Answer
Verified
414.9k+ views
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.
Complete step-by-step answer:
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.
Complete step-by-step answer:
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.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Two charges are placed at a certain distance apart class 12 physics CBSE
Difference Between Plant Cell and Animal Cell
What organs are located on the left side of your body class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
The planet nearest to earth is A Mercury B Venus C class 6 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is BLO What is the full form of BLO class 8 social science CBSE