Question

If the data ${{x}_{1}},{{x}_{2}},.....,{{x}_{10}}$ is such that the mean of the first four of these is 11, the mean of the remaining six is 16 and the sum of the square of all these is 2000; the standard deviation of this data is?
(a) 4
(b) 2
(c) $\sqrt{2}$
(d) $2\sqrt{2}$

Answer 1

Find the sum of the first four numbers using the information given about the mean of first four numbers. Use the formula: - Mean of first four data = $\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}}{4}$. Now, find the sum of the remaining six numbers using the information given about the mean of remaining six numbers. Use the formula: - Mean of remaining six numbers = $\dfrac{{{x}_{5}}+{{x}_{6}}+{{x}_{7}}+{{x}_{8}}+{{x}_{9}}+{{x}_{10}}}{6}$. In the next step find the mean of all the 10 numbers using the relation: - $\overline{x}=\dfrac{1}{10}\sum\limits_{i=1}^{10}{{{x}_{i}}}$, where $\overline{x}$ is the mean of 10 numbers. Finally, apply the formula for standard deviation given by: - $\sigma =\sqrt{\dfrac{1}{10}\sum\limits_{i=1}^{10}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}$, to get the answer. Here, ‘$\sigma$’ is the notation of standard deviation.

Complete step-by-step solution:
We have been provided with 10 data ${{x}_{1}},{{x}_{2}},.....,{{x}_{10}}$.
It is given that the mean of the first four numbers is 11. Therefore, applying the formula for mean, we get,
Mean of first four numbers = $\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}}{4}$
$\Rightarrow 44={{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}$ - (i)
Also, it is given that the mean of the remaining six numbers is 16. Therefore, applying the formula for mean in this case, we get,
Mean of remaining six numbers = $\dfrac{{{x}_{5}}+{{x}_{6}}+{{x}_{7}}+{{x}_{8}}+{{x}_{9}}+{{x}_{10}}}{6}$
$\Rightarrow 96={{x}_{5}}+{{x}_{6}}+{{x}_{7}}+{{x}_{8}}+{{x}_{9}}+{{x}_{10}}$ - (ii)
Now, adding equation (i) and (ii), we get,
$\Rightarrow {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}+{{x}_{6}}+{{x}_{7}}+{{x}_{8}}+{{x}_{9}}+{{x}_{10}}=96+44$
This can be written as: -
$\Rightarrow \sum\limits_{i=1}^{10}{{{x}_{i}}}=140$
Dividing both sides by 10, we get,
$\Rightarrow \dfrac{1}{10}\sum\limits_{i=1}^{10}{{{x}_{i}}}=\dfrac{140}{10}$
$\Rightarrow \overline{x}=14$ - (iii)
Here, $\overline{x}$ is the mean of all the ten observations ${{x}_{1}},{{x}_{2}},.....,{{x}_{10}}$.
Now, we know that standard deviation for 10 observations is by: - $\sigma =\sqrt{\dfrac{1}{10}\sum\limits_{i=1}^{10}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}$, where ‘$\sigma$’ is the standard deviation.
Therefore, substituting the value of $\overline{x}$ in the above formula, from equation (iii), we get,
$\Rightarrow \sigma =\sqrt{\dfrac{1}{10}\sum\limits_{i=1}^{10}{{{\left( {{x}_{i}}-14 \right)}^{2}}}}$
Applying the identity, ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, we get,

& \Rightarrow \sigma =\sqrt{\dfrac{1}{10}\sum\limits_{i=1}^{10}{\left( {{x}_{i}}^{2}+{{14}^{2}}-28{{x}_{i}} \right)}} \\\ & \Rightarrow \sigma =\sqrt{\dfrac{1}{10}\left[ \sum\limits_{i=1}^{10}{{{x}_{i}}^{2}+}\sum\limits_{i=1}^{10}{196}-28\sum\limits_{i=1}^{10}{{{x}_{i}}} \right]} \\\ \end{aligned}$$ Substituting the value of $$\sum\limits_{i=1}^{10}{{{x}_{i}}}=140$$, we get, $$\begin{aligned} & \Rightarrow \sigma =\sqrt{\dfrac{1}{10}\left[ \sum\limits_{i=1}^{10}{{{x}_{i}}^{2}+}\sum\limits_{i=1}^{10}{196}\times 10-28\times 140 \right]} \\\ & \Rightarrow \sigma =\sqrt{\dfrac{1}{10}\left[ \sum\limits_{i=1}^{10}{{{x}_{i}}^{2}+}1960-3920 \right]} \\\ \end{aligned}$$ $$\Rightarrow \sigma =\sqrt{\dfrac{1}{10}\left[ \sum\limits_{i=1}^{10}{{{x}_{i}}^{2}-}1960 \right]}$$ - (iv) Now, in the question we have been provided with another information. It is given that sum of square of $${{x}_{1}},{{x}_{2}},.....,{{x}_{10}}$$ is 2000. Therefore, we have, $$\begin{aligned} & \Rightarrow x_{1}^{2}+x_{2}^{2}+.....+x_{10}^{2}=2000 \\\ & \Rightarrow \sum\limits_{i=1}^{10}{x_{i}^{2}}=2000 \\\ \end{aligned}$$ Therefore, substituting this value in equation (iv), we get, $$\begin{aligned} & \Rightarrow \sigma =\sqrt{\dfrac{1}{10}\left[ 2000-1960 \right]} \\\ & \Rightarrow \sigma =\sqrt{\dfrac{1}{10}\times 40} \\\ & \Rightarrow \sigma =\sqrt{4} \\\ & \Rightarrow \sigma =2 \\\ \end{aligned}$$ **Hence, option (b) is the correct answer.** **Note:** One may note that, we must calculate the mean of all the 10 data $${{x}_{1}},{{x}_{2}},.....,{{x}_{10}}$$. This is because we need this information to calculate the standard deviation. Now, it is not possible to find the mean of 10 numbers by directly adding the mean of four numbers and mean of six numbers separately. This will be the wrong approach. That is why it is necessary to find the sum of the four numbers and six numbers separately.