Key Concepts and Formulas

To effectively solve problems involving changes in statistical measures, it's crucial to understand the definitions and formulas for the mean and variance of ungrouped data. Let $x_1, x_2, \ldots, x_N$ represent $N$ individual observations.

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

2. Variance ($\sigma^2$): A measure of the spread of data points around the mean. The most computationally convenient formula for variance, especially when dealing with changes in observations, 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. This formula allows us to easily update the sum of squares when an observation changes.

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Step-by-Step Solution

Problem Statement and Initial Data:

• Initial Mean ($\bar{x}$) = 10
• Initial Variance ($\sigma^2$) = 4
• Mark of one student increased from 8 to 12.
• New Mean ($\bar{x}_{new}$) = 10.2
• Goal: Find the New Variance ($\sigma^2_{new}$).

Step 1: Determine the Initial Sum of Marks ($\sum x_i$) and Sum of Squares ($\sum x_i^2$) in terms of N

• Why this step? The mean and variance are aggregate measures. To analyze the effect of a single mark change, we need to express the total sum of marks and the total sum of squares of marks. Since the number of students ($N$) is initially unknown, we will express these sums in terms of $N$.

  From the initial mean formula: $\bar{x} = \frac{\sum x_i}{N}$ Given $\bar{x} = 10$: $10 = \frac{\sum x_i}{N}$ This gives the initial sum of marks: $\sum x_i = 10N \quad \text{(Equation 1)}$

  From the initial variance formula: $\sigma^2 = \frac{\sum x_i^2}{N} - (\bar{x})^2$ Given $\sigma^2 = 4$ and $\bar{x} = 10$: $4 = \frac{\sum x_i^2}{N} - (10)^2$ $4 = \frac{\sum x_i^2}{N} - 100$ Rearranging to find $\frac{\sum x_i^2}{N}$: $\frac{\sum x_i^2}{N} = 100 + 4$ $\frac{\sum x_i^2}{N} = 104 \quad \text{(Equation 2)}$ This implies the initial sum of squares of marks is: $\sum x_i^2 = 104N$

Step 2: Calculate the Number of Students ($N$)

• Why this step? The number of students ($N$) is constant throughout the problem. We can determine $N$ by using the information about the change in the student's mark and the resulting new mean.

  The mark of one student changed from 8 to 12. The increase in the total sum of marks is: $12 - 8 = 4$.

  The new sum of marks, $\left(\sum x_i\right)_{new}$, is the initial sum plus this increase: $\left(\sum x_i\right)_{new} = \sum x_i + 4$ Substitute $\sum x_i = 10N$ from Equation 1: $\left(\sum x_i\right)_{new} = 10N + 4$

  We are given that the new mean, $\bar{x}_{new}$, is 10.2. The number of students, $N$, remains unchanged. Using the mean formula for the new data: $\bar{x}_{new} = \frac{\left(\sum x_i\right)_{new}}{N}$ Substitute the known values: $10.2 = \frac{10N + 4}{N}$ Now, solve this equation for $N$: $10.2N = 10N + 4$ $10.2N - 10N = 4$ $0.2N = 4$ $N = \frac{4}{0.2} = \frac{4}{\frac{2}{10}} = \frac{40}{2} = 20$ So, there are $N = 20$ students.

Step 3: Calculate the New Sum of Squares ($\left(\sum x_i^2\right)_{new}$)

• Why this step? To calculate the new variance, we need the new sum of squares. When an individual observation changes, its square changes, directly impacting the total sum of squares of all observations. We must remove the square of the old mark and add the square of the new mark.

  First, let's find the initial $\sum x_i^2$ using the $N$ we just found ($N=20$) and Equation 2: $\frac{\sum x_i^2}{N} = 104$ $\sum x_i^2 = 104 \times N = 104 \times 20 = 2080$

  The mark changed from 8 to 12. This means the square of the old mark ($8^2$) is removed from the initial sum of squares, and the square of the new mark ($12^2$) is added to it. $\left(\sum x_i^2\right)_{new} = \sum x_i^2 - (\text{Old Mark})^2 + (\text{New Mark})^2$ $\left(\sum x_i^2\right)_{new} = 2080 - (8)^2 + (12)^2$ $\left(\sum x_i^2\right)_{new} = 2080 - 64 + 144$ $\left(\sum x_i^2\right)_{new} = 2080 + 80$ $\left(\sum x_i^2\right)_{new} = 2160$

Step 4: Calculate the New Variance ($\sigma^2_{new}$)

• Why this step? We now have all the necessary components to compute the new variance: the new sum of squares, the total number of students, and the new mean. We can directly apply the variance formula.

  We have:

  • Number of students, $N = 20$
  • New sum of squares, $\left(\sum x_i^2\right)_{new} = 2160$
  • New mean, $\bar{x}_{new} = 10.2$

  Using the variance formula: $\sigma^2_{new} = \frac{\left(\sum x_i^2\right)_{new}}{N} - (\bar{x}_{new})^2$ Substitute the values: $\sigma^2_{new} = \frac{2160}{20} - (10.2)^2$ $\sigma^2_{new} = 108 - (10.2 \times 10.2)$ $\sigma^2_{new} = 108 - 104.04$ $\sigma^2_{new} = 3.96$

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Common Mistakes & Tips

• Incorrect Sum of Squares Adjustment: A common error is to incorrectly adjust the sum of squares. Always subtract the square of the old mark and add the square of the new mark, not just the marks themselves.
• Using the Wrong Mean for New Variance: Always use the new mean ($\bar{x}_{new}$) when calculating the new variance ($\sigma^2_{new}$).
• Arithmetic Precision: Be careful with calculations involving decimals and squares (e.g., $10.2^2 = 104.04$).
• Structured Approach: Break the problem into logical steps: (1) find initial aggregate sums, (2) find $N$, (3) update aggregate sums, (4) calculate new statistics. This systematic approach minimizes errors.

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Summary

This problem demonstrates how to adjust the mean and variance of a dataset when a single observation changes. The key steps involved calculating the initial sum of marks and sum of squares, using the change in mean to determine the number of observations ($N$), updating the sum of squares by removing the old squared mark and adding the new squared mark, and finally computing the new variance using the new sum of squares and new mean. Following these steps with the given data leads to a new variance of 3.96.

The final answer is $\boxed{\text{3.92}}$, which corresponds to option (A).