Let x̄ be the mean of x1 , x2 , ... , xn and ȳ the mean of y1 , y2 , ... , yn . If z̄ is the mean of x1 , x2 , ... , xn , y1 , y2 , ... , yn , then z̄ is equal to
a. x̄ + ȳ
b. (x̄ + ȳ)/2
c. (x̄ + ȳ)/n
d. (x̄ + ȳ)/2n

Solution:

It is given that

\(\sum_{i=1}^{n}x_{i}=n\overline{x}\: and\: \sum_{i=1}^{n}y_{i}=n\overline{y}\) …. (1)

As \(\overline{x}=\frac{\sum_{i=1}^{n}x_{i}}{n}\)

z̄ = (x1 + x2 + ….. +xn) + (y1 + y2 + …. + yn)/n + n

So we get

z̄ = \(\frac{\sum_{i=1}^{n}x_{i}+\sum_{j=1}^{n}y_{i}}{2n}\)

z̄ = (nx̄ + nȳ)/2n [From equation (1)]

Taking out n as common

z̄ = (x̄ + ȳ)/2

Therefore, z̄ is equal to (x̄ + ȳ)/2.

✦ Try This: The mean of five numbers is 16. If one number is excluded, their mean becomes 12. The excluded number is :

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 14

NCERT Exemplar Class 9 Maths Exercise 14.1 Problem 15

Let x̄ be the mean of x1 , x2 , ... , xn and ȳ the mean of y1 , y2 , ... , yn . If z̄ is the mean of x1 , x2 , ... , xn , y1 , y2 , ... , yn , then z̄ is equal to a. x̄ + ȳ, b. (x̄ + ȳ)/2, c. (x̄ + ȳ)/n, d. (x̄ + ȳ)/2n

Summary:

Let x̄ be the mean of x1 , x2 , ... , xn and ȳ the mean of y1 , y2 , ... , yn . If z̄ is the mean of x1 , x2 , ... , xn , y1 , y2 , ... , yn , then z̄ is equal to (x̄ + ȳ)/2

☛ Related Questions:

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