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(C) Given data: $6, 10, 7, 13, a, 12, b, 12$. Total number of observations $n = 8$.Mean $\bar{x} = \frac{6+10+7+13+a+12+b+12}{8} = 9$.$60 + a + b = 72

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$16$

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(C) Given data: $6, 10, 7, 13, a, 12, b, 12$. Total number of observations $n = 8$.Mean $\bar{x} = \frac{6+10+7+13+a+12+b+12}{8} = 9$.$60 + a + b = 72 \implies a + b = 12$  $(1)$.Variance $\sigma^{2} = \frac{\sum x_{i}^{2}}{n} - (\bar{x})^{2} = \frac{37}{4}$.$\sum x_{i}^{2} = 6^{2} + 10^{2} + 7^{2} + 13^{2} + a^{2} + 12^{2} + b^{2} + 12^{2} = 36 + 100 + 49 + 169 + a^{2} + 144 + b^{2} + 144 = a^{2} + b^{2} + 642$.$\frac{a^{2} + b^{2} + 642}{8} - (9)^{2} = \frac{37}{4}$.$\frac{a^{2} + b^{2} + 642}{8} - 81 = \frac{37}{4}$.$\frac{a^{2} + b^{2} + 642}{8} = \frac{37}{4} + 81 = \frac{37 + 324}{4} = \frac{361}{4}$.$a^{2} + b^{2} + 642 = 2 	imes 361 = 722$.$a^{2} + b^{2} = 722 - 642 = 80$  $(2)$.We know that $(a-b)^{2} = (a+b)^{2} - 4ab$.Also,$(a+b)^{2} = a^{2} + b^{2} + 2ab \implies 12^{2} = 80 + 2ab \implies 144 - 80 = 2ab \implies 2ab = 64$.Therefore,$(a-b)^{2} = a^{2} + b^{2} - 2ab = 80 - 64 = 16$.

If the mean and variance of the following data: $6, 10, 7, 13, a, 12, b, 12$ are $9$ and $\frac{37}{4}$ respectively,then $(a-b)^{2}$ is equal to:

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