frac{\sum x _{i, n e w}^{2}}{n} - (\overset{x}{ˉ}_{n e w})^{2}

Substituting the values:

σ_{n e w}^{2} = \frac{2120}{20} - (10.1)^{2}

This simplifies to:

σ_{n e w}^{2} = 106 - 102.01 = 3.99

The corrected variance is $3.99$ .

Common Traps & FAQs

Wait, why do we have to calculate the sum of squares? Can't we just adjust the variance directly?

That is a common trap! Variance is a non-linear measure. It depends on the sum of squares, not just the mean. Because the variance formula involves squaring the observations, you must update the sum of squares first to maintain mathematical integrity.

Why do we subtract

9^{2}

and add

1 1^{2}

instead of just adding

(11 - 9)^{2}

Great question! The variance formula uses the sum of the squares of the individual observations,

\sum x_{i}^{2}

. It does not use the square of the sum. Therefore, we must remove the square of the incorrect value and add the square of the correct value to the total sum of squares.

Is the number of observations

n

always constant?

In this specific problem, yes. We are replacing one value with another, so the total count

n = 20

remains unchanged. Always check if the problem implies adding or removing a data point, which would change

n

Animated Solution for Mathematics - Statistics: 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:

Select Answer:

Visualized Solution

Initial Parameters: $n, \overset{x}{ˉ}, σ^{2}$

Calculate Old Sum: $\sum x_{i, o l d}$

Calculate Old Sum of Squares: $\sum x_{i, o l d}^{2}$

Identify the Error: $x_{w r o n g} \to x_{r i g h t}$

Calculate New Sum: $\sum x_{i, n e w}$

Calculate New Mean: $\overset{x}{ˉ}_{n e w}$

Calculate New Sum of Squares: $\sum x_{i, n e w}^{2}$

Setup Correct Variance: $σ_{n e w}^{2}$

Final Computation: $σ_{n e w}^{2} = 3.99$

The Sigma Insight: Measures of Dispersion

Analyzing the Setup

Deconstructing the Data

Surgical Precision

The Final Reveal

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 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

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

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.

In an experiment with 15 observations on x, the following results were available: $\sum x^{2} = 2830, \sum x = 170$ . One observation that was 20 was found to be wrong and was replaced by the correct value 30. The corrected variance is

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 10 observations are 9 and 34.2, respectively. If 8 of these observations are $2, 3, 5, 10, 11, 13, 15, 21$ , then the mean deviation about the median of all the 10 observations is

If the mean and variance of five observations are $\frac{24}{5}$ and $\frac{194}{25}$ respectively and the mean of first four observations is $\frac{7}{2}$ , then the variance of the first four observations in equal to

The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are 3, 4 and 4; then the absolute value of the difference of the other two observations, is :

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 :

Animated Solution for Mathematics - Statistics: 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:

Select Answer:

Visualized Solution

Initial Parameters: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n,xˉ,σ2

Calculate Old Sum: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑xi,old​

Calculate Old Sum of Squares: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑xi,old2​

Identify the Error: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xwrong​→xright​

Calculate New Sum: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑xi,new​

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

Calculate New Sum of Squares: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑xi,new2​

Setup Correct Variance: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } σnew2​

Final Computation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } σnew2​=3.99

The Sigma Insight: Measures of Dispersion

Analyzing the Setup

Deconstructing the Data

Surgical Precision

The Final Reveal

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 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

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

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 the variance of five observations are 4 and 5.20, respectively. If three of the observations are 3, 4 and 4; then the absolute value of the difference of the other two observations, is :

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 :

Initial Parameters: $n, \overset{x}{ˉ}, σ^{2}$

Calculate Old Sum: $\sum x_{i, o l d}$

Calculate Old Sum of Squares: $\sum x_{i, o l d}^{2}$

Identify the Error: $x_{w r o n g} \to x_{r i g h t}$

Calculate New Sum: $\sum x_{i, n e w}$

Calculate New Mean: $\overset{x}{ˉ}_{n e w}$

Calculate New Sum of Squares: $\sum x_{i, n e w}^{2}$

Setup Correct Variance: $σ_{n e w}^{2}$

Final Computation: $σ_{n e w}^{2} = 3.99$