Key Concepts and Formulas

This problem requires us to work with the fundamental concepts of mean and variance, specifically how to combine these statistics for two separate groups into a single overall group. Understanding these formulas is crucial for solving problems involving grouped data.

1. Combined Mean Formula: When we have two distinct groups, say Group 1 with $n_1$ observations and an average (mean) of $\overline{X}_1$, and Group 2 with $n_2$ observations and an average of $\overline{X}_2$, the overall average (combined mean, $\overline{X}$) of all $N = n_1 + n_2$ observations is calculated as a weighted average: $\overline{X} = \frac{n_1 \overline{X}_1 + n_2 \overline{X}_2}{n_1 + n_2}$ This formula makes intuitive sense: the total sum of all observations is divided by the total number of observations. The numerator ($n_1 \overline{X}_1 + n_2 \overline{X}_2$) represents the sum of marks for Group 1 plus the sum of marks for Group 2.

2. Combined Variance Formula: If we have two groups with $n_1$ observations, mean $\overline{X}_1$, and variance $\sigma_1^2$, and $n_2$ observations, mean $\overline{X}_2$, and variance $\sigma_2^2$, then the combined variance ($\sigma^2$) of the total $N = n_1 + n_2$ observations is given by: $\sigma^2 = \frac{n_1 \sigma_1^2 + n_2 \sigma_2^2}{n_1 + n_2} + \frac{n_1 n_2}{(n_1 + n_2)^2} (\overline{X}_1 - \overline{X}_2)^2$ This formula is more complex because variance measures the spread of data. It has two main components:

   • The first term, $\frac{n_1 \sigma_1^2 + n_2 \sigma_2^2}{n_1 + n_2}$, represents the average of the variances within each group, weighted by their respective sizes. This accounts for the inherent variability within each group.
   • The second term, $\frac{n_1 n_2}{(n_1 + n_2)^2} (\overline{X}_1 - \overline{X}_2)^2$, accounts for the variability arising from the difference between the means of the two groups. If the group means are very different, this term will be large, indicating greater overall spread. If the group means are identical, this term becomes zero. This term is often called the variance due to the difference in means.

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Step-by-Step Solution

Let's organize the given information and define variables clearly for systematic problem-solving:

• Total number of candidates, $N = 50$.

• Let's denote boys as Group 1 and girls as Group 2.

• For Boys (Group 1):

  • Number of boys, $n_1 = 20$.
  • Average marks of boys, $\overline{X}_1 = 12$.
  • Variance of marks of boys, $\sigma_1^2 = 2$.

• For Girls (Group 2):

  • Number of girls, $n_2 = 30$.
  • Variance of marks of girls, $\sigma_2^2 = 2$.
  • Average marks of girls, $\mu = \overline{X}_2$ (this is what we need to find).

• For All Candidates (Combined Group):

  • Total number of candidates, $N = n_1 + n_2 = 20 + 30 = 50$.
  • Average marks of all 50 candidates, $\overline{X} = 15$.
  • Variance of marks of all 50 candidates, $\sigma^2$ (this is what we need to find).

Our ultimate goal is to calculate the value of $\mu + \sigma^2$.

Step 1: Determine the average marks of girls ($\mu$)

• What we are doing: We need to find the average marks obtained by girls, which we have denoted as $\mu$ or $\overline{X}_2$.
• Why we are doing it: The average marks of girls are unknown but required for calculating the overall variance in a later step and for the final sum. We can determine this using the given overall average marks and the statistics for the boys.
• Method: We will use the Combined Mean Formula.
• Calculations: The combined mean formula is: $\overline{X} = \frac{n_1 \overline{X}_1 + n_2 \overline{X}_2}{n_1 + n_2}$ Substitute the known values into the formula: $15 = \frac{(20 \times 12) + (30 \times \overline{X}_2)}{20 + 30}$ First, perform the multiplications in the numerator and the sum in the denominator: $15 = \frac{240 + 30 \overline{X}_2}{50}$ To isolate the term containing $\overline{X}_2$, multiply both sides of the equation by 50: $15 \times 50 = 240 + 30 \overline{X}_2$ $750 = 240 + 30 \overline{X}_2$ Next, subtract 240 from both sides to gather the constant terms: $750 - 240 = 30 \overline{X}_2$ $510 = 30 \overline{X}_2$ Finally, divide by 30 to solve for $\overline{X}_2$: $\overline{X}_2 = \frac{510}{30}$ $\overline{X}_2 = 17$ Thus, the average marks of girls, $\mu = 17$.

Step 2: Calculate the variance of marks of all 50 candidates ($\sigma^2$)

• What we are doing: We need to find the variance of marks for all 50 candidates combined, which is denoted as $\sigma^2$.

• Why we are doing it: This is one of the quantities required for the final sum $\mu + \sigma^2$. We now have all the necessary individual group statistics (number of observations, mean, and variance) to apply the combined variance formula.

• Method: We will use the Combined Variance Formula.

• Calculations: The combined variance formula is: $\sigma^2 = \frac{n_1 \sigma_1^2 + n_2 \sigma_2^2}{n_1 + n_2} + \frac{n_1 n_2}{(n_1 + n_2)^2} (\overline{X}_1 - \overline{X}_2)^2$ Substitute the known values into the formula:

  • $n_1 = 20, \sigma_1^2 = 2, \overline{X}_1 = 12$
  • $n_2 = 30, \sigma_2^2 = 2, \overline{X}_2 = 17$ (from Step 1)
  • $n_1 + n_2 = 50$

  $\sigma^2 = \frac{(20 \times 2) + (30 \times 2)}{20 + 30} + \frac{20 \times 30}{(20 + 30)^2} (12 - 17)^2$

  Let's calculate each term separately for clarity:

  First Term (Weighted average of within-group variances): $\frac{(20 \times 2) + (30 \times 2)}{50} = \frac{40 + 60}{50} = \frac{100}{50} = 2$ This term quantifies the average variability within the boys' and girls' groups.

  Second Term (Variance due to difference in group means): $\frac{20 \times 30}{(50)^2} (12 - 17)^2$ Calculate the components:

  • Denominator: $(50)^2 = 2500$.
  • Numerator part: $20 \times 30 = 600$.
  • Difference in means squared: $(12 - 17)^2 = (-5)^2 = 25$. Substitute these values back: $\frac{600}{2500} \times 25$ We can simplify this expression. Notice that $2500 = 100 \times 25$: $\frac{600}{100 \times 25} \times 25 = \frac{600}{100} = 6$ This term accounts for the additional spread in the overall data because the average marks of boys and girls are different.

  Combine the two terms: Now, add the results of the first and second terms to get the total combined variance: $\sigma^2 = 2 + 6$ $\sigma^2 = 8$ Thus, the variance of marks of all 50 candidates, $\sigma^2 = 8$.

Step 3: Calculate the final value $\mu + \sigma^2$

• What we are doing: We are computing the sum of the average marks of girls ($\mu$) and the variance of marks of all 50 candidates ($\sigma^2$).
• Why we are doing it: This is the final value requested by the problem statement.
• Calculations: Using the values obtained in Step 1 and Step 2: $\mu = 17$ $\sigma^2 = 8$ $\mu + \sigma^2 = 17 + 8$ $\mu + \sigma^2 = 25$

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Common Mistakes & Tips

1. Omitting the Second Term in Combined Variance: A very common error is to assume the combined variance is simply the weighted average of individual variances ($\frac{n_1 \sigma_1^2 + n_2 \sigma_2^2}{n_1 + n_2}$). This is only true if the means of the two groups are identical. Always remember the second term, $\frac{n_1 n_2}{(n_1 + n_2)^2} (\overline{X}_1 - \overline{X}_2)^2$, which accounts for the variability due to differing group means.
2. Algebraic Errors: Pay close attention to arithmetic, especially when squaring negative numbers (e.g., $(-5)^2 = 25$, not $-25$) and simplifying fractions. Double-check each step of calculation.
3. Incorrect Variable Assignment: Ensure you correctly identify and substitute values for $n_1, \overline{X}_1, \sigma_1^2, n_2, \overline{X}_2, \sigma_2^2$, and the overall $\overline{X}$. A clear initial setup helps prevent this.

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Summary

This problem provided a comprehensive application of combined statistics, requiring us to merge data from two distinct groups (boys and girls) to find overall statistics. We first utilized the combined mean formula to determine the average marks of girls ($\mu = 17$). Subsequently, we applied the combined variance formula, meticulously calculating both the within-group variance component and the between-group variance component, to find the overall variance of marks for all 50 candidates ($\sigma^2 = 8$). Finally, we summed these two calculated values to arrive at $\mu + \sigma^2 = 25$. This problem highlights the importance of understanding and correctly applying both components of the combined variance formula.

The final answer is \boxed{25}.