Question

Find the standard deviation of 10 observations: $111,211,311,......,1011$.
A) $100\sqrt 3$
B) $250$
C) $300$
D) $50\sqrt {33}$

Answer 1

In this question, we are given 10 observations and we have been asked to find the standard deviation of these observations. First, find the mean of the given observation. Then, find the deviation of these observations from the mean. Square all the deviations and then add them up.
Put all the values in the formula and simplify them. You will get your answer.

Formula used: $\sigma = \sqrt {\dfrac{1}{N}\sum\limits_{i = 1}^N {{{\left( {{x_i} - \bar X} \right)}^2}} }$, where $N =$ number of observations, ${x_i} =$ observations and, $\bar X =$ Mean of the observations.

Complete step-by-step solution:
We are given 10 observations - $111,211,311,......,1011$. We have to find the standard deviation of the given observations. We will use the formula $\sigma = \sqrt {\dfrac{1}{N}\sum\limits_{i = 1}^N {{{\left( {{x_i} - \bar X} \right)}^2}} }$ for finding the standard deviation. So, first we will find the mean first.
Mean = $\dfrac{{{\text{Sum of observations}}}}{{{\text{No}}{\text{. of observations}}}}$
Putting in the formula,
$\Rightarrow \bar X = \dfrac{{111 + 211 + 311 + .... + 1011}}{{10}}$
Using sum of A.P formula to find the sum,
$\Rightarrow {S_{10}} = \dfrac{{10}}{2}\left( {111 + 1011} \right)$
$\Rightarrow {S_{10}} = 5 \times 1122 = 5610$
Putting in the formula of mean,
$\Rightarrow \bar X = \dfrac{{5610}}{{10}}$
$\Rightarrow \bar X = 561$
Now, we have our mean. Using this mean, let us find the deviations from these observations. Once you have found the observations, square them and then add them.

${x_i}$	${x_i} - \bar X$	${\left( {{x_i} - \bar X} \right)^2}$
$111$	$- 450$	$202500$
$211$	$- 350$	$122500$
$311$	$- 250$	$62500$
$411$	$- 150$	$22500$
$511$	$- 50$	$2500$
$611$	$50$	$2500$
$711$	$150$	$22500$
$811$	$250$	$62500$
$911$	$350$	$122500$
$1011$	$450$	$202500$
	$\sum {{{\left( {{x_i} - \bar X} \right)}^2} = 825000}$

Now, we have all the values. Let us put them in the formula –
$\Rightarrow \sigma = \sqrt {\dfrac{1}{N}\sum\limits_{i = 1}^N {{{\left( {{x_i} - \bar X} \right)}^2}} }$
Putting N=10 and other values,
$\Rightarrow \sigma = \sqrt {\dfrac{1}{{10}} \times 825000}$
Simplifying further, we get,
$\Rightarrow \sigma = \sqrt {82500}$
On solving the under root, we will get,
$\Rightarrow \sigma = 50\sqrt {33}$

Hence, our answer is option D) $50\sqrt {33}$.

Note: The last step of finding the final answer has not been shown correctly as it is not required. But, it has been shown below for your understanding.
At first, you find the prime factors of $82500$. Then you make pairs of their factors. The factors whose pairs are available, are multiplied together and written outside the root. Then, those numbers whose pair does not exist are multiplied together and written inside the root.
${\text{ }}2\left| \\!{\underline {\, {82500} \,}} \right. \\\ {\text{ }}2\left| \\!{\underline {\, {41250} \,}} \right. \\\ {\text{ }}3\left| \\!{\underline {\, {20625} \,}} \right. \\\ {\text{ 5}}\left| \\!{\underline {\, {6875} \,}} \right. \\\ {\text{ }}5\left| \\!{\underline {\, {1375} \,}} \right. \\\ {\text{ }}5\left| \\!{\underline {\, {275} \,}} \right. \\\ {\text{ }}5\left| \\!{\underline {\, {55} \,}} \right. \\\ 11\left| \\!{\underline {\, {11} \,}} \right. \\\ {\text{ }}\left| \\!{\underline {\, 1 \,}} \right. \\\$
$\Rightarrow \sqrt {82500} = \sqrt {2 \times 2 \times 3 \times 5 \times 5 \times 5 \times 5 \times 11}$
Hence,
$\Rightarrow \sqrt {82500} = 2 \times 5 \times 5\sqrt {3 \times 11} = 50\sqrt {33}$
As we can see in this, 3 and 11 cannot form pairs. So, we will write 33 inside the root. And others which find pairs will be written outside the root.
Hence, we will get $50\sqrt {33}$.