First slide

Introduction to statistics

Question

The mean of $5$ observations is $5$ and their variance is $9.2 .$ If three of the observations are $1, 2,$ and $6,$ then the mean deviation from the mean of the data, is

Moderate

A

$2.5$
B

$2.8$
C

$2.6$
D

$2.4$

Solution

Let $x_{1} = 1, x_{2} = 2, x_{3} = 6, x_{4}, x_{5}$ be five observations. It is given that
Mean $= 5$ and Variance $= 9.2$
$\Rightarrow \frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5}}{5} = 5$
and, $\frac{x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} + x_{5}^{2}}{5} - 5^{2} = 9.2$
$\Rightarrow \frac{1 + 2 + 6 + x_{4} + x_{5}}{5} = 5$
and, $\frac{1^{2} + 2^{2} + 6^{2} + x_{4}^{2} + x_{5}^{2}}{5} - 25 = 9.2$
$\Rightarrow x_{4} + x_{5} = 16$ and $x_{4}^{2} + x_{5}^{2} = 130$
$\Rightarrow {(x_{4} + x_{5})}^{2} = 256$ and $x_{4}^{2} + x_{5}^{2} = 130$
$\Rightarrow 130 + 2 x_{4} x_{5} = 256$ and $x_{4}^{2} + x_{5}^{2} = 130$
$\Rightarrow x_{4} x_{5} = 63$ and    $x_{4}^{2} + x_{5}^{2} = 130$
$\begin{array}{l} ∴ {(x_{4} - x_{5})}^{2} = {(x_{4} + x_{5})}^{2} - 4 x_{4} x_{5} = 256 - 252 = 4 \\ x_{4} - x_{5} = 2 \end{array}$
Thus, we have
   $x_{4} + x_{5} = 16$ and, $x_{4} - x_{5} = 2$
$\Rightarrow x_{4} = 9$ and,    $x_{5} = 7$
The mean deviation $M$ from the mean is given by
   $M = \frac{| 1 - 5 | + | 2 - 5 | + | 6 - 5 | + | 9 - 5 | + | 7 - 5 |}{5}$
$\Rightarrow M = \frac{4 + 3 + 1 + 4 + 2}{5} = 2.8$

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