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JEE Main 2009
Exam
JEE Main
Year
2009
Subject
Mathematics
Chapter
Statistics

JEE Main 2009 · Mathematics · Statistics

If the mean deviation of number 1,1+d, 1+2 d, ., 1+100 d from their...

If the mean deviation of number 1,1+d,1+2d,….,1+100d1,1+d, 1+2 d, \ldots ., 1+100 d from their mean is 255 , then the dd is equal to

Options

  1. A10.010.0
  2. B20.020.0
  3. C10.110.1
  4. D20.220.2

Correct answer

C. 10.110.1

Step-by-step solution

 Mean (xˉ)= sum of  quantities n=n2(a+1)n̸=12[1+1+100d]=1+50d M.D. =1n∑∣xi−xˉ∣⇒255=1101[50d+49d+48d+….+d+0+d+……+50d]=2d101[50×512]⇒d=255×10150×51=10.1\begin{aligned}& \text{ Mean }(\bar{x})=\frac{\text{ sum\text{ of } quantities }}{n}=\frac{\frac{n}{2}(a+1)}{\not n}=\frac{1}{2}[1+1+100 d]=1+50 d \\ & \text{ M.D. }=\frac{1}{n} \sum\left|x_i-\bar{x}\right| \Rightarrow 255=\frac{1}{101}[50 d+49 d+48 d+\ldots .+d+0+d+\ldots \ldots+50 d]=\frac{2 d}{101} \left[\frac{50 \times 51}{2}\right]\\ & \Rightarrow d=\frac{255 \times 101}{50 \times 51}=10.1 \end{aligned}
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