Question

C = 7.00 + 0.0742N. where N denotes the number of units, N is modelled as a random variable with mean 600 and variance 250. fine the mean and variance of the charges C in their respective units

Answer 1

N=(C-7.00)/0.0742.

Mean of N=S(Nsubj)/n=600, where S is the sum of elements Nsubj for j between 1 and n, where n is the number of elements of N={ N1 N2 N3 ...}. So S(Csubj-7.00)/0.0742n=600=(mean of C)/0.0742-7.00n/0.0742n.

(Mean of C)/0.0742=600+7.00/0.0742=694.33962; mean of C=51.52.

This is the same as C=7.00+0.0742*600=51.52, just substituting the mean of N for N in the equation. I just showed the logic behind it.

The logic for the variance is similar.

Var of N = 250=S((Nsubj-600)^2)/n=S(((Csubj-7.00)/0.0742 - 600)^2)/n.

250=S((Csubj-51.52)^2)/0.0742^2n=(var of C)/0.0742^2, so var of C=250*0.0742^2=1.38 approx.