Question

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

 

Answer 1

Let \[x \text{ and } y\]  be the other two observations.

Mean is 4.4. 

\[\therefore \frac{1 + 2 + 6 + x + y}{5} = 4 . 4\]

\[\Rightarrow 9 + x + y = 22\]

\[ \Rightarrow x + y = 13 . . . (1)\]

Let Var (X) be the variance of these observations, which is given to be 8.24.

If \[\bar{x}\] is the mean, then we have:

\[8 . 24 = \frac{1}{5}\left( 1^2 + 2^2 + 6^2 + x^2 + y^2 \right) - \left( x \right)^2 \]

\[ = \frac{1}{5}\left( 1 + 4 + 36 + x^2 + y^2 \right) - \left( 4 . 4 \right)^2 \]

\[ = \frac{1}{5}\left( 41 + x^2 + y^2 \right) - 19 . 36\]

\[ \Rightarrow x^2 + y^2 = 97 . . . (2)\]

\[ \left( x + y \right)^2 + \left( x - y \right)^2 = 2\left( x^2 + y^2 \right) \]

\[ \Rightarrow {13}^2 + \left( x - y \right)^2 = 2 \times 97 \left[ \text{ using eq (1) and eq } (2) \right]\]

\[ \Rightarrow \left( x - y \right)^2 = 194 - 169 = 25\]

\[ \Rightarrow x - y = \pm 5 . . . . (3)\]

\[ \text{ Solving eq } (1)\text{  and  eq (3) for x - y = - 5 and }  x + y = 13\]

\[ 2x = 18 \]

\[ \Rightarrow x = 9\]

\[ \Rightarrow y = 4\]

Thus, the other two observations are 9 and 4.