Let $S$ be the set of all values of $a_1$ for which the mean deviation about the mean of 100 consecutive positive integers $a_1, a_2, a_3, \ldots ., a_{100}$ is 25 . Then $S$ is :
Solution
<p>Let ${a_1} = a \Rightarrow {a_2} = a + 1,\,.....\,{a_{100}} = a + 90$</p>
<p>$\mu = {{{{100a + (99 \times 100)} \over 2}} \over {100}} = a + {{99} \over 2}$</p>
<p>M.D $$ = {{\sum {|{a_1} - \mu |} } \over {100}} = {{{{\left( {{{99} \over 2} + {{97} \over 2}\, + \,...\, + \,{1 \over 2}} \right)}^2}} \over {100}} = {{2500} \over {100}} = 25$$</p>
<p>$\therefore$ $a \to z$</p>
About this question
Subject: Mathematics · Chapter: Statistics · Topic: Measures of Central Tendency
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