TS EAMCET · Maths · Statistics
If 10 is the mean deviation of ' \(n\) ' observations \(x_1, x_2\), \(x_3, \ldots, x_n\) then the mean deviation of the observations \(\frac{2 x_1+5}{3}, \frac{2 x_2+5}{3}, \frac{2 x_3+5}{3}, \ldots ., \frac{2 x_n+5}{3}\) is
- A \(\frac{25}{3}\)
- B \(\frac{40}{9}\)
- C \(\frac{20}{3}\)
- D 15
Answer & Solution
Correct Answer
(C) \(\frac{20}{3}\)
Step-by-step Solution
Detailed explanation
Given observation are \(x_1, x_2, x_3 \ldots \ldots x_n\). \[ \bar{x}=\frac{x_1+x_2+x_3+\ldots . .+x_n}{n} \] New \(\bar{x}=\frac{1}{n}\left[\frac{2 x_1+5}{3}+\frac{2 x_2+5}{3}+\ldots . .+\frac{2 x_n+5}{3}\right]\) New…
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