Let the observations xi(1 \leq I £ 10) satisfy the equations, \sum_{i = 1}^{10} \left(\right. x_{i} - 5 \left.\right) = 10 \textrm{ }\textrm{ } and \textrm{ }\textrm{ } \sum_{i = 1}^{10} \left(\left(\right. x_{i} - 5 \left.\right)\right)^{2} = 40 . If μ and λ are the mean and the variance of the observations, x1 – 3, x2 – 3, …… x10 – 3, then the ordered pair (μ, λ) is equal to :

Engineering

Mathematics

Measure of Dispersion

Question

Let the observations xi(1 $\leq$ I £ 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 the variance of the observations, x₁ – 3, x₂ – 3, …… x₁₀– 3, then the ordered pair (μ, λ) is equal to :

(3, 6)

(6, 6)

(6, 3)

(3, 3)

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$Mean (x_{i} - 5) = \frac{Σ (x_{i} - 5)}{10} = 1 ∴ λ = {Mean (x_{i} - 5)} + 2 = 3$

$μ = var (x_{i} - 5) = \frac{Σ {(x_{i} - 5)}^{2}}{10} - \frac{Σ (x_{i} - 5)}{10} = 3$