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If x̄ is the mean of x1 , x2 , ... , xn , then for a ≠ 0, the mean of ax1 , ax2 , ..., axn , x1/a, x2/a, ….. , xn/a is
a. (a + 1/a) x̄
b. (a + 1/a) x̄/2
c. (a + 1/a) x̄/n
d. [(a + 1/a) x̄]/2n

Solution:

It is given that

Mean of x1, x2, ….., xn is x̄

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

Consider the mean of ax1 , ax2 , ..., axn , x1/a, x2/a, ….. , xn/a is z̄

We get

z̄ = [(ax1 + ax2 + ….. + axn) + (x1/a + x2/a + …. + xn/a)]/ n + n

Taking out a as common

z̄ = [a (x1 + x2 + …. + xn) + 1/a (x1 + x2 + ….. + xn)]/2n

It can be written as

z̄ = [(a + 1/a) (x1 + x2 + …. + xn)]/ 2n

So we get

\(\overline{z}=\frac{(a+1/a)\sum_{i=1}^{n}x_{i}}{2n}\)

Using equation (1)

z̄ = [(a + 1/a)nx̄]/2n

z̄ = [(a + 1/a)x̄]/2

Therefore, the mean is (a + 1/a) x̄/2.

✦ Try This: The mean of five numbers is 14. If one number is excluded, their mean becomes 18. The excluded number is :

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

NCERT Exemplar Class 9 Maths Exercise 14.1 Problem 16

If x̄ is the mean of x1 , x2 , ... , xn , then for a ≠ 0, the mean of ax1 , ax2 , ..., axn , x1/a, x2/a, ….. , xn/a is a. (a + 1/a) x̄, b. (a + 1/a) x̄/2, c. (a + 1/a) x̄/n, d. [(a + 1/a) x̄]/2n

Summary:

If x̄ is the mean of x1 , x2 , ... , xn , then for a ≠ 0, the mean of ax1 , ax2 , ..., axn , x1/a, x2/a, ….. , xn/a is (a + 1/a) x̄/2

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