Statistics-And-Probability Question 598
Question: If the mean deviation of the numbers $ 1,1+d, $ $ 1+2d,…1+100d $ from their mean is 255, then d is equal to:
Options:
A) 20.0
B) 10.1
C) 20.2
D) 10.0
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Mean $ =\frac{101+d(1+2+3+…+100)}{101} $ $ =1+\frac{d\times 100\times 101}{101\times 2}=1+50d $ $ \because $ Mean deviation from the mean = 255
$ \Rightarrow \frac{1}{101}[| 1-(1+50d) |+| (1+d)-(1+50d) |+| (1+2d)-(1+5d) | $ $ -(1+50d)|+…+| (1+100d)-(1+50d) |]=255 $
$ \Rightarrow 2d[1+2+3+…+50]=101\times 255 $
$ \Rightarrow 2d\times \frac{50\times 51}{2}=101\times 255 $
$ \Rightarrow d=\frac{101\times 255}{50\times 51}=10.1 $
BETA
