InterviewSolution
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InterviewSolution
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The arithmetic mean of the following frequency distribution: `{:("Variate(X)" ,:,0,1,2,3, ..., n),("Frequency" (f),:,""^(n)C_(0),""^(n)C_(1),""^(n)C_(2),""^(n)C_(3),...,""^(n)C_(n)):}` isA. `(2^(n))/(n)`B. `(2^(n))/(n+1)`C. `(n)/(2)`D. `(2)/(n)` |
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Answer» Correct Answer - C Let `bar(X)` denote the required mean, Then, `bar(X)=(sum_(r=0)^(n)r.""^(n)C_(r))/(sum_(r=0)^(n)""^(n)C_(r))=(sum_(r=1)^(n)r.(n)/(r)""^(n-1)C_(r-1))/(sum_(r=0)^(n)r""^(n)C_(r))=(n sum_(r=1)^(n)""^(n-1)C_(r-1))/(2^(n))` ` rArr bar(X)=(n xx 2^(n-1))/(2^(n))=(n)/(2) " "[ therefore sum_(r=1)^(n)""^(n-1)C_(r-1)=2^(n-1)]` |
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