If the mean of the data 7, 8, 9, 7, 8, 7, lam | vedclass

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(D) Given the mean $\bar{x} = 8$ for the data set $7, 8, 9, 7, 8, 7, \lambda, 8$.$\bar{x} = \frac{7 + 8 + 9 + 7 + 8 + 7 + \lambda + 8}{8} = 8$$\Righta

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$1$

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(D) Given the mean $\bar{x} = 8$ for the data set $7, 8, 9, 7, 8, 7, \lambda, 8$.$\bar{x} = \frac{7 + 8 + 9 + 7 + 8 + 7 + \lambda + 8}{8} = 8$$\Rightarrow \frac{54 + \lambda}{8} = 8$$\Rightarrow 54 + \lambda = 64$$\Rightarrow \lambda = 10$The data set is $7, 8, 9, 7, 8, 7, 10, 8$.Variance $\sigma^2 = \frac{1}{n} \sum (x_i - \bar{x})^2$$\sigma^2 = \frac{(7-8)^2 + (8-8)^2 + (9-8)^2 + (7-8)^2 + (8-8)^2 + (7-8)^2 + (10-8)^2 + (8-8)^2}{8}$$\sigma^2 = \frac{(-1)^2 + 0^2 + 1^2 + (-1)^2 + 0^2 + (-1)^2 + 2^2 + 0^2}{8}$$\sigma^2 = \frac{1 + 0 + 1 + 1 + 0 + 1 + 4 + 0}{8} = \frac{8}{8} = 1$.Thus,the variance is $1$.

If the mean of the data $7, 8, 9, 7, 8, 7, \lambda, 8$ is $8$,then the variance of this data is

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