# Question: In the proof of Theorem 6 6 we twice differentiated the

In the proof of Theorem 6.6 we twice differentiated the moment– generating function of the normal distribution with respect to t to show that E(X) = µ and var(X) = σ2. Differentiating twice more and using the formula of Exercise 4.25 on page 129, find expressions for µ3 and µ4.

## Answer to relevant Questions

If X is a random variable having a normal distribution with the mean µ and the standard deviation s, use the third part of Theorem 4.10 on page 128 and Theorem 6.6 to show that the moment– generating function of Z = X ...With reference to Exercise 6.39, show that for nor–mal distributions k2= σ2 and all other cumulants are zero. In exercise If we let KX(t) = lnMX – µ(t), the coefficient of tr/r! in the Maclaurin’s series of KX(t) is ...If X and Y have a bivariate normal distribution, it can be shown that their joint moment– generating function (see Exercise 4.48 on page 139) is given by Verify that (a) The first partial derivative of this function with ...The number of bad checks that a bank receives during a 5-hour business day is a Poisson random variable with λ = 2. What is the probability that it will not receive a bad check on any one day during the first 2 hours of ...Find z if the standard- normal-curve area (a) Between 0 and z is 0.4726; (b) To the left of z is 0.9868; (c) To the right of z is 0.1314; (d) Between – z and z is 0.8502.Post your question