If the data is x1,x2, …x10 such that the mean of first four of these is 11, the mean of the remaining six is 16 and the sum of squares of all of these is 2000; then the standard deviation of this data is

# If the data is  such that the mean of first four of these is 11, the mean of the remaining six is 16 and the sum of squares of all of these is 2000; then the standard deviation of this data is

1. A

2

2. B

$2\sqrt{2}$

3. C

4

4. D

$\sqrt{2}$

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### Solution:

Given, $\frac{{x}_{1}+{x}_{2}+{x}_{3}+{x}_{4}}{4}=11$

$⇒{x}_{1}+{x}_{2}+{x}_{3}+{x}_{4}=44\to \left(i\right)$

Also, $\frac{{x}_{5}+{x}_{6}+{x}_{7}+{x}_{8}+{x}_{9}+{x}_{10}}{6}=96\to \left(ii\right)$

from (i) and (ii),$\sum _{i=1}^{10} {x}_{i}=44+96=140$

Now, Variance $\left({\sigma }^{2}\right)=\frac{\sum _{i=1}^{10} {x}_{i}^{2}}{10}-{\left(\frac{\sum _{i=1}^{10} {x}_{i}}{10}\right)}^{2}$

$\therefore$ Standard deviation $\left(\sigma \right)=2$

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