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. (2024)

Hint: In these types of questions remember to use the mean and variance i.e. mean = $\dfrac{{sum{\text{ }}of{\text{ }}observation\;\;}}{{number{\text{ }}of{\text{ }}observation}}$ and variance = $\dfrac{1}{n}\sum {({x_i} - \bar x)^2}$ formulas to find out the values of remaining observations.

Let the other two observations be x and y.
Therefore, our observations are 1, 2, x, y, and 6.
Given mean = 4.4 i.e. (sum of observation) / (number of observation) = 4.4
\[\dfrac{{1 + 2 + 6 + x + y\;}}{5} = 4.4\]
9 + x + y = 4.4 $ \times $5
x + y = 22 – 9
x + y = 13 (equation 1)
Also, given variance = 8.24 i.e.
$\dfrac{1}{n}\sum {({x_i} - \bar x)^2} = 8.24$, where ${x_i}$ is the value of the one observation, $\bar x$ is the mean value of all observation and n is the number of observations
$ \Rightarrow $$\dfrac{1}{5}\sum {({x_i} - 4.4)^2} = 8.24$
$ \Rightarrow $\[\dfrac{1}{5}[{(1 - 4.4)^2} + {(2 - 4.4)^2} + {(6 - 4.4)^2} + {(x - 4.4)^2} + {(y - 4.4)^2}] = 8.24\]
$ \Rightarrow $\[\dfrac{1}{5}[{( - 3.4)^2} + {( - 2.4)^2} + {(1.6)^2} + {(x - 4.4)^2} + {(y - 4.4)^2}] = 8.24\]
$ \Rightarrow $\[[19.88 + {x^2} + 19.36 - 8.8x + {y^2} + 19.36 - 8.8y] = 8.24 \times 5\]
$ \Rightarrow $\[[58.6 + {x^2} + {y^2} - 8.8(x + y)] = 41.2\]
Putting the value of x + y form the equation 1
$ \Rightarrow $\[[58.6 + {x^2} + {y^2} - 8.8(13)] = 41.2\]
$ \Rightarrow $\[58.6 + {x^2} + {y^2} - 114.4 = 41.2\]
$ \Rightarrow $\[{x^2} + {y^2} = 41.2 + 114.4 - 58.6\]
$ \Rightarrow $\[{x^2} + {y^2} = 97\] (Equation 2)
From equation 1
x + y = 13
Squaring both sides
${(x + y)^2} = {13^2}$
\[{x^2} + {y^2} + 2xy = 169\]
Putting the value of \[{x^2} + {y^2}\]form equation 2
97 + 2xy = 169
$ \Rightarrow $2xy = 169 – 97$ \Rightarrow $xy = 36
$x = \dfrac{{36}}{y}$ (Equation 3)
Putting the value of x from equation 3 in equation 1
x + y = 13
$\dfrac{{36}}{y}$ + y = 13
$\Rightarrow$ 36 + y(y) = 13(y)
$\Rightarrow$ 36 + ${y^2}$= 13y
$\Rightarrow$ - 13y + 36 = 0
$\Rightarrow$ ${y^2}$- 9y – 4y + 36 = 0
$\Rightarrow$ y(y – 9) – 4(y – 9) =0
$\Rightarrow$ (y – 4) (y – 9) = 0
So, y = 4 and y = 9
For y = 4
$x = \dfrac{{36}}{y} = \dfrac{{36}}{4} = 9$
Hence x = 9, y=4 are the remaining two observations
Thus, remaining observations are 4 and 9.

Note: In these types of questions first let x and y be the other observations then use the mean formula to find the value of x + y and assume it as equation 1 then use the variance formula to find the value of \[{x^2} + {y^2}\]and assume it as equation 2 then use square root on equation 1 and with the equation 2 find out the value of x and take it as equation 3 and use the value of equation 3 in equation 1 and find the value of y and you will get the 2 values of y now use the values of y and find out the value of x hence you get the values of the remaining values.