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.

• A. $4$, $9$
• B. $3$, $9$
• C. $4$, $4$
• D. $9$, $9$

Answer 1

Answer: (A)

$4$, $9$

Answer 2

Let the other two observations be $x$ and $y$. Therefore, the series is $1$, $2$, $6$, $x$, $y$. Now, Mean $\left(\bar{x} \right) = 4.4 = \frac{1+2+6+x+y}{5}$ or $22 = 9 + x + y$ Therefore, $x + y = 13\quad\ldots\left(i\right)$ Also, variance $= 8.24$ $= \frac{1}{n} \sum\limits^{5}_{i=1}\left(x_{i}-\bar{x}\right)^{2}$ i.e. $8.24 = \frac{1}{5}\left[\left(3.4\right)^{2} + \left(2.4\right)^{2} +\left(1.6\right)^{2} +x^{2}+y^{2} - 2 \times 4.4\left(x + y \right)+ 2 \times \left(4.4\right)^{2}\right]$ or $41.20 = 11.56 + 5.76 + 2.56 + x^{2} + y^{2} - 8.8 \times 13 + 38.72$ Therefore $x^{2} + y^{2} = 97 \quad\ldots\left(ii\right)$ But from $\left(i\right)$, we have $x^{2} + y^{2} + 2xy =169\quad\ldots\left(iii\right)$ From (ii) and (iii), we have $2xy = 72$ $...(iv)$ Subtracting (iv) from (ii), we get $x^{2}+ y^{2} - 2xy = 97 - 72$ i.e. $(x - y)^{2} = 25$ or $x - y = ? 5$ $...(v)$ So, from $(i)$ and $(v)$, we get $x = 9$, $y = 4$ when $x-7 = 5$ or $x = 4$, $y = 9$ when $x-y = -5$ Thus, the remaining observations are $4$ and $9$.