if x has the discrete uniform distribution f(x)=1/k for x=1,2,...,k, show that
(a)its mean is μ= (K+1)/2 Proof:μ=E(x)=[x=1 to k]Σxf(x)= [x=1 to k]Σx*1/k=1/k*[x=1 to k]Σx=1/k*（1+k）*k/2=(k+1)/2
(b)its variance is σ^2=(K^2-1)/12Proof:E(x^2)=[x=1 to k]Σx^ 2f (x)= [x=1 to k]Σx^2*1/k=1/k*[x=1 to k]Σx^2=1/k*k*（k+1）(2k+1)/6=（k+1）(2k+1)/6σ^2=E(x^2)- μ^2=（k+1）(2k+1)/6-[(k+1)/2]^2=(k+1)[(2k+1)/6-(k+1)/4]= (k+1)[2(2k+1)/12-3(k+1)/12]=(k+1)[(k-1)/12]=(k^2-1)/12
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if x has the discrete uniform distribution f(x)=1/k for x=1,2,...,k, show that
(a)its mean is μ= (K+1)/2 Proof:μ=E(x)=[x=1 to k]Σxf(x)= [x=1 to k]Σx*1/k=1/k*[x=1 to k]Σx=1/k*（1+k）*k/2=(k+1)/2
(b)its variance is σ^2=(K^2-1)/12Proof:E(x^2)=[x=1 to k]Σx^ 2f (x)= [x=1 to k]Σx^2*1/k=1/k*[x=1 to k]Σx^2=1/k*k*（k+1）(2k+1)/6=（k+1）(2k+1)/6σ^2=E(x^2)- μ^2=（k+1）(2k+1)/6-[(k+1)/2]^2=(k+1)[(2k+1)/6-(k+1)/4]= (k+1)[2(2k+1)/12-3(k+1)/12]=(k+1)[(k-1)/12]=(k^2-1)/12