Answer
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Hint: In the given problem, \[\sum\limits_{i = 1}^n {x_i^2} = 400\] and \[\sum\limits_{i = 1}^n {x_1^{}} = 80\]
As already given, We have to calculate which of the given options is the possible value that \[n\] could have where ‘\[n\]’ is the number of observations.
In this problem, root mean square and the arithmetic mean are going to be used. The value of the root means the square of the given observations is always greater than or equal to the arithmetic mean.
This relation is going to be used in the solution.
Complete step-by-step answer:
In the given problem, the number of observations are given to be ‘\[n\]’.
Also, it is given that, \[\sum\limits_{i = 1}^n {x_i^2} = 400...........(1)\]
and \[\sum\limits_{i = 1}^n {x_1^{}} = 80...................(2)\]
Using equation (1), the root mean square value becomes \[\sqrt {\dfrac{1}{n}\sum\limits_{i = 1}^n {x_i^2} } = \sqrt {\dfrac{{400}}{n}} \]
i.e., R.M.S \[ = \dfrac{{20}}{{\sqrt n }}....................(3)\]
Using equation (2), The arithmetic mean becomes
$\Rightarrow$ \[\dfrac{1}{n}\sum\limits_{i = 1}^n {x_1^{}} = \dfrac{{80}}{n}\]
i.e., A.M \[ = \dfrac{{80}}{n}.....................(4)\]
Now, according to the fact, the root mean square value of \[n\] numbers is greater than or equal to the arithmetic mean of the numbers.
i.e., \[R.M.S\; \geqslant A.M\]
i.e., using equations \[(3)\] and \[\left( 4 \right)\],
\[\dfrac{{20}}{{\sqrt n }} \geqslant \dfrac{{80}}{n}\]
On rearranging and solving the terms, we get:
$\Rightarrow$ \[\sqrt n \geqslant 4\]
On squaring both sides, we get:
$\Rightarrow$ \[n \geqslant 16\]
Hence, according to the given options, \[n = 18\] is the correct answer.
Note: In the given problem, root mean square value and the arithmetic mean are used. By using the relation between them, we found the possible value of \[n\] amongst the given options i.e., \[18\].
Because, option (A) i.e., \[9 < 16\]
Similarly, option (B) i.e., \[12 < 16\]
And, option (C) i.e., \[15 < 16\]
Only option (D) i.e., \[18 > 16\]
Hence, the possible value could be \[18\] among the given options.
So, the correct option is (D) i.e., \[18\].
As already given, We have to calculate which of the given options is the possible value that \[n\] could have where ‘\[n\]’ is the number of observations.
In this problem, root mean square and the arithmetic mean are going to be used. The value of the root means the square of the given observations is always greater than or equal to the arithmetic mean.
This relation is going to be used in the solution.
Complete step-by-step answer:
In the given problem, the number of observations are given to be ‘\[n\]’.
Also, it is given that, \[\sum\limits_{i = 1}^n {x_i^2} = 400...........(1)\]
and \[\sum\limits_{i = 1}^n {x_1^{}} = 80...................(2)\]
Using equation (1), the root mean square value becomes \[\sqrt {\dfrac{1}{n}\sum\limits_{i = 1}^n {x_i^2} } = \sqrt {\dfrac{{400}}{n}} \]
i.e., R.M.S \[ = \dfrac{{20}}{{\sqrt n }}....................(3)\]
Using equation (2), The arithmetic mean becomes
$\Rightarrow$ \[\dfrac{1}{n}\sum\limits_{i = 1}^n {x_1^{}} = \dfrac{{80}}{n}\]
i.e., A.M \[ = \dfrac{{80}}{n}.....................(4)\]
Now, according to the fact, the root mean square value of \[n\] numbers is greater than or equal to the arithmetic mean of the numbers.
i.e., \[R.M.S\; \geqslant A.M\]
i.e., using equations \[(3)\] and \[\left( 4 \right)\],
\[\dfrac{{20}}{{\sqrt n }} \geqslant \dfrac{{80}}{n}\]
On rearranging and solving the terms, we get:
$\Rightarrow$ \[\sqrt n \geqslant 4\]
On squaring both sides, we get:
$\Rightarrow$ \[n \geqslant 16\]
Hence, according to the given options, \[n = 18\] is the correct answer.
Note: In the given problem, root mean square value and the arithmetic mean are used. By using the relation between them, we found the possible value of \[n\] amongst the given options i.e., \[18\].
Because, option (A) i.e., \[9 < 16\]
Similarly, option (B) i.e., \[12 < 16\]
And, option (C) i.e., \[15 < 16\]
Only option (D) i.e., \[18 > 16\]
Hence, the possible value could be \[18\] among the given options.
So, the correct option is (D) i.e., \[18\].
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