Is it necessary to calculate the exact value of $\sigma^2$ before plugging it into the final expression?

Absolutely not! In JEE Advanced, the final expression is often a trap designed to test your algebraic foresight. By keeping $\sigma^2$ as an expression involving $\mu^2$, you allow the terms to cancel out, saving time and preventing calculation errors.

JEE Main 2024 (27 Jan Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Statistics: The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12. If $μ$ and $σ^{2}$ denote the mean and variance of the correct observations respectively, then $15 (μ + μ^{2} + σ^{2})$ is equal to

Enter Numerical Value:

Progressive Hints

Hint 1

Think of the mean and variance not as static numbers, but as aggregate properties derived from the sum of observations and the sum of their squares. When one data point changes, you do not need to re-examine the entire set; you only need to update these two fundamental aggregates: the sum and the sum of squares.

Hint 2

Recall the fundamental definitions:

μ = \frac{\sum x _{i}}{n}

and

σ^{2} = \frac{\sum x _{i}^{2}}{n} - μ^{2}

. Your toolkit for this problem is the 'correction' logic:

\sum x_{cor r} = \sum x_{in c} - x_{w r o n g} + x_{r i g h t}

and

\sum x_{cor r}^{2} = \sum x_{in c}^{2} - (x_{w r o n g})^{2} + (x_{r i g h t})^{2}

. Use these to find the true parameters.

Hint 3

Before you start calculating decimals, look at the final expression:

15 (μ + μ^{2} + σ^{2})

. Substitute the expression for

σ^{2}

in terms of the sum of squares and

μ^{2}

. You will find that the

μ^{2}

terms cancel out beautifully, leaving you with a much simpler calculation.

Visualized Solution

Initial Data Analysis

Number of observations $n = 15$
Incorrect mean $μ^{'} = 12$
Incorrect standard deviation $σ^{'} = 3$
Incorrect observation $= 10$
Correct observation $= 12$

Formula for Mean

$μ = \frac{\sum x _{i}}{n}$
Sum of observations $\sum x_{i} = n \times μ$

Calculating Incorrect $\sum x_{i}$

Incorrect sum $\sum x_{in c} = 15 \times 12$
$\sum x_{in c} = 180$

Correcting the Sum $\sum x_{i}$

Correct sum $\sum x_{cor r} = \sum x_{in c} - 10 + 12$
$\sum x_{cor r} = 180 - 10 + 12 = 182$
Correct mean $μ = \frac{182}{15}$

Formula for Variance

Variance $σ^{2} = \frac{\sum x _{i}^{2}}{n} - μ^{2}$
We need to correct the sum of squares $\sum x_{i}^{2}$

Calculating Incorrect $\sum x_{i}^{2}$

Incorrect variance $(σ^{'})^{2} = 3^{2} = 9$
$9 = \frac{\sum x _{in c}^{2}}{15} - 1 2^{2}$
$\frac{\sum x _{in c}^{2}}{15} = 9 + 144 = 153$
$\sum x_{in c}^{2} = 153 \times 15 = 2295$

Correcting the Sum of Squares $\sum x_{i}^{2}$

Correct sum of squares $\sum x_{cor r}^{2} = \sum x_{in c}^{2} - 1 0^{2} + 1 2^{2}$
$\sum x_{cor r}^{2} = 2295 - 100 + 144$
$\sum x_{cor r}^{2} = 2339$

Defining Correct Variance $σ^{2}$

Correct variance $σ^{2} = \frac{\sum x _{cor r}^{2}}{15} - μ^{2}$
$σ^{2} = \frac{2339}{15} - μ^{2}$

Setting up the Final Expression

Required value $= 15 (μ + μ^{2} + σ^{2})$

Substituting Variance

Substitute $σ^{2} = \frac{2339}{15} - μ^{2}$
Expression: $15 (μ + μ^{2} + \frac{2339}{15} - μ^{2})$

Simplifying the Expression

The $μ^{2}$ terms cancel out: $μ^{2} - μ^{2} = 0$
Expression becomes: $15 (μ + \frac{2339}{15})$
Expanding the bracket: $15 μ + 2339$

Final Atomic Computation

Substitute $μ = \frac{182}{15}$
$= 15 (\frac{182}{15}) + 2339$
$= 182 + 2339$
$= 2521$

Summary and Takeaway

Key Takeaway: Always correct $\sum x_{i}$ and $\sum x_{i}^{2}$ first.
Calculation Tip: Keep fractions until the final step to allow for cancellations.
Final Answer: $2521$

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The Sigma Insight: Variance and Standard Deviation

Solution Diagram

Analyzing the Dataset DNA

In this scenario, we are dealing with a dataset of

n = 15

observations. The initial mean is

μ_{in c} = 12

and the initial variance is

σ_{in c}^{2} = 9

The mean is defined as the sum of observations divided by

n

, while the variance is given by the relation:

σ^{2} = \frac{\sum x ^{2}}{n} - μ^{2}

Surgical Correction of the Sum

First, we calculate the incorrect sum of observations:

\sum x_{in c} = 15 \times 12 = 180

We perform the surgical correction by subtracting the incorrect value (

10

) and adding the correct value (

12

\sum x_{cor r} = 180 - 10 + 12 = 182

The corrected mean is therefore

μ = \frac{182}{15}

Surgical Correction of the Sum of Squares

Using the variance formula, we extract the incorrect sum of squares:

9 = \frac{\sum x _{in c}^{2}}{15} - 1 2^{2}

\sum x_{in c}^{2} = 15 (9 + 144) = 15 (153) = 2295

We correct this by subtracting the square of the wrong value (

1 0^{2} = 100

) and adding the square of the right value (

1 2^{2} = 144

\sum x_{cor r}^{2} = 2295 - 100 + 144 = 2339

The Master Equation

The problem asks us to evaluate the expression

15 (μ + μ^{2} + σ^{2})

. By substituting the definition of variance

σ^{2} = \frac{\sum x ^{2}}{n} - μ^{2}

, the expression simplifies elegantly:

15 (μ + μ^{2} + (\frac{\sum x ^{2}}{15} - μ^{2}))

The

μ^{2}

terms cancel out, leaving us with:

15 (μ + \frac{\sum x ^{2}}{15}) = 15 μ + \sum x^{2}

Final Calculation

Substituting our corrected values

μ = \frac{182}{15}

and

\sum x^{2} = 2339

into the simplified expression:

15 (\frac{182}{15}) + 2339

182 + 2339 = 2521

The final result of the calculation is 2521.

Common Traps & FAQs

Why can't I just adjust the mean and variance directly without finding the sums?

Because mean and variance are not linear operators. You cannot simply add or subtract from the mean or variance to correct a single data point. You must revert to the 'sum of observations' and 'sum of squares' to perform the correction, as these are the fundamental building blocks of the statistical parameters.

Is it necessary to calculate the exact value of

σ^{2}

before plugging it into the final expression?

Absolutely not! In JEE Advanced, the final expression is often a trap designed to test your algebraic foresight. By keeping

σ^{2}

as an expression involving

μ^{2}

, you allow the terms to cancel out, saving time and preventing calculation errors.

I keep getting the wrong answer. Did I mess up the sum of squares correction?

A common pitfall is subtracting the wrong value instead of the square of the wrong value. Remember, variance is about the spread of the squares of the data. Always square the observation before modifying the sum of squares.

Enter Numerical Value:

Visualized Solution

Initial Data Analysis

Formula for Mean

Calculating Incorrect $\sum x_{i}$

Correcting the Sum $\sum x_{i}$

Formula for Variance

Calculating Incorrect $\sum x_{i}^{2}$

Correcting the Sum of Squares $\sum x_{i}^{2}$

Defining Correct Variance $σ^{2}$

Setting up the Final Expression

Substituting Variance

Simplifying the Expression

Final Atomic Computation

Summary and Takeaway

The Sigma Insight: Variance and Standard Deviation

Analyzing the Dataset DNA

Surgical Correction of the Sum

Surgical Correction of the Sum of Squares

The Master Equation

Final Calculation

Similar Questions

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 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

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

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 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

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 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 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 $50$ 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 :

Enter Numerical Value:

Visualized Solution

Initial Data Analysis

Formula for Mean

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

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

Formula for Variance

Calculating Incorrect .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑xi2​

Correcting the Sum of Squares .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑xi2​

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

Setting up the Final Expression

Substituting Variance

Simplifying the Expression

Final Atomic Computation

Summary and Takeaway

The Sigma Insight: Variance and Standard Deviation

Analyzing the Dataset DNA

Surgical Correction of the Sum

Surgical Correction of the Sum of Squares

The Master Equation

Final Calculation

Similar Questions

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 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

Calculating Incorrect $\sum x_{i}$

Correcting the Sum $\sum x_{i}$

Calculating Incorrect $\sum x_{i}^{2}$

Correcting the Sum of Squares $\sum x_{i}^{2}$

Defining Correct Variance $σ^{2}$