If x is the mean of x₁,x₂............xⁿ and y is the mean of y₁, y₂,…….yn and z is the mean of x₁,x₂............xn , y₁, y₂,……….yⁿ then z =? Write the correct alternative answer for the question.
(A)x +y/2
(B)x+y
(C)x+y/n
(D)x+y/2n
Answers
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Your answer is (C) x+y/n
Answered by
2
Solution :
i ) mean of x1,x2,x3 ,...,xn = x
=> (x1+x²+....+ xn )/n = x ---( 1 )
ii ) mean of y1,y2,y3,...,yn = y
=> (y1+y2+y3+...+yn)/n = y----( 2 )
=> mean of x1,x2,x3,...,xn and
y1,y2,y3,...,yn
= [(x1+x2+x3+...+xn) +(y1+y2+...+yn )]2n
=(1/2)[(x1+x2+x3+...+xn)/n +(y1+y2+...+yn)/n]
=(1/2)[ x + y ] { from ( 1 ) & ( 2 ) }
= ( x + y )/2
Therefore ,
Option ( A ) is correct.
••••
i ) mean of x1,x2,x3 ,...,xn = x
=> (x1+x²+....+ xn )/n = x ---( 1 )
ii ) mean of y1,y2,y3,...,yn = y
=> (y1+y2+y3+...+yn)/n = y----( 2 )
=> mean of x1,x2,x3,...,xn and
y1,y2,y3,...,yn
= [(x1+x2+x3+...+xn) +(y1+y2+...+yn )]2n
=(1/2)[(x1+x2+x3+...+xn)/n +(y1+y2+...+yn)/n]
=(1/2)[ x + y ] { from ( 1 ) & ( 2 ) }
= ( x + y )/2
Therefore ,
Option ( A ) is correct.
••••
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