If the mean of a set of observations x_1,x_2,x_3,....,x_n is x¯, then the mean of observations x_i+2i, ∀i=1,2,3,....n is equal to

Correct answer: (B) x¯+n+1. Given, nx¯=x_1+x_2+x_3+...+x_n Now, new observations are x_1+2,x_2+2.2,....x_n+2.n ∴ new mean =(x_1+2+x_2+2.2+x_3+2.3+....+x_n+2.n)/n =(x_1+x_2+x_3+...+x_n)/n+(21+2+3+...+n)/n =x¯+(2n+1)/2

Q27 / 77 · NTA Abhyas (JEE Main) 2020 · Mathematics · Statistics

If the mean of a set of observations $x_{1}, x_{2}, x_{3}, . . . ., x_{n}$ is $\bar{x},$ then the mean of observations $x_{i} + 2 i, \forall i = 1,2, 3, . . . . n$ is equal to

A
$\bar{x} + 2 (n + 1)$
Correct answer
B
$\bar{x} + (n + 1)$
C
$\bar{x} + \frac{n + 1}{2 n}$
D
None of these

Solution

Given,

n \bar{x} = x_{1} + x_{2} + x_{3} + . . . + x_{n}

Now, new observations are

x_{1} + 2, x_{2} + 2.2, . . . . x_{n} + 2 . n

∴

new mean

= \frac{(x_{1} + 2) + (x_{2} + 2.2) + (x_{3} + 2.3) + . . . . + (x_{n} + 2 . n)}{n}

= \frac{x_{1} + x_{2} + x_{3} + . . . + x_{n}}{n} + \frac{2 (1 + 2 + 3 + . . . + n)}{n}

= \bar{x} + \frac{2 (n + 1)}{2}

Hence, the correct answer is option: 'B'.

If the mean of a set of observations x1,x2,x3,....,xn is x¯, then the mean of observations xi+2i, ∀i=1,2,3,....n is equal to

Solution

More Statistics Questions

If the mean of a set of observations $x_{1}, x_{2}, x_{3}, . . . ., x_{n}$ is $\bar{x},$ then the mean of observations $x_{i} + 2 i, \forall i = 1,2, 3, . . . . n$ is equal to