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 \underline\hspace2cm .

Statistics PYQ — JEE Main 2021 · 27 Aug · Shift 2 · Q40 with Solution

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Correct answer: 25

For boys:

For girls:

For all 50 candidates:

Using the combined mean formula:

\frac{20\cdot 12+30\cdot \mu}{50}=15

So,

240+30\mu=750

30\mu=510

\mu=17

For combined variance, use:

\sigma^2=\frac{n_1\left(\sigma_1^2+(\mu_1-\bar x)^2\right)+n_2\left(\sigma_2^2+(\mu_2-\bar x)^2\right)}{N}

Substitute the values:

\sigma^2=\frac{20\left(2+(12-15)^2\right)+30\left(2+(17-15)^2\right)}{50}

=\frac{20(2+9)+30(2+4)}{50}

=\frac{20\cdot 11+30\cdot 6}{50}

=\frac{220+180}{50}

=\frac{400}{50}=8

Thus,

\sigma^2=8

\mu+\sigma^2=17+8=25

\boxed{25}

View written solutionFree

Correct answer: 25

For boys:

For girls:

For all 50 candidates:

Using the combined mean formula:

\frac{20\cdot 12+30\cdot \mu}{50}=15

So,

240+30\mu=750

30\mu=510

\mu=17

For combined variance, use:

\sigma^2=\frac{n_1\left(\sigma_1^2+(\mu_1-\bar x)^2\right)+n_2\left(\sigma_2^2+(\mu_2-\bar x)^2\right)}{N}

Substitute the values:

\sigma^2=\frac{20\left(2+(12-15)^2\right)+30\left(2+(17-15)^2\right)}{50}

=\frac{20(2+9)+30(2+4)}{50}

=\frac{20\cdot 11+30\cdot 6}{50}

=\frac{220+180}{50}

=\frac{400}{50}=8

Thus,

\sigma^2=8

\mu+\sigma^2=17+8=25

\boxed{25}