JEE Main 2023 (08 April Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Statistics: Let the mean and variance of 12 observations be $\frac{9}{2}$ and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is $\frac{m}{n}$ , where $m$ and $n$ are coprime, then $m + n$ is equal to

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the data as a collection of points. The mean is the balance point, and the variance is the spread. When you replace incorrect values with correct ones, you are essentially shifting the balance point and changing the spread. You must first find the total sum of the original observations to identify the error.

Hint 2

Hint 2: To find the correct variance, you need two key quantities: the sum of the observations and the sum of the squares of the observations. Use the incorrect mean and variance to reverse-engineer the incorrect sum of squares, then apply the correction:

\sum x_{n e w}^{2} = \sum x_{o l d}^{2} - (wrong values)^{2} + (correct values)^{2}

Hint 3

Hint 3: Once you have the correct sum of observations and the correct sum of squares, use the formula

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

to calculate the new variance. Remember to use the updated mean

\overset{x}{ˉ}_{n e w} = \frac{\sum x _{n e w}}{n}

Visualized Solution

Understanding the Problem

Given: $n = 12$ observations.
Incorrect Mean $(\overset{x}{ˉ}_{o l d}) = \frac{9}{2}$ .
Incorrect Variance $(σ_{o l d}^{2}) = 4$ .
Incorrect values recorded: $9, 10$ .
Actual values required: $7, 14$ .

Calculating the Incorrect Sum $\sum x_{i}$

Mean formula: $\overset{x}{ˉ} = \frac{\sum x _{i}}{n}$
$\Rightarrow \sum x_{o l d} = n \times \overset{x}{ˉ}_{o l d}$
$\Rightarrow \sum x_{o l d} = 12 \times \frac{9}{2} = 54$

Correcting the Sum of Observations

Correct Sum $(\sum x_{n e w}) = \sum x_{o l d} - (wrong values) + (correct values)$
$\sum x_{n e w} = 54 - (9 + 10) + (7 + 14)$
$\sum x_{n e w} = 54 - 19 + 21 = 56$

Finding the Correct Mean $\overset{x}{ˉ}_{n e w}$

Correct Mean $(\overset{x}{ˉ}_{n e w}) = \frac{\sum x _{n e w}}{n}$
$\overset{x}{ˉ}_{n e w} = \frac{56}{12} = \frac{14}{3}$

The Variance Formula Strategy

Variance formula: $σ^{2} = \frac{\sum x _{i}^{2}}{n} - (\overset{x}{ˉ})^{2}$
We need to find $\sum x_{o l d}^{2}$ first.

Calculating Incorrect Sum of Squares $\sum x_{o l d}^{2}$

$4 = \frac{\sum x _{o l d}^{2}}{12} - (\frac{9}{2})^{2}$
$4 = \frac{\sum x _{o l d}^{2}}{12} - \frac{81}{4}$
$\frac{\sum x _{o l d}^{2}}{12} = 4 + \frac{81}{4} = \frac{16 + 81}{4} = \frac{97}{4}$
$\sum x_{o l d}^{2} = 12 \times \frac{97}{4} = 3 \times 97 = 291$

Correcting the Sum of Squares

Correct Sum of Squares $(\sum x_{n e w}^{2}) = \sum x_{o l d}^{2} - (9^{2} + 1 0^{2}) + (7^{2} + 1 4^{2})$
$\sum x_{n e w}^{2} = 291 - (81 + 100) + (49 + 196)$
$\sum x_{n e w}^{2} = 291 - 181 + 245 = 110 + 245 = 355$

Setting up the Correct Variance

Correct Variance $(σ_{n e w}^{2}) = \frac{\sum x _{n e w}^{2}}{n} - (\overset{x}{ˉ}_{n e w})^{2}$
$σ_{n e w}^{2} = \frac{355}{12} - (\frac{14}{3})^{2}$
$σ_{n e w}^{2} = \frac{355}{12} - \frac{196}{9}$

Fraction Arithmetic for Variance

LCM of $12$ and $9$ is $36$ .
$σ_{n e w}^{2} = \frac{355 \times 3 - 196 \times 4}{36}$
$σ_{n e w}^{2} = \frac{1065 - 784}{36} = \frac{281}{36}$

Final Answer: $m + n$

Given: $σ_{n e w}^{2} = \frac{m}{n}$
$m = 281, n = 36$
Check: $gcd (281, 36) = 1$ (Coprime)
$m + n = 281 + 36 = 317$

00:00 / 00:00

The Sigma Insight: Variance and Standard Deviation

Solution Diagram

Analyzing the Data Integrity

We are given

n = 12

observations with an incorrect mean

\overset{x}{ˉ}_{o l d} = \frac{9}{2}

and an incorrect variance

σ_{o l d}^{2} = 4

. The two incorrect observations are

9

and

10

, while the correct values are

7

and

14

The first step is to recover the original sum of the data. Since

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

, we calculate the total sum as:

\sum x_{o l d} = 12 \times \frac{9}{2} = 54

Now, we perform the correction by removing the erroneous values and adding the correct ones:

\sum x_{n e w} = 54 - (9 + 10) + (7 + 14) = 54 - 19 + 21 = 56

The correct mean is therefore:

\overset{x}{ˉ}_{n e w} = \frac{56}{12} = \frac{14}{3}

The Hidden Complexity of Variance

To find the correct variance, we must first determine the sum of squares. We use the variance formula

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

to reverse-engineer the incorrect sum of squares:

4 = \frac{\sum x _{o l d}^{2}}{12} - (\frac{9}{2})^{2}

Solving for

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

\frac{\sum x _{o l d}^{2}}{12} = 4 + \frac{81}{4} = \frac{97}{4} \Rightarrow \sum x_{o l d}^{2} = 12 \times \frac{97}{4} = 291

Next, we calculate the correct sum of squares by adjusting for the specific values:

\sum x_{n e w}^{2} = \sum x_{o l d}^{2} - (9^{2} + 1 0^{2}) + (7^{2} + 1 4^{2})

\sum x_{n e w}^{2} = 291 - (81 + 100) + (49 + 196) = 291 - 181 + 245 = 355

Final Calculation

We now substitute the correct sum of squares and the correct mean into the variance formula:

σ_{n e w}^{2} = \frac{355}{12} - (\frac{14}{3})^{2} = \frac{355}{12} - \frac{196}{9}

To subtract these, we use the least common multiple of

12

and

9

, which is

36

σ_{n e w}^{2} = \frac{355 \times 3 - 196 \times 4}{36} = \frac{1065 - 784}{36} = \frac{281}{36}

Given that the variance is expressed as

\frac{m}{n}

where

m

and

n

are coprime, we identify

m = 281

and

n = 36

The final result is:

$m + n = 281 + 36 = 317$

Common Traps & FAQs

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

That is a common trap! Variance is not a linear property; it depends on the sum of squares, not just the sum of the values. Because the variance formula

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

involves the square of the mean and the sum of squares, you must calculate both the new sum and the new sum of squares independently to find the correct variance.

Why do we subtract the squares of the wrong values and add the squares of the correct ones?

Think of it as cleaning up your data. The incorrect sum of squares

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

includes the squares of the wrong values (

9^{2}

and

1 0^{2}

). To remove their influence, you subtract them. Then, to include the correct data, you add the squares of the actual values (

7^{2}

and

1 4^{2}

). This leaves you with the true sum of squares for the dataset.

How do I know if

m

and

n

are coprime?

Two numbers are coprime if their greatest common divisor (GCD) is 1. In this problem, after calculating the variance as

\frac{281}{36}

, you check if 281 and 36 share any factors. Since 281 is a prime number and not a factor of 36, their GCD is indeed 1, confirming they are coprime.

Select Answer:

Visualized Solution

Understanding the Problem

Calculating the Incorrect Sum $\sum x_{i}$

Correcting the Sum of Observations

Finding the Correct Mean $\overset{x}{ˉ}_{n e w}$

The Variance Formula Strategy

Calculating Incorrect Sum of Squares $\sum x_{o l d}^{2}$

Correcting the Sum of Squares

Setting up the Correct Variance

Fraction Arithmetic for Variance

Final Answer: $m + n$

The Sigma Insight: Variance and Standard Deviation

Analyzing the Data Integrity

The Hidden Complexity of Variance

Final Calculation

Similar Questions

The mean and standard deviation of $15$ observations are found to be $8$ and $3$ respectively. On rechecking it was found that, in the observations, $20$ was misread as $5$ . Then, the correct variance is equal to _______.

The mean and variance of a data of 10 observations are 10 and 2, respectively. If an observations $α$ in this data is replaced by $β$ , then the mean and variance become 10.1 and 1.99, respectively. Then $α + β$ equals

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

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 :

The mean and standard deviation of 100 observations are 40 and 5.1, respectively. By mistake one observation is taken as 50 instead of 40. If the correct mean and the correct standard deviation are $μ$ and $σ$ respectively, then $10 (μ + σ)$ is equal to

The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted and $a$ and $b$ are respectively mean and variance of remaining 6 observation, then $a + 3 b - 5$ is equal to ______.

The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is

Select Answer:

Visualized Solution

Understanding the Problem

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

Correcting the Sum of Observations

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

The Variance Formula Strategy

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

Correcting the Sum of Squares

Setting up the Correct Variance

Fraction Arithmetic for Variance

Final Answer: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m+n

The Sigma Insight: Variance and Standard Deviation

Analyzing the Data Integrity

The Hidden Complexity of Variance

Final Calculation

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

The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is

Calculating the Incorrect Sum $\sum x_{i}$

Finding the Correct Mean $\overset{x}{ˉ}_{n e w}$

Calculating Incorrect Sum of Squares $\sum x_{o l d}^{2}$

Final Answer: $m + n$