Question

The S.D. of the numbers 31, 32, 33, .... 47 is A. $2\sqrt{6}$
B. 4\sqrt{3}$$$$$ C. \sqrt{\dfrac{{{47}^{2}}-1}{12}}$$$$$
D. None of these$$$$

Answer 1

We find the number of terms $n$from the given data by observing that the numbers in the data are in AP and using the formula ${{a}_{n}}=a+(n-1)d$ . Where a is the first term and d is the common difference We find the mean of the observation using the formula$\overline{x}=\dfrac{\sum{{{x}_{i}}}}{n}=\dfrac{\dfrac{n}{2}\left( 2\times a+(n-1)d \right)}{n}$ where ${{x}_{i}}$ are the data points. We find the standard deviation using the formula $\sigma =\sqrt{\dfrac{\sum {{({{x}_{i}}-\overline{x})}^{2}}}{n}}$.$$$$

Complete step by step answer: Now, here we know that the series we have is in arithmetic progression.
As the terms of the series are in arithmetic progression, hence, we can find the total number of terms through the general formula as follows
${{a}_{n}}=a+(n-1)d$
(Where a is the first term and d is the common difference and n is the term)
So, as we know the value of the last term as 47, we can use the above formula to get the total number of terms as follows

& \Rightarrow {{a}_{n}}=a+(n-1)d \\\ & \Rightarrow 47=31+(n-1)1 \\\ & \Rightarrow 47=31+(n-1) \\\ & \Rightarrow 47-31+1=n \\\ & \Rightarrow n=17 \\\ \end{aligned}$$ Hence, the total number of terms is 17. Now, the formula for calculating the standard deviation is as follows $$\sigma =\sqrt{\dfrac{\sum {{({{x}_{i}}-\overline{x})}^{2}}}{n}}$$ (Where ${{x}_{i}}$ is the term at position i, $\overline{x}$ is the mean for the elements of the series and $n$ is the total elements in the series) Now, here, we will first calculate the value of ${{({{x}_{i}}-\overline{x})}^{2}}$ first and then we will find its summation. Then we will divide it with the total number of terms in the series and then we will take the root of the value that we get. As mentioned in the question, we have to find the value of the standard deviation. Firstly, we will find the value of the mean for the series as follows $$\begin{aligned} & \Rightarrow \overline{x}=\dfrac{\sum{{{x}_{i}}}}{n} \\\ & \Rightarrow \overline{x}=\dfrac{\dfrac{n}{2}\left( 2\times a+(n-1)d \right)}{n} \\\ \end{aligned}$$ (Because the series is in arithmetic progression) $$\begin{aligned} & \Rightarrow \overline{x}=\dfrac{\dfrac{17}{2}\left( 2\times 31+(17-1)1 \right)}{17} \\\ & \Rightarrow \overline{x}=\dfrac{\dfrac{17}{2}\left( 62+16 \right)}{17} \\\ & \Rightarrow \overline{x}=\dfrac{1}{2}\left( 78 \right) \\\ & \Rightarrow \overline{x}=39 \\\ \end{aligned}$$ Hence, we get that the mean of the series is 39. Now, as given in the hint, we will first find the value of ${{({{x}_{i}}-\overline{x})}^{2}}$ for ${{x}_{1}}=31$ as follows $$\begin{aligned} & {{({{x}_{1}}-\overline{x})}^{2}}={{\left( 31-39 \right)}^{2}} \\\ & \Rightarrow {{({{x}_{1}}-\overline{x})}^{2}}={{\left( -8 \right)}^{2}}=64 \\\ \end{aligned}$$ Similarly, we will get the value of every term ${x_2} = 32,{x_3} = 33,...,{x_{16}} = 46,{x_{17}} = 47$ given in the question as follows $$\begin{aligned} & {{\left( 32-39 \right)}^{2}}={{\left( -7 \right)}^{2}} \\\ & {{\left( 33-39 \right)}^{2}}={{\left( -6 \right)}^{2}} \\\ & {{\left( 34-39 \right)}^{2}}={{\left( -5 \right)}^{2}} \\\ & \vdots \\\ & {{\left( 39-39 \right)}^{2}}={{\left( 0 \right)}^{2}} \\\ & \vdots \\\ & {{\left( 46-39 \right)}^{2}}={{\left( 7 \right)}^{2}} \\\ & {{\left( 47-39 \right)}^{2}}={{\left( 8 \right)}^{2}} \\\ \end{aligned}$$ Now, there summation can be written as follows $$\begin{aligned} & \Rightarrow \sum {{({{x}_{i}}-\overline{x})}^{2}}={{\left( -8 \right)}^{2}}+{{(-7)}^{2}}+{{(-6)}^{2}}+{{(-5)}^{2}}+{{(-4)}^{2}}+.......{{(7)}^{2}}+{{(8)}^{2}} \\\ & \Rightarrow \sum {{({{x}_{i}}-\overline{x})}^{2}}=\dfrac{2\times 8\times \left( 8+1 \right)\times \left( 2\times 8+1 \right)}{6}=24\times 17=408 \\\ \end{aligned}$$ Here in above calculation formula of sum of squares of natural numbers is used. S= $$\dfrac{n(n+1)(2n+1)}{6}$$ Now, on dividing this number with the total number of elements in the series and then taking the square root , we get the following $$\begin{aligned} & \Rightarrow \sigma =\sqrt{\dfrac{\sum {{({{x}_{i}}-\overline{x})}^{2}}}{n}} \\\ & \Rightarrow \sigma =\sqrt{24}=2\sqrt{6} \\\ \end{aligned}$$ **So, the correct answer is “Option A”.** **Note:** We note that we can standard deviation of an AP series using the formula $\sigma =\left| d \right|\sqrt{\dfrac{\left( n-1 \right)\left( n-2 \right)}{12}}$. The standard deviation signifies how the data is deviated from the centre or mean. The square of standard deviation is called variance denoted ${{\sigma }^{2}}$. Coefficient of variation is the ratio of SD to mean which is a very useful quantity measure risk in decision making .