First slide

Introduction to statistics

Question

The mean of 5 observations is 4.4 and the variance is 8.24. If three of the five observations are 1, 2 and 6, the two values are

Moderate

A

4 and 9
B

3 and 5
C

2 and 6
D

4 and 6

Solution

$Σ x_{i} = n \bar{x} = 5 \times 4.4 = 22$
So    $1 + 2 + 6 + x_{4} + x_{5} = 22$
   $x_{4} + x_{5} = 13$ (1)
   $\begin{array}{l} σ^{2} = \frac{1}{n} ∣ \sum x_{1}^{2} - {(\frac{1}{n} \sum x_{i})}^{2} \Rightarrow \sum x_{i}^{2} = n (σ^{2} + {\bar{x}}^{2}) \\ = 5 (8.24 + 19.36) \\ = 138 . \\ \Rightarrow 1 + 4 + 36 + x_{4}^{2} + x_{5}^{2} = 138 \\ x_{4}^{2} + x_{5}^{2} = 97 \end{array}$
So $x_{4}^{2} + {(13 - x_{4})}^{2} = 97 \Rightarrow x_{4}^{2} + 169 + x_{4}^{2} - 26 x_{4} = 97$
$\begin{array}{l} \Rightarrow 2 x_{4}^{2} - 26 x_{4} + 72 = 0 \\ \Rightarrow x_{4}^{2} - 13 x_{4} + 36 = 0 \\ x_{4} = \frac{13 \pm \sqrt{169 - 144}}{2} = \frac{13 \pm 5}{2} = 9, 4 \end{array}$
So $x_{5} = 13 - x_{4}^{2} = 4, 9$
Hence, the other two nembers are 4 and 9.

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