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  • Let sets A and B have 5 elements each. Let mean of the elements in sets A and B be 5 and 8 respectively and the variance of the elements in sets A and B be 12 and 20 respectively. A new set

Let sets A and B have 5 elements each. Let mean of the elements in sets A and B be 5 and 8 respectively and
the variance of the elements in sets A and B be 12 and 20 respectively. A new set C of 10 elements is formed
by subtracting 3 from each element of A and adding 2 to each element of B. Then the sum of the mean and
variance of the elements of C is ______.

Option: 1

36


Option: 2

40


Option: 3

32


Option: 4

38


Answers (1)

best_answer

\begin{aligned} & \omega \mathrm{A}=\left\{\mathrm{a}_1, \mathrm{a}_2, \mathrm{a}_3, \mathrm{a}_4, \mathrm{a}_5\right\} \\ & \mathrm{B}=\left\{\mathrm{b}_1, \mathrm{~b}_2, \mathrm{~b}_3, \mathrm{~b}_4, \mathrm{~b}_5\right\} \\ & \text { Given, } \sum_{\mathrm{i}=1}^5 \mathrm{ai}=25, \sum_{\mathrm{i}=1}^5 \mathrm{bi}=40 \\ & \frac{\sum_{\mathrm{i}=1}^5 \mathrm{a}_{\mathrm{i}}^2}{5}-\left(\frac{\sum_{i=1}^5 \mathrm{a}_{\mathrm{i}}}{5}\right)^2=12, \frac{\sum_{\mathrm{i}=1}^5 \mathrm{~b}_{\mathrm{i}}^2}{5}-\left(\frac{\sum_{\mathrm{i}=1}^5 \mathrm{~b}_{\mathrm{i}}}{5}\right)^2=20 \\ & \Rightarrow \sum_{\mathrm{i}=1}^5 \mathrm{a}_{\mathrm{i}}^2=185, \sum_{\mathrm{i}=1}^5 \mathrm{~b}_{\mathrm{i}}^2=420 \\ & \text { Now, } \mathrm{C}=\left\{\mathrm{C}_1, \mathrm{C}_2, \ldots \mathrm{C}_{10}\right\} \end{aligned}

\begin{aligned} &\begin{aligned} & \therefore \text { Mean of } \mathrm{C}, \overline{\mathrm{C}}=\frac{\left(\sum \mathrm{a}_{\mathrm{i}}-15\right)+\left(\sum \mathrm{b}_{\mathrm{i}}-10\right)}{10} \\ & \overline{\mathrm{C}}=\frac{10+50}{10}=6 \end{aligned}\\ &\begin{aligned} & \therefore \sigma^2=\frac{\sum_{i=1}^{10} C_i^2}{10}=(\overline{\mathrm{C}})^2 \\ & =\frac{\sum\left(a_i-3\right)^2+\sum\left(b_i-2\right)^2+}{10}-(6)^2 \\ & =\frac{\sum a_i^2+\sum b_i^2-6 \sum a_i+4 \sum b_i+65}{10}-36 \\ & =\frac{185+420-150+160+65}{10}-36 \\ & =32 \\ & \therefore \text { Mean }+ \text { Variance }=\overline{\mathrm{C}}+\sigma^2=6+32=38 \end{aligned} \end{aligned}

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Rakesh

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