QJEEM20Ma8c11: JEE Main 2020 Mathematics - Class 11 JEEM20Ma8c11 Solution: Let the observations x i (1 i 10) satisfy the equations, = 10…

Question

Verified Solution • 2025–26 Session moderate

Let the observations x_i (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

\sum\limits_{i = 1}^{10} {\left( {{x_1} - 5} \right)}

= 10

\Rightarrow

x₁ + x₂ + .... + x₁₀ = 60 ....(1)

\sum\limits_{i = 1}^{10} {{{\left( {{x_1} - 5} \right)}^2}}

= 40

\Rightarrow

(

x_1^2 + x_2^2 + ... + x_{10}^2

) + 25

\times

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} \right) + \left( {{x_2} - 3} \right) + ... + \left( {{x_{10}} - 3} \right)} \over {10}}

=

{{60 - 3 \times 10} \over {10}}

= 3

And

\lambda

= variance =

{{\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - 3} \right)}^2}} } \over {10}}

-

\mu

²

=

{{\left( {x_1^2 + x_2^2 + ... + x_{10}^2} \right) + 90 - 6\left( {\sum {{x_i}} } \right)} \over {10}}

- 9

=

{{390 + 90 - 360} \over {10}}

- 9

= 12 - 9 = 3

Step-by-Step Verified Solution