This problem requires us to adjust the mean and variance of a dataset after a recording error is identified. We'll use the fundamental formulas for mean and variance to first determine the original sum and sum of squares of observations, then correct these sums, and finally calculate the new (correct) mean and variance.

1. Key Concepts and Formulas

   For a set of $n$ observations, $x_1, x_2, \ldots, x_n$:

   • Mean ($\bar{x}$): The average of the observations. $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$ where $\sum x_i$ is the sum of all observations.

   • Variance ($\sigma^2$): A measure of the spread of data points around the mean. The most convenient computational formula is: $\sigma^2 = \frac{\sum_{i=1}^{n} x_i^2}{n} - (\bar{x})^2$ where $\sum x_i^2$ is the sum of the squares of all observations.

   • Standard Deviation ($\sigma$): The positive square root of the variance. $\sigma = \sqrt{\sigma^2}$

2. Step-by-Step Solution

   We are given:

   • Number of observations, $n = 10$
   • Original Mean, $\bar{x}_{\text{orig}} = 20$
   • Original Standard Deviation, $\sigma_{\text{orig}} = 8$
   • Original Variance, $\sigma_{\text{orig}}^2 = 8^2 = 64$
   • Incorrectly recorded observation = 50
   • Correct observation = 40

   Step 1: Calculate the Original Sum of Observations ($\sum x_i^{\text{orig}}$)

   • Why this step? The mean is defined by the sum of observations. To correct for an error in one observation, we first need to know the total sum that was calculated using the incorrect value. This sum will be our basis for correction.
   • Using the formula for the mean, $\bar{x} = \frac{\sum x_i}{n}$, we can find the sum: $\sum x_i^{\text{orig}} = n \times \bar{x}_{\text{orig}}$
   • Substituting the given values: $\sum x_i^{\text{orig}} = 10 \times 20 = 200$

   Step 2: Calculate the Original Sum of Squares of Observations ($\sum (x_i^2)^{\text{orig}}$)

   • Why this step? The variance formula requires the sum of squares of observations. Similar to the sum of observations, we need the original sum of squares (calculated with the incorrect value) to adjust it for the correction.
   • Using the formula for variance, $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$, we can find the sum of squares: $\sum x_i^2 = n \left( \sigma^2 + (\bar{x})^2 \right)$
   • Substituting the original given values ($\sigma_{\text{orig}}^2 = 64$ and $\bar{x}_{\text{orig}} = 20$): $\sum (x_i^2)^{\text{orig}} = 10 \left( 64 + (20)^2 \right)$ $\sum (x_i^2)^{\text{orig}} = 10 (64 + 400)$ $\sum (x_i^2)^{\text{orig}} = 10 \times 464 = 4640$

   Step 3: Calculate the Correct Sum of Observations ($\sum x_i^{\text{correct}}$)

   • Why this step? We need to update the total sum of observations by removing the incorrect value and adding the correct value.
   • The formula for the corrected sum is: $\sum x_i^{\text{correct}} = \sum x_i^{\text{orig}} - \text{Incorrect Value} + \text{Correct Value}$
   • Substituting the values: $\sum x_i^{\text{correct}} = 200 - 50 + 40$ $\sum x_i^{\text{correct}} = 190$

   Step 4: Calculate the Correct Sum of Squares of Observations ($\sum (x_i^2)^{\text{correct}}$)

   • Why this step? Similarly, the sum of squares needs to be corrected. When we remove an observation, we subtract its square. When we add an observation, we add its square.
   • The formula for the corrected sum of squares is: $\sum (x_i^2)^{\text{correct}} = \sum (x_i^2)^{\text{orig}} - (\text{Incorrect Value})^2 + (\text{Correct Value})^2$
   • Substituting the values: $\sum (x_i^2)^{\text{correct}} = 4640 - (50)^2 + (40)^2$ $\sum (x_i^2)^{\text{correct}} = 4640 - 2500 + 1600$ $\sum (x_i^2)^{\text{correct}} = 2140 + 1600$ $\sum (x_i^2)^{\text{correct}} = 3740$

   Step 5: Calculate the Correct Mean ($\bar{x}_{\text{correct}}$)

   • Why this step? The correct variance depends on the correct mean. We must calculate the new mean using the corrected sum of observations.
   • Using the formula for mean: $\bar{x}_{\text{correct}} = \frac{\sum x_i^{\text{correct}}}{n}$
   • Substituting the corrected sum and number of observations: $\bar{x}_{\text{correct}} = \frac{190}{10}$ $\bar{x}_{\text{correct}} = 19$

   Step 6: Calculate the Correct Variance ($\sigma_{\text{correct}}^2$)

   • Why this step? This is our final goal. We now have all the necessary components: the correct sum of squares, the correct mean, and the number of observations.
   • Using the variance formula: $\sigma_{\text{correct}}^2 = \frac{\sum (x_i^2)^{\text{correct}}}{n} - (\bar{x}_{\text{correct}})^2$
   • Substituting the calculated values: $\sigma_{\text{correct}}^2 = \frac{3740}{10} - (19)^2$ $\sigma_{\text{correct}}^2 = 374 - 361$ $\sigma_{\text{correct}}^2 = 13$

3. Common Mistakes & Tips

   • Confusing Standard Deviation with Variance: Always remember that variance is the square of the standard deviation. A common mistake is to use the standard deviation value directly in the variance formula without squaring it.
   • Incorrectly Adjusting Sum of Squares: When an observation $x$ is removed or added, its square $x^2$ must be removed or added from the sum of squares, not just $x$.
   • Forgetting to Recalculate Mean: The mean changes when an observation is corrected. Always calculate the new (correct) mean before calculating the new variance, as variance depends on the mean.
   • Using $n-1$ for Variance: For JEE problems, unless specified as sample variance or an unbiased estimator, assume population variance (division by $n$).

4. Summary

   We systematically corrected the given statistical measures. First, we derived the original sum of observations and the original sum of squares using the initial mean and standard deviation. Then, we adjusted these sums by removing the incorrectly recorded value and adding the correct value. Finally, using these corrected sums, we calculated the new mean and then the correct variance of the observations. The correct variance was found to be 13.

5. Final Answer

   The final answer is $\boxed{13}$ which corresponds to option (C).