The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechec

Q: Wait, why do we use the old mean when calculating the old sum of squares? Shouldn't we use the new mean?

That is a fantastic question! Think of it this way: the old variance was calculated using the old, incorrect data. Therefore, the old sum of squares must be consistent with the old mean. We are essentially 'reverse-engineering' the original dataset to find the true sum of squares before we apply the correction. If you use the new mean with the old variance, you will get a completely wrong sum of squares.

Q: Why can't I just adjust the standard deviation directly? Like, add the difference between the correct and wrong values?

I understand the temptation to look for a shortcut! However, standard deviation is not a linear operator. It is derived from the variance, which is a measure of the *average squared deviation* from the mean. Because the mean itself changes when you correct an observation, and because the variance involves squaring the values, you cannot simply add or subtract a constant. You must go back to the fundamental sums: the sum of observations and the sum of squares.

Q: Is there a way to avoid calculating the sum of squares? It feels like a lot of extra work.

In statistics, the sum of squares is the 'DNA' of the variance. Without it, you cannot reconstruct the variance after a data point changes. While it feels like extra work, it is the most robust and error-proof method. Think of it as the price of accuracy. Once you master this process, you will never be tricked by these 'correction' problems again!

Question

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

Options:

(A) 1.8

(B) 1.94

(C) $\sqrt{3.96}$

(D) $\sqrt{3.86}$

Sigma Percentile · Accepted Answer

### The Data Detective: Mastering Statistical Corrections

Welcome, future engineers! Today, we are stepping into the shoes of a data detective. In the world of JEE Advanced, statistics problems often test your ability to handle data integrity.

You are given a set of observations, a mean, and a standard deviation, but there is a catch: a mistake was made during data entry. Your mission is to recover the truth.

### Phase 1: The Anatomy of the Error

We are given $n = 20$ observations. The initial, incorrect mean is $\bar{x}_{old} = 10$, and the incorrect standard deviation is $\sigma_{old} = 2$.

We know that one observation was recorded as $x_{wrong} = 8$, but it should have been $x_{correct} = 12$. To fix this, we must first understand the relationship between the mean and the sum of observations.

The mean is defined as:

$$ \bar{x} = \frac{\sum x_i}{n} $$

Therefore, the total sum is simply $\sum x_{i, old} = n 	imes \bar{x}_{old}$. With $n = 20$ and $\bar{x}_{old} = 10$, our incorrect sum is $200$.

To correct this, we perform a simple surgery:

$$ \sum x_{i, new} = \sum x_{i, old} - x_{wrong} + x_{correct} = 200 - 8 + 12 = 204 $$

Now, our new mean is:

$$ \bar{x}_{new} = \frac{204}{20} = 10.2 $$

We have successfully corrected the mean!

### Phase 2: The Hidden Engine

Now, here is where most students stumble. They think correcting the mean is enough. But the standard deviation is a measure of spread, and it relies on the sum of squares, $\sum x_i^2$.

The variance formula is our guiding light:

$$ \sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 $$

We need to find the correct $\sum x_i^2$. First, let's find the old sum of squares using the incorrect parameters. We know $\sigma_{old}^2 = 2^2 = 4$.

Plugging this into the formula:

$$ 4 = \frac{\sum x_{i, old}^2}{20} - (10)^2 $$

Solving for the old sum of squares:

$$ 4 + 100 = \frac{\sum x_{i, old}^2}{20} \Rightarrow \sum x_{i, old}^2 = 104 	imes 20 = 2080 $$

This is the 'hidden' sum of squares that was used to calculate the incorrect variance.

### Phase 3: The Final Synthesis

Now that we have the old sum of squares, we can correct it just like we did with the simple sum. We subtract the square of the wrong value and add the square of the correct value:

$$ \sum x_{i, new}^2 = \sum x_{i, old}^2 - (x_{wrong})^2 + (x_{correct})^2 $$

Substituting our values:

$$ \sum x_{i, new}^2 = 2080 - 8^2 + 12^2 = 2080 - 64 + 144 = 2160 $$

We are almost there! We now have the correct sum of squares and the correct mean. Let's calculate the new variance:

$$ \sigma_{new}^2 = \frac{\sum x_{i, new}^2}{n} - (\bar{x}_{new})^2 = \frac{2160}{20} - (10.2)^2 $$

Calculating this:

$$ 108 - 104.04 = 3.96 $$

Finally, the standard deviation is the square root of the variance:

$$ \sigma_{new} = \sqrt{3.96} $$

### Conclusion: The Precision Mindset

Look at the elegance of that result! By carefully separating the correction of the sum and the sum of squares, we navigated the trap perfectly.

Remember, in JEE Advanced, the difference between a good score and a great score is often this kind of attention to detail. You have now mastered the art of statistical correction.

Keep practicing, stay curious, and keep pushing the boundaries of your understanding!

Animated Solution for Mathematics - Statistics: 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

Select Answer:

Visualized Solution

Given Parameters

Formula for Mean

Calculate Incorrect Sum

Correct the Sum

Calculate Correct Mean

The Variance Formula

Find Old Sum of Squares

Solve for $\sum x_{i, o l d}^{2}$

Correct the Sum of Squares

Setup New Variance

Calculate New Variance

Final Correct Standard Deviation

The Sigma Insight: Variance and Standard Deviation

The Data Detective

Phase 1

Phase 2

Phase 3

Conclusion

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