The mean and standard deviation of 100 observations are 40 and 5.1, respectively. By mistake one obs

Q: Why can't we just subtract the difference from the standard deviation directly?

Standard deviation is not a linear measure like the mean. It is derived from the variance, which depends on the sum of squares—a non-linear operation. Because the variance is the average of the squared deviations from the mean, you must go back to the raw sum of squares to perform the correction accurately.

Q: Why do we need to calculate the sum of squares?

The variance formula $Var(X) = \frac{\sum x_i^2}{n} - (\bar{x})^2$ is the bridge between the mean and the spread. Without the sum of squares, you cannot account for how the individual values contribute to the variance. It is the only way to 'undo' the mistake and 'redo' the calculation with the correct value.

Q: Is there a shortcut for this?

While there are algebraic shortcuts for specific types of errors, for JEE Advanced, it is safer and more reliable to follow the fundamental steps: correct the sum, correct the mean, correct the sum of squares, and finally, correct the variance. This systematic approach prevents errors and ensures you understand the underlying statistics.

Question

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 $\mu$ and $\sigma$ respectively, then $10(\mu + \sigma)$ is equal to

Options:

(A) 445

(B) 451

(C) 447

(D) 449

Sigma Percentile · Accepted Answer

### The Statistical Correction: A Journey of Precision

Welcome, future engineer. Today, we are not just solving a statistics problem; we are performing a surgical correction on a dataset.

Imagine you are a data scientist who has just discovered a flaw in a report. You have $n = 100$ observations, a mean of $\bar{x} = 40$, and a standard deviation of $s = 5.1$.

There is a ghost in the machine: one value was recorded as $50$ when it should have been $40$. Let us walk through the restoration process step-by-step.

### Phase 1: The Mean Correction

The mean, $\bar{x}$, is the center of gravity of your data. It is defined by the following relation:

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

With $n = 100$ and $\bar{x} = 40$, the total sum of your observations is $\sum x_i = 100 	imes 40 = 4000$.

We have a mistake: one value is $50$, but it should be $40$. The correction is intuitive; we subtract the error and add the truth:

$$
\sum x_{new} = 4000 - 50 + 40 = 3990
$$

Now, the correct mean $\mu$ is simply:

$$
\mu = \frac{3990}{100} = 39.9
$$

We have successfully shifted our center of gravity.

### Phase 2: The Variance Trap

The standard deviation is a measure of spread based on the sum of squares. We use the variance formula:

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

Our incorrect variance is $s^2 = (5.1)^2 = 26.01$. We must find the incorrect sum of squares, $\sum x_{old}^2$.

Rearranging the formula, we get:

$$
\frac{\sum x_{old}^2}{100} = 26.01 + 40^2
$$

This leads us to:

$$
\sum x_{old}^2 = 100 	imes (26.01 + 1600) = 162601
$$

This number represents the total energy of the spread in our incorrect dataset.

### Phase 3: The Final Restoration

Just as we corrected the sum, we must correct the sum of squares. We remove the contribution of the wrong value and add the contribution of the correct one:

$$
\sum x_{new}^2 = 162601 - 50^2 + 40^2
$$

Calculating this, we get $162601 - 2500 + 1600 = 161701$. Now, we find the correct variance $\sigma^2$:

$$
\sigma^2 = \frac{161701}{100} - (39.9)^2
$$

This simplifies to $1617.01 - 1592.01 = 25$. The correct standard deviation $\sigma$ is the square root of $25$, which is $5$.

Finally, the question asks for $10(\mu + \sigma)$. Plugging in our values:

$$
10(39.9 + 5) = 10(44.9) = 449
$$

You have navigated the trap, corrected the data, and arrived at the truth. The final result is **449**.

Select Answer:

Visualized Solution

Given Data and Error Identification

Visualizing the Correction

Calculating the Incorrect Sum

Correcting the Sum of Observations

Calculating Correct Mean $μ$

The Variance Formula

Calculating Incorrect Sum of Squares

Correcting the Sum of Squares

Calculating Correct Variance $σ^{2}$

Calculating Correct Standard Deviation $σ$

Final Value of $10 (μ + σ)$

The Sigma Insight: Variance and Standard Deviation

The Statistical Correction

Phase 1

Phase 2

Phase 3

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