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Question

Find the mean and variance for the first n natural numbers.

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Solution

Here x = 1, 2, 3, 4,.....,n

x=1+2+3+4+...+n=n(n+1)2

Mean(¯¯¯x)=n(n+1)2n=(n+1)2

x2=(1)2+(2)2+(3)2+(4)2+...+n2

=n(n+1)(2n+1)6

Variance (σ)2=Nx2(x)2N2

=n×n(n+1)(2n+1)6[n(n+1)2]2n2

= n×n(n+1)(2n+1)6[(n+1)24]n2

=4n2+6n+23n26n312=n2112


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