JEE MAIN - Mathematics (2021 - 27th July Morning Shift - No. 1)
If the mean and variance of the following data : 6, 10, 7, 13, a, 12, b, 12 are 9 and $${{37} \over 4}$$
respectively, then (a $$-$$ b)2 is equal to :
respectively, then (a $$-$$ b)2 is equal to :
24
12
32
16
Explanation
Mean = $${{6 + 10 + 7 + 13 + a + 12 + b + 12} \over 8} = 9$$
60 + a + b = 72
a + b = 12 .....(1)
variance $$ = {{\sum {x_i^2} } \over n} - {\left( {{{\sum {x_i^{}} } \over n}} \right)^2} = {{37} \over 4}$$
$$\sum {x_i^2} = {6^2} + {10^2} + {7^2} + {13^2} + {a^2} + {b^2} + {12^2} + {12^2} = {a^2} + {b^2} + 642$$
$${{{a^2} + {b^2} + 642} \over 8} - {(9)^2} = {{37} \over 4}$$
$${{{a^2} + {b^2}} \over 8} + {{321} \over 4} - 81 = {{37} \over 4}$$
$${{{a^2} + {b^2}} \over 8} = 81 + {{37} \over 4} - {{321} \over 4}$$
$${{{a^2} + {b^2}} \over 8} = 81 - 71$$
$$\therefore$$ a2 + b2 + 2ab = 144
80 + 2ab = 144
$$\therefore$$ 2ab = 64
$$ \therefore $$ $${(a - b)^2} = {a^2} + {b^2} - 2ab = 80 - 64 = 16$$
60 + a + b = 72
a + b = 12 .....(1)
variance $$ = {{\sum {x_i^2} } \over n} - {\left( {{{\sum {x_i^{}} } \over n}} \right)^2} = {{37} \over 4}$$
$$\sum {x_i^2} = {6^2} + {10^2} + {7^2} + {13^2} + {a^2} + {b^2} + {12^2} + {12^2} = {a^2} + {b^2} + 642$$
$${{{a^2} + {b^2} + 642} \over 8} - {(9)^2} = {{37} \over 4}$$
$${{{a^2} + {b^2}} \over 8} + {{321} \over 4} - 81 = {{37} \over 4}$$
$${{{a^2} + {b^2}} \over 8} = 81 + {{37} \over 4} - {{321} \over 4}$$
$${{{a^2} + {b^2}} \over 8} = 81 - 71$$
$$\therefore$$ a2 + b2 + 2ab = 144
80 + 2ab = 144
$$\therefore$$ 2ab = 64
$$ \therefore $$ $${(a - b)^2} = {a^2} + {b^2} - 2ab = 80 - 64 = 16$$
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