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Question: If variance of ten observations \[10,20,30,....100\] is \[A\] and variance of ten another observatio...

If variance of ten observations 10,20,30,....10010,20,30,....100 is AA and variance of ten another observations 22,42,62,......,20222,42,62,......,202 is BB then find the value of BA\dfrac{B}{A}.
A. 00
B. 11
C. 22
D. 44

Explanation

Solution

We solve by finding the mean of two given observations separately and calculate the variance of two observations by substituting the value of mean in the formula of variance. And at the end, we find the ratio of the variance of two observations.

  • Formula for mean of nn number of observations where xi{x_i} is the ith{i^{th}}term of the observation is given by x‾=1n∑i=1nxi\overline x = \dfrac{1}{n}\sum\limits_{i = 1}^n {{x_i}}
  • Formula for variance of n number of observations having a mean is given by Var(X)=1n∑i=1n[xi−x‾]2Var(X) = \dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left[ {{x_i} - \overline x } \right]}^2}}

Complete step by step solution:
Firstly, we will calculate mean (x‾\overline x ) by making use of the formula x‾=1n∑i=1nxi\overline x = \dfrac{1}{n}\sum\limits_{i = 1}^n {{x_i}} where xi{x_i} is the ith{i^{th}} term and number of observations is nn.
We will substitute the values n=10,x1=10,x2=20,.....x10=100n = 10,{x_1} = 10,{x_2} = 20,.....{x_{10}} = 100 in the above formula
x‾=110∑i=110xi\overline x = \dfrac{1}{{10}}\sum\limits_{i = 1}^{10} {{x_i}}
Since we know that summation opens up as ∑a=1bxa=x1+.....+xb\sum\limits_{a = 1}^b {{x_a}} = {x_1} + ..... + {x_b} , so we substitute a=i,b=10a = i,b = 10
Which means we add all the values having subscript from one to ten.

x‾=110(x1+x2+x3+......x10)\overline x = \dfrac{1}{{10}}\left( {{x_1} + {x_2} + {x_3} + ......{x_{10}}} \right)

x‾=110(10+20+30+40+50+60+70+80+90+100)\overline x = \dfrac{1}{{10}}\left( {10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 + 100} \right)
Now we will add up all the numbers in the round bracket
x‾=110(550)\overline x = \dfrac{1}{{10}}(550)
Now we will reduce 550 & 10 by dividing both by their HCF 10
x‾=55\overline x = 55 …… (1)
Now we will calculate Var(X) by substituting the given values and the value from the equation 1 in the formula Var(X)=1n∑i=1n[xi−x‾]2Var(X) = \dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left[ {{x_i} - \overline x } \right]}^2}}
Here n=10,x‾=55,1⩽i⩽10n = 10,\overline x = 55,1 \leqslant i \leqslant 10
Var(X)=1n[(10−55)2+(20−55)2+(30−55)2+(40−55)2+(50−55)2+(60−55)2+(70−55)2+Var(X) = \dfrac{1}{n}[{\left( {10 - 55} \right)^2} + {\left( {20 - 55} \right)^2} + {\left( {30 - 55} \right)^2} + {\left( {40 - 55} \right)^2} + {\left( {50 - 55} \right)^2} + {\left( {60 - 55} \right)^2} + {\left( {70 - 55} \right)^2} +
(80−55)2+(90−55)2+(100−55)2]{\left( {80 - 55} \right)^2} + {\left( {90 - 55} \right)^2} + {\left( {100 - 55} \right)^2}]
Now we will solve the fundamental operations inside the round bracket
Var(X)=110[(−45)2+(−35)2+(−25)2+(−15)2+(−5)2+(5)2+(15)2+(25)2+(35)2+(45)2]Var(X) = \dfrac{1}{{10}}[{( - 45)^2} + {( - 35)^2} + {( - 25)^2} + {( - 15)^2} + {( - 5)^2} + {(5)^2} + {(15)^2} + {(25)^2} + {(35)^2} + {(45)^2}]
Now we will take 52{5^2} common from the bracket
Var(X)=110×52[(−9)2+(−7)2+(−5)2+(−3)2+(−1)2+(1)2+(3)2+(5)2+(7)2+(9)2]Var(X) = \dfrac{1}{{10}} \times {5^2}[{( - 9)^2} + {( - 7)^2} + {( - 5)^2} + {( - 3)^2} + {( - 1)^2} + {(1)^2} + {(3)^2} + {(5)^2} + {(7)^2} + {(9)^2}]
Now we will find the square of all the numbers inside the bracket
Var(X)=110×52[81+49+25+9+1+1+9+25+49+81]Var(X) = \dfrac{1}{{10}} \times {5^2}[81 + 49 + 25 + 9 + 1 + 1 + 9 + 25 + 49 + 81]
Now we will add all the numbers inside the bracket
Var(X)=110×52[330]Var(X) = \dfrac{1}{{10}} \times {5^2}\left[ {330} \right]
Now we will reduce 10 & 5 by dividing both by 5
Var(X)=52[330]Var(X) = \dfrac{5}{2}\left[ {330} \right]
Now we will reduce 330 & 2 by dividing both by 2
Var(X)=5×165Var(X) = 5 \times 165
Now we will find the product of 5 & 165
Var(X)=825Var(X) = 825 …… (2)
Now it is given that variance of ten observations 10,20,30,40,50, ……100 is A and if we compare hint and equation 2 we will find that A= Var(X) = 825 …… (3)
Now, we will calculate mean (y‾\overline y ) by making use of the formula y‾=1n∑i=1nyi\overline y = \dfrac{1}{n}\sum\limits_{i = 1}^n {{y_i}} where yi{y_i} is the ith{i^{th}} term and number of observations is nn.
We will substitute the values n=10,y1=22,y2=42,.....y10=202n = 10,{y_1} = 22,{y_2} = 42,.....{y_{10}} = 202 in the above formula y‾=1n∑i=1nyi\overline y = \dfrac{1}{n}\sum\limits_{i = 1}^n {{y_i}}
Since we know that summation opens up as ∑a=1bya=y1+.....+yb\sum\limits_{a = 1}^b {{y_a}} = {y_1} + ..... + {y_b} , so we substitute a=i,b=10a = i,b = 10
Which means we add all the values having subscript from one to ten.
y‾=110(y1+y2+y3+......y10)\overline y = \dfrac{1}{{10}}\left( {{y_1} + {y_2} + {y_3} + ......{y_{10}}} \right)
We will substitute the given values in the above formula
y‾=110(22+42+62+82+102+122+142+162+182+202)\overline y = \dfrac{1}{{10}}\left( {22 + 42 + 62 + 82 + 102 + 122 + 142 + 162 + 182 + 202} \right)
Now we will add up all the numbers in the round bracket
y‾=110(1120)\overline y = \dfrac{1}{{10}}(1120)
Now we will reduce 1120 & 10 by dividing both by their HCF 10
y‾=112\overline y = 112 …… (4)
Now we will calculate Var(Y) by substituting the given values and the value from the equation 4 in the formulaVar(Y)=1n∑i=1n[yi−y‾]2Var(Y) = \dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left[ {{y_i} - \overline y } \right]}^2}} where yi{y_i} is the ith{i^{th}} term and number of observations is nn.
Here, n=10,y‾=112,1⩽i⩽10n = 10,\overline y = 112,1 \leqslant i \leqslant 10
Var(Y)=1n[(22−112)2+(42−112)2+(62−112)2+(82−112)2+(102−112)2+(122−112)2+(142−112)2+Var(Y) = \dfrac{1}{n}[{\left( {22 - 112} \right)^2} + {\left( {42 - 112} \right)^2} + {\left( {62 - 112} \right)^2} + {\left( {82 - 112} \right)^2} + {\left( {102 - 112} \right)^2} + {\left( {122 - 112} \right)^2} + {\left( {142 - 112} \right)^2} +
(162−112)2+(182−112)2+(202−112)2]{\left( {162 - 112} \right)^2} + {\left( {182 - 112} \right)^2} + {\left( {202 - 112} \right)^2}]

Now we will solve the fundamental operations inside the round bracket
Var(Y)=110[(−90)2+(−70)2+(−50)2+(−30)2+(−10)2+(10)2+(30)2+(50)2+(70)2+(90)2]Var(Y) = \dfrac{1}{{10}}[{( - 90)^2} + {( - 70)^2} + {( - 50)^2} + {( - 30)^2} + {( - 10)^2} + {(10)^2} + {(30)^2} + {(50)^2} + {(70)^2} + {(90)^2}]
Now we will take 102{10^2} common from the bracket
Var(Y)=110×102[(−9)2+(−7)2+(−5)2+(−3)2+(−1)2+(1)2+(3)2+(5)2+(7)2+(9)2]Var(Y) = \dfrac{1}{{10}} \times {10^2}[{( - 9)^2} + {( - 7)^2} + {( - 5)^2} + {( - 3)^2} + {( - 1)^2} + {(1)^2} + {(3)^2} + {(5)^2} + {(7)^2} + {(9)^2}]
Now we will find the square of all the numbers inside the bracket
Var(Y)=110×102[81+49+25+9+1+1+9+25+49+81]Var(Y) = \dfrac{1}{{10}} \times {10^2}[81 + 49 + 25 + 9 + 1 + 1 + 9 + 25 + 49 + 81]
Now we will add all the numbers inside the bracket
Var(Y)=110×102[330]Var(Y) = \dfrac{1}{{10}} \times {10^2}\left[ {330} \right]
Now we will reduce 10 & 10 by dividing both by 10
Var(Y)=10[330]Var(Y) = 10\left[ {330} \right]
Now we will find the product of 10 & 330
Var(Y)=3300Var(Y) = 3300 …… (5)
Now it is given that variance of ten observations 22,42,62,82,102, ……202 is B, and if we compare hint and equation 5 we will find that B = Var(Y) =3300 …… (6)
Now we will find the value of BA\dfrac{B}{A} by substituting the value of B & A from equation 5 & 6
BA=3300825\dfrac{B}{A} = \dfrac{{3300}}{{825}}
Now we will reduce the RHS by dividing both the numbers by 825 in order to get the value of BA\dfrac{B}{A}
BA=41\dfrac{B}{A} = \dfrac{4}{1}
BA=4\dfrac{B}{A} = 4

Therefore, option D is correct.

Note:
Students are likely to make the mistake of calculating the square values in the variance part of the question directly which gives us very complex calculations, instead, we should try to take out as many common factors as we can so we have less confusing calculations.