Question:
An online exam is attempted by 50 candidates out of which 20 are boys. The average marks obtained by boys is 12 with a variance 2. The variance of marks obtained by 30 girls is also 2. The average marks of all 50 candidates is 15. If \(\mu\) is the average marks of girls and \(\sigma^2\) is the variance of marks of 50 candidates, then \(\mu + \sigma^2\) is equal to _________.
An online exam is attempted by 50 candidates out of which 20 are boys. The average marks obtained by boys is 12 with a variance 2. The variance of marks obtained by 30 girls is also 2. The average marks of all 50 candidates is 15. If \(\mu\) is the average marks of girls and \(\sigma^2\) is the variance of marks of 50 candidates, then \(\mu + \sigma^2\) is equal to _________.
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The combined variance formula is essential for this type of problem. Remember that it's not a simple weighted average of the individual variances. It includes terms (\(d^2\)) that account for the spread between the means of the subgroups and the overall combined mean.
Updated On: Feb 16, 2026
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Verified By Collegedunia
Correct Answer: 25
Solution and Explanation
Step 1: Calculate \(\mu\), the average marks of girls.
Let \(n_B=20\) be the number of boys and \(n_G=30\) be the number of girls. Let \(\bar{x}_B=12\) be the boys' average and \(\bar{x}_G=\mu\) be the girls' average. The combined average \(\bar{x}_C=15\). Using the formula for combined mean: \[ \bar{x}_C = \frac{n_B \bar{x}_B + n_G \bar{x}_G}{n_B + n_G} \] \[ 15 = \frac{20(12) + 30(\mu)}{20 + 30} \] \[ 15 = \frac{240 + 30\mu}{50} \] \[ 750 = 240 + 30\mu \] \[ 510 = 30\mu \implies \mu = 17 \] Step 2: Calculate \(\sigma^2\), the combined variance.
The formula for the combined variance of two groups is: \[ \sigma^2 = \frac{n_B(\sigma_B^2 + d_B^2) + n_G(\sigma_G^2 + d_G^2)}{n_B+n_G} \] where \(\sigma_B^2=2\) and \(\sigma_G^2=2\) are the variances of the groups, and \(d_B, d_G\) are the deviations of the group means from the combined mean. - \(d_B = \bar{x}_B - \bar{x}_C = 12 - 15 = -3\), so \(d_B^2 = 9\). - \(d_G = \bar{x}_G - \bar{x}_C = \mu - 15 = 17 - 15 = 2\), so \(d_G^2 = 4\). Now substitute all the values into the formula: \[ \sigma^2 = \frac{20(2 + 9) + 30(2 + 4)}{50} \] \[ \sigma^2 = \frac{20(11) + 30(6)}{50} \] \[ \sigma^2 = \frac{220 + 180}{50} = \frac{400}{50} = 8 \] Step 3: Calculate the final value \(\mu + \sigma^2\).
\[ \mu + \sigma^2 = 17 + 8 = 25 \]
Let \(n_B=20\) be the number of boys and \(n_G=30\) be the number of girls. Let \(\bar{x}_B=12\) be the boys' average and \(\bar{x}_G=\mu\) be the girls' average. The combined average \(\bar{x}_C=15\). Using the formula for combined mean: \[ \bar{x}_C = \frac{n_B \bar{x}_B + n_G \bar{x}_G}{n_B + n_G} \] \[ 15 = \frac{20(12) + 30(\mu)}{20 + 30} \] \[ 15 = \frac{240 + 30\mu}{50} \] \[ 750 = 240 + 30\mu \] \[ 510 = 30\mu \implies \mu = 17 \] Step 2: Calculate \(\sigma^2\), the combined variance.
The formula for the combined variance of two groups is: \[ \sigma^2 = \frac{n_B(\sigma_B^2 + d_B^2) + n_G(\sigma_G^2 + d_G^2)}{n_B+n_G} \] where \(\sigma_B^2=2\) and \(\sigma_G^2=2\) are the variances of the groups, and \(d_B, d_G\) are the deviations of the group means from the combined mean. - \(d_B = \bar{x}_B - \bar{x}_C = 12 - 15 = -3\), so \(d_B^2 = 9\). - \(d_G = \bar{x}_G - \bar{x}_C = \mu - 15 = 17 - 15 = 2\), so \(d_G^2 = 4\). Now substitute all the values into the formula: \[ \sigma^2 = \frac{20(2 + 9) + 30(2 + 4)}{50} \] \[ \sigma^2 = \frac{20(11) + 30(6)}{50} \] \[ \sigma^2 = \frac{220 + 180}{50} = \frac{400}{50} = 8 \] Step 3: Calculate the final value \(\mu + \sigma^2\).
\[ \mu + \sigma^2 = 17 + 8 = 25 \]
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