If x is the mean of x1 , x2 , ……. , xn and y is the mean of y1 , y2 , ….. yn and z is the mean of x 1,x2 , …… , xn , y1 , y2 , …. yn , then z = ?

Question

If  x is the mean of x₁ , x₂ , ……. , x_n and is the mean of y₁ , y₂ , ….. y_n and is the mean of x₁ ,x₂ , …… , x_n , y_{1 ,} y₂ , …. y_n , then z = ?

(A) \(\frac{\bar x+\bar y}{2}\)

(B) \({\bar x+\bar y}\)

(C) \(\frac{\bar x+\bar y}{n}\)

 (D) \(\frac{\bar x+\bar y}{2n}\)

Answer 1

x₁, x₂ , x₃ , ……. , x_n

 ∴ \(\bar{x}\) = \(\frac{\sum x}{n}\)

∴ n = ∑x  

Similarly, n\(\bar{y}\) = ∑y 

Now,

Correct option is  (A) \(\frac{\bar x+\bar y}{2}\)

Answer 2

look if we apply mean formula so x = x1 + x2 + x3 .....   + xn / n

so x1 + x2 + x3 .....   + xn = nx      (Eq 1 )

similarly find the same mean as y1 + y2 + y3 .... + yn= ny     (Eq 2 )

add eq 1 and eq 2 , also divide it by 2n as now no. of entities is n + n =2n

z = x + y / 2