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Question
If the mean of observation \[x_1 , x_2 , . . . . , x_n is x\] then the mean of x1 + a, x2 + a, ....., xn + a is
Options
a`overlineX`
`overlineX -a`
`overlineX +a`
`overlineX/a`
Solution
The mean of \[x_1 , x_2 , . . . , x_n\text { is } \]`overlineX`.
\[\therefore \frac{x_1 + x_2 + x_3 + . . . + x_n}{n}\] = `overlineX`
\[ \Rightarrow x_1 + x_2 + x_3 + . . . + x_n = n\]`overlineX`
Mean of x1 + a, x2 + a, ... , xn + a
\[= \frac{\left( x_1 + a \right) + \left( x_2 + a \right) + \left( x_3 + a \right) + . . . + \left( x_n + a \right)}{n}\]
\[ = \frac{\left( x_1 + x_2 + x_3 + . . . + x_n \right) + \left( a + a + a + . . . + a \right)}{n}\]
`= (noverlineX + na)/n`
`= overlineX + a`
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