Let the observations xi (1 \le i \le 10) satisfy the equations, \sum\limits_{i = 1}^{10} {\left( {{x_1} - 5} \right)} = 10 and \sum\limits_{i = 1}^{10} {{{\left( {{x_1} - 5} \right)}^2}} = 40. If \mu and \lambda are the mean and the variance of the observations, x1 – 3, x2 – 3, ...., x10 – 3, then t

Question

Let the observations xi (1 \le i \le 10) satisfy the equations, \sum\limits_{i = 1}^{10} {\left( {{x_1} - 5} \right)} = 10 and \sum\limits_{i = 1}^{10} {{{\left( {{x_1} - 5} \right)}^2}} = 40. If \mu and \lambda are the mean and the variance of the observations, x1 – 3, x2 – 3, ...., x10 – 3, then the ordered pair (\mu , \lambda ) is equal to :

Accepted Answer

Answer: B. (3, 3) \sum\limits_{i = 1}^{10} {\left( {{x_1} - 5} ight)} = 10 \Rightarrow x1 + x2 + .... + x10 = 60 ....(1) \sum\limits_{i = 1}^{10} {{{\left( {{x_1} - 5} ight)}^2}} = 40 \Rightarrow (x_1^2 + x_2^2 + ... + x_{10}^2) + 25 	imes 10 - 10( x1 + x2 + .... + x10) = 40 \Rightarrow x_1^2 + x_2^2 + ... + x_{10}^2 = 390 .....(2) From question, \mu = {{\left( {{x_1} - 3} ight) + \left( {{x_2} - 3} ight) + ... + \left( {{x_{10}} - 3} ight)} \over {10}} = {{60 - 3 	imes 10} \over {10}} = 3 And \lambda = variance = {{\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - 3} ight)}^2}} } \over {10}} - \mu 2 = {{\left( {x_1^2 + x_2^2 + ... + x_{10}^2} ight) + 90 - 6\left( {\sum {{x_i}} } ight)} \over {10}} - 9 = {{390 + 90 - 360} \over {10}} - 9 = 12 - 9 = 3

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