The arithmetic mean of the following frequency distribution: Variable (X) : 0 1 2 3 …. nFrequency (f) : nC0 nC1 nC2 nC3 …. nCnis

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2nn

2nn+1

answer is C.

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Detailed Solution

Let X¯ denote the required mean. Then,X¯=∑r=0n r⋅nCr∑r=0n nCr=∑r=1n r⋅nrn−1Cr−1∑r=0n rnCr=n∑r=1n n−1Cr−12n⇒ X¯=n×2n−12n=n2 ∵∑r=1n n−1Cr−1=2n−1

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