Let the observations xi(1≤i≤10) satisfy the equations, ∑i=110xi-5=10 and ∑i=110xi-52=40. If μ and λ are the mean and variance of the observations, x1-3,x2-3,…x10-3, then the ordered pair (μ,λ) is equal to:

Given ∑i=110xi=60 and ∑i=110xi2-10∑i=110xi+250=40⇒∑i=110xi2=390 For the observations x1-3,x2-3,…x10-3 Mean =∑i=110xi-310=3 Variance =∑i=110xi-3-3210=∑i=110xi-6210=∑i=110xi2-12(60)+36010=3

Let the observations xi(1≤i≤10) satisfy the equations, ∑i=110xi-5=10 and ∑i=110xi-52=40. If μ and λ are the mean and variance of the observations, x1-3,x2-3,…x10-3, then the ordered pair (μ,λ) is equal to:

Given ∑i=110xi=60 and ∑i=110xi2-10∑i=110xi+250=40⇒∑i=110xi2=390 For the observations x1-3,x2-3,…x10-3 Mean =∑i=110xi-310=3 Variance =∑i=110xi-3-3210=∑i=110xi-6210=∑i=110xi2-12(60)+36010=3

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$Let the observations x_{i} (1 \leq i \leq 10) satisfy the equations, \sum_{i = 1}^{10} (x_{i} - 5) = 10 and$
$\sum_{i = 1}^{10} {(x_{i} - 5)}^{2} = 40 . If μ and λ are the mean and variance of the observations,$
$x_{1} - 3, x_{2} - 3, \dots x_{10} - 3, then the ordered pair (μ, λ) is equal to:$

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$(3, 6)$

$(6, 3)$

$(3, 3)$

$(6, 6)$

answer is C.

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Detailed Solution

$\begin{array}{l} Given \sum_{i = 1}^{10} x_{i} = 60 and \sum_{i = 1}^{10} x_{i}^{2} - 10 \sum_{i = 1}^{10} x_{i} + 250 = 40 \Rightarrow \sum_{i = 1}^{10} x_{i}^{2} = 390 \\ For the observations x_{1} - 3, x_{2} - 3, \dots x_{10} - 3 \\ Mean = \frac{\sum_{i = 1}^{10} (x_{i} - 3)}{10} = 3 \\ Variance = \frac{\sum_{i = 1}^{10} {((x_{i} - 3) - 3)}^{2}}{10} = \frac{\sum_{i = 1}^{10} {(x_{i} - 6)}^{2}}{10} = \frac{\sum_{i = 1}^{10} x_{i}^{2} - 12 (60) + 360}{10} = 3 \end{array}$