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, n ¯ 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 = ( 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 + (2(1 + 2 + 3 +... + n))/(n) = ¯ x + (2(n + 1))/(2)

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Statistics

NTA Abhyas 2020 · Mathematics · Statistics

If the mean of a set of observations x _ 1 , x _ 2 , x _ 3 ,.... ,...

If the mean of a set of observations

{x}_{1}, {x}_{2}, {x}_{3},.... , {x}_{n}

\overset{¯}{x},

then the mean of observations

{x}_{i} + 2i, \forall i = 1,2, 3,.... n

& is equal to

Options

A $\overset{¯}{x} + 2 \left(n + 1\right)$
B $\overset{¯}{x} + \left(n + 1\right)$
C $\overset{¯}{x} + \dfrac{n + 1}{2n}$
DNone of these

Correct answer

B. $\overset{¯}{x} + \left(n + 1\right)$

Step-by-step solution

Given,

n\overset{¯}{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

= \dfrac{\left({x}_{1} + 2\right) + \left({x}_{2} + 2.2\right) + \left({x}_{3} + 2.3\right) +.... + \left({x}_{n} + 2. n\right)}{n}

= \dfrac{{x}_{1} + {x}_{2} + {x}_{3} +... + {x}_{n}}{n} + \dfrac{2\left(1 + 2 + 3 +... + n\right)}{n}

= \overset{¯}{x} + \dfrac{2\left(n + 1\right)}{2}

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