JEE Main 2022 (26 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Statistics: The mean and standard deviation of 40 observations are 30 and 5 respectively. It was noticed that two of these observations 12 and 10 were wrongly recorded. If $σ$ is the standard deviation of the data after omitting the two wrong observations from the data, then $38 σ^{2}$ is equal to

Enter Numerical Value:

Progressive Hints

Hint 1

The Spark: Visualize the dataset as a collection of 40 points. The core trap is realizing that variance is not just about the mean, but about the sum of squares. You must isolate the original sum of observations and the original sum of squares before you can even think about removing the incorrect data.

Hint 2

The Arsenal: You need the fundamental relationship between variance, the mean, and the sum of squares:

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

. Remember that when you remove data, both the sum of observations

\sum x_{i}

and the sum of squares

\sum x_{i}^{2}

change, and the count

n

decreases.

Hint 3

The Execution: Start by calculating the original sum

\sum x_{i} = n \times \overset{x}{ˉ}

and the original sum of squares

\sum x_{i}^{2} = n (σ^{2} + (\overset{x}{ˉ})^{2})

. Once you have these, subtract the values of the wrong observations (12 and 10) from the sum, and subtract their squares (

1 2^{2}

and

1 0^{2}

) from the sum of squares. Finally, update

n

to 38 and apply the variance formula.

Visualized Solution

Understanding the Initial Data

Given: Number of observations $n = 40$
Original Mean $\overset{x}{ˉ} = 30$
Original Standard Deviation $σ = 5$
Wrong observations to be omitted: $12$ and $10$

Defining the Mean Formula $\overset{x}{ˉ}$

The formula for the mean is: $\overset{x}{ˉ} = \frac{\sum x _{i}}{n}$
Where $\sum x_{i}$ is the sum of all observations.

Calculating the Original Sum

Substitute the known values: $30 = \frac{\sum x _{i}}{40}$
Calculate the sum: $\sum x_{i} = 30 \times 40 = 1200$

Defining the Variance Formula $σ^{2}$

The formula for variance is: $σ^{2} = \frac{\sum x _{i}^{2}}{n} - (\overset{x}{ˉ})^{2}$
Given standard deviation $σ = 5$ , so variance $σ^{2} = 25$ .

Setting up the Variance Equation

Substitute values: $25 = \frac{\sum x _{i}^{2}}{40} - (30)^{2}$
Rearrange: $25 + 900 = \frac{\sum x _{i}^{2}}{40}$
So, $925 = \frac{\sum x _{i}^{2}}{40}$

Calculating Original Sum of Squares

Calculate the sum of squares: $\sum x_{i}^{2} = 925 \times 40 = 37000$

Removing the Wrong Observations

New number of observations: $n^{'} = 40 - 2 = 38$
Observations to remove: $12$ and $10$

Calculating the New Sum $\sum x_{i}^{'}$

New sum: $\sum x_{i}^{'} = 1200 - 12 - 10$
$\sum x_{i}^{'} = 1178$

Calculating the New Mean $\overset{x}{ˉ}^{'}$

New mean: $\overset{x}{ˉ}^{'} = \frac{1178}{38}$
$\overset{x}{ˉ}^{'} = 31$

Calculating New Sum of Squares

New sum of squares: $\sum x_{i}^{'2} = 37000 - (12)^{2} - (10)^{2}$
$\sum x_{i}^{'2} = 37000 - 144 - 100 = 36756$

Setting up the New Variance

New variance: $σ^{2} = \frac{\sum x _{i}^{'2}}{38} - (\overset{x}{ˉ}^{'})^{2}$
Multiply by 38: $38 σ^{2} = \sum x_{i}^{'2} - 38 (\overset{x}{ˉ}^{'})^{2}$

Final Calculation for $38 σ^{2}$

Substitute values: $38 σ^{2} = 36756 - 38 (31)^{2}$
$38 σ^{2} = 36756 - 38 (961)$
$38 σ^{2} = 36756 - 36518 = 238$

Conclusion and Key Takeaway

Key Takeaway: Always update the count $n$ when observations are omitted or added.
Final Result: $38 σ^{2} = 238$
Next Challenge: What if the observations were replaced instead of omitted? How would the denominator change?

00:00 / 00:00

The Sigma Insight: Variance and Standard Deviation

Solution Diagram

Analyzing the Setup

In the world of statistics, data is often prone to errors. We are given a dataset of

n = 40

observations with a mean

\overset{x}{ˉ} = 30

and a standard deviation

σ = 5

Our objective is to remove two incorrect observations,

12

and

10

, and calculate the new variance of the remaining

38

observations.

The Detective Work

Uncovering the Hidden Sums

We utilize the fundamental formulas for mean and variance:

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

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

From the mean formula, we find the total sum of the observations:

\sum x_{i} = 30 \times 40 = 1200

Next, we determine the original sum of squares using the variance formula where

σ^{2} = 25

25 = \frac{\sum x _{i}^{2}}{40} - (30)^{2}

\frac{\sum x _{i}^{2}}{40} = 25 + 900 = 925

\sum x_{i}^{2} = 37000

The Cleanup

Surgical Removal

We now remove the imposters

12

and

10

. The new count of observations is

n^{'} = 38

The new sum of observations is:

\sum x_{i}^{'} = 1200 - 12 - 10 = 1178

The new mean

\overset{x}{ˉ}^{'}

is:

\overset{x}{ˉ}^{'} = \frac{1178}{38} = 31

We must also update the sum of squares by subtracting the squares of the removed values:

\sum x_{i}^{'2} = 37000 - 1 2^{2} - 1 0^{2}

\sum x_{i}^{'2} = 37000 - 144 - 100 = 36756

The Final Stretch

Calculating the Result

We are tasked with finding the value of

38 σ^{2}

. Using the variance formula for the updated dataset:

σ^{2} = \frac{\sum x _{i}^{'2}}{38} - (\overset{x}{ˉ}^{'})^{2}

Multiplying by

38

, we isolate the term:

38 σ^{2} = \sum x_{i}^{'2} - 38 (\overset{x}{ˉ}^{'})^{2}

Substituting our calculated values:

38 σ^{2} = 36756 - 38 (31)^{2}

38 σ^{2} = 36756 - 38 (961)

38 σ^{2} = 36756 - 36518

The final result is: $38 σ^{2} = 238$

Common Traps & FAQs

Wait, why do we subtract

1 2^{2}

and

1 0^{2}

instead of just

12

and

10

when calculating the new sum of squares?

That is a fantastic question! Remember, the variance formula relies on the sum of the *squares* of the observations,

\sum x_{i}^{2}

. When you remove an observation

x_{k}

, you are removing its contribution to the total sum of squares, which is

x_{k}^{2}

. If you only subtracted the value itself, you would be ignoring the geometric nature of variance, which measures the spread based on squared deviations. Always treat the sum of squares as a distinct entity from the sum of observations.

Can I just calculate the new mean and then find the variance from scratch using the remaining 38 numbers?

Technically, yes, but you don't know the individual values of the other 38 observations! You only know the aggregate properties (mean and standard deviation) of the original 40. The algebraic method we used is the only way to solve this because it allows us to manipulate the aggregate sums without needing to know the specific values of every single data point.

What if the question asked for the new standard deviation instead of

38 σ^{2}

Great intuition! If the question asked for the standard deviation

σ

, you would simply take the square root of the variance you calculated. In this case, you would find

σ^{2} = \frac{238}{38}

, and then

σ = \frac{238}{38}

. Always check what the final expression asks for—sometimes, like here, it asks for a multiple of the variance to keep the numbers clean and avoid messy square roots until the very end.

Enter Numerical Value:

Visualized Solution

Understanding the Initial Data

Defining the Mean Formula $\overset{x}{ˉ}$

Calculating the Original Sum

Defining the Variance Formula $σ^{2}$

Setting up the Variance Equation

Calculating Original Sum of Squares

Removing the Wrong Observations

Calculating the New Sum $\sum x_{i}^{'}$

Calculating the New Mean $\overset{x}{ˉ}^{'}$

Calculating New Sum of Squares

Setting up the New Variance

Final Calculation for $38 σ^{2}$

Conclusion and Key Takeaway

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Detective Work

The Cleanup

The Final Stretch

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 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 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 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:

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 set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and $σ^{2}$ respectively. If the variance of all the 30 numbers in the two sets is 13, then $σ^{2}$ is equal to

Enter Numerical Value:

Visualized Solution

Understanding the Initial Data

Defining the Mean Formula .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xˉ

Calculating the Original Sum

Defining the Variance Formula .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } σ2

Setting up the Variance Equation

Calculating Original Sum of Squares

Removing the Wrong Observations

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

Calculating the New Mean .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xˉ′

Calculating New Sum of Squares

Setting up the New Variance

Final Calculation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 38σ2

Conclusion and Key Takeaway

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Detective Work

The Cleanup

The Final Stretch

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

Defining the Mean Formula $\overset{x}{ˉ}$

Defining the Variance Formula $σ^{2}$

Calculating the New Sum $\sum x_{i}^{'}$

Calculating the New Mean $\overset{x}{ˉ}^{'}$

Final Calculation for $38 σ^{2}$