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  • Help me understand! 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 :

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

 

Answers (1)

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 

\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 :

\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.

Posted by

Vakul

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