Question

Let ${{x}_{1}},{{x}_{2}},\ldots \ldots {{x}_{n}}$ be n observation such that $\sum{{{x}_{i}}^{2}}=400\text{ and }\sum{{{x}_{i}}}=80$. Then a possible value of n among the following is
\[\begin{align}
  & A.9 \\
 & B.12 \\
 & C.15 \\
 & D.18 \\
\end{align}\]

Answer 1

Hint: In this question, we are given an observation along with their sum and the sum of their squares. We need to find the value of n. For this, we will use the formula of variance which is given by:
$\text{Variance}\left( \sigma \right)=\dfrac{\sum{{{x}_{i}}^{2}}}{n}-{{\left( \dfrac{\sum{{{x}_{i}}}}{n} \right)}^{2}}$
We will use the property that variance cannot be negative and then solve the formula by putting the variance greater than or equal to zero. At last, we will obtain a value of n and then match it with our options.

Complete step-by-step solution
Here, we are given the n observations as ${{x}_{1}},{{x}_{2}},\ldots \ldots {{x}_{n}}$. The sum of these n terms denoted by $\sum{{{x}_{i}}}$ is equal to 80. The sum of the squares of these n terms denoted by \[\sum{{{x}_{i}}^{2}}\] is equal to 400. Now we know that, formula for calculating variance is given by $\sigma =\dfrac{\sum{{{x}_{i}}^{2}}}{n}-{{\left( \dfrac{\sum{{{x}_{i}}}}{n} \right)}^{2}}$ where n is the number of observations, $\sigma $ is variance.
As we know, the variance cannot be negative so we can say that $\sigma \ge 0$.
Since $\sigma $ was equal to $\dfrac{\sum{{{x}_{i}}^{2}}}{n}-{{\left( \dfrac{\sum{{{x}_{i}}}}{n} \right)}^{2}}$ so we get:
$\dfrac{\sum{{{x}_{i}}^{2}}}{n}-{{\left( \dfrac{\sum{{{x}_{i}}}}{n} \right)}^{2}}\ge 0$.
Now putting in the value of \[\sum{{{x}_{i}}^{2}}\] as 400 as the value of $\sum{{{x}_{i}}}$ as 80, we get:
$\dfrac{400}{n}-{{\left( \dfrac{80}{n} \right)}^{2}}\ge 0$.
Square of 80 can be calculated as $8\times 8\times 10\times 10=64\times 100=6400$. Hence, we get:
$\dfrac{400}{n}-\dfrac{6400}{{{n}^{2}}}\ge 0$.
Taking negative term to the other side, we get:
$\dfrac{400}{n}\ge \dfrac{6400}{{{n}^{2}}}$.
As we can see, n is in the denominator of both sides, so it cancels out. Hence we get:
$400\ge \dfrac{6400}{n}$.
Cross multiplying we get:
$400n\ge 6400$.
Dividing both sides by 400 we get:
$\dfrac{400n}{400}\ge \dfrac{6400}{400}$.
Now 400 cancels out on the left side and $\dfrac{6400}{400}$ is equal to 16. So we get:
$n\ge 16$.
Hence, we get a value of n that is 16 or greater than 16. Now taking a look at our options, we see that only 18 is greater than 16. So 18 is our correct answer.
Hence option D is the correct answer.

Note: Students can make mistakes in formulas of variance. They can get confused between the terms $\dfrac{\sum{{{x}_{i}}^{2}}}{n}\text{ and }{{\left( \dfrac{\sum{{{x}_{i}}}}{n} \right)}^{2}}$. First, we have added squared terms and then divided by n. In second, we have taken the sum of terms, divided it by n, and then taken the whole square. This question has multiple answers if options are not given.