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AIEEE 2011 Solved Paper

Section: Mathematics

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Question : 91 of 91

Marks: +1, -0

If the mean deviation about the median of the numbers a,

2 a

50 a

\left|a\right|

4
5
2
3

Solution:

NOTE :

If total no of terms are even then median

=\;\frac{1}{2} [\;\frac{n}{2}.

th term

+ \left(\;\frac{n}{2}+1\right)

th term]

Here total terms

=50

, which is even

\therefore

Median

=\;\frac{1}{2} [\;\frac{50}{2}.

th term

+ \left(\;\frac{50}{2}+1\right)

th term]

=\;\frac{1}{2}[25

th term

+26

th term

]

=\;\frac{1}{2}[25 a+26 a]

=25.5 a

Mean deviation (M.D.) about the median

\;=\;\frac{\sum\limits_{i=1}^{50} \left| x_i-\; \text{ Median }\; \right|}{N}=50 \; \text{ (given) }\;

\; \therefore \; \frac{1}{50} [2 \times |a| \times (0.5+1.5+2.5+\dots+24.5)]=50

\; \Rightarrow 2|a| \times \; \frac{25}{2} \times 25 = 2500

\; \Rightarrow |a| = \frac{2500}{25 \times 25} = 4

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