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Statistics Part 1

Section: Mathematics

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Question : 55 of 100

Marks: +1, -0

Let the observations

x_i(1 \leq i \leq 10)

satisfy the equations,

\sum\limits_{i=1}^{10}(x_i-5)=10

\sum\limits_{i=1}^{10} (x_i-5)^2=40

. If µ and λ are the mean and the variance of the observations,

x_1 - 3, x_2 - 3, \dots, x_{10} - 3

, then the ordered pair (µ, A,) isequal to :

[9 Jan 2020 Shift 1]

(6, 6)
(3, 6)
(6, 3)
(3, 3)

Solution:

\sum\limits_{i=1}^{10}(x_i-5)=10

⇒ Mean of observation

x_i-5=\frac{1}{10} = \frac{1}{10}\sum\limits_{i=1}^{3}(x_i-5)=10

⇒ µ = mean of observation

(x_i - 3)

= (mean of observation

(x_i-5)) + 2

=1+2=3

Variance of observation

x_i-5=\frac{1}{10} \sum\limits_{i=1}^{10}(x_i-5)^2

- (\text{Mean of}(x_i-5))^2=3

⇒ λ = variance of observation

(x_i - 3)

= variance of observation

(x_i - 5) = 3

\therefore (\mu , \lambda) = (3, 3)

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