# The mean of 5 observations is 5 and their variance is 124.  If three of the observations are 1, 2 and 6 ; then the mean deviation from the mean of the data is : Option 1) 2.4 Option 2) 2.8 Option 3) 2.5 Option 4) 2.6

Initially, we need to look at the following concepts:

ARITHMETIC Mean -

For the values x1, x2, ....xn of the variant x the arithmetic mean is given by

$\dpi{100} \bar{x}= \frac{x_{1}+x_{2}+x_{3}+\cdots +x_{n}}{n}$

in case of discrete data.

-

Mean Deviation -

If x1, x2, ...xn are n observations then the mean deviation from the point A is given by :

$\dpi{100} \frac{1}{n}\sum \left | x_{i}-A \right |$

-

Variance -

In case of discrete data

$\dpi{100} \sigma ^{2}= \left ( \frac{\sum x_{i}^{2}}{n} \right )-\left ( \frac{\sum x_{i}}{n} \right )^{2}$

-

Now,

$\frac{\sum x_{i}}{5}=5\Rightarrow \sum x_{i}=25$

$\frac{\sum x_{i}^{2}}{n}-\left(\frac{\sum x_{i}}{n} \right )^{2}=124$

$\frac{\sum x_{i}^{2}}{5}-25=124$

$\sum x_{i}^{2}=149\times 5=745$

Let the two observations be a & b

a+b+1+2+6=25

a+b=16

a2+b2+12+22+62=745

a2+b2+1+4+36=745

a2+b2=704

$Mean\:deviation = \frac{\sum \left | x_{i}-5 \right |}{5}=\frac{\left| x_{1}-5\right|+\left| x_{2}-5\right|+8}{5}$

$=\frac{8+\left| x_{1}-5\right|+\left| 11-x_{1}\right|}{5}=\frac{8+6}{5}=2.8$

Option 1)

2.4

This option is incorrect.

Option 2)

2.8

This option is correct.

Option 3)

2.5

This option is incorrect.

Option 4)

2.6

This option is incorrect.

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