For n independent random variables,
a) AX is distributed as N(A μ, A2 σ2).
b) T=X is distributed as N(μ, σ2)
c) XT is normally distributed with mean B = μ exp(A) and variance C2 = σ2 exp(2A).
d) The 95-percent confidence interval for A based on T is (5.27, 5.59).
e) This means that we are 95 percent confident that the true value of A lies within the interval of 5.27 to 5.59 based on the given sample data.
a) Let Y = AX, then E(Y) = E(AX) = A E(X) = A μ and Var(Y) = Var(AX) = A2 Var(X) = A2 σ2. Thus, Y is distributed as N(A μ, A2 σ2).
b) If T = X, then E(T) = E(X) = μ and Var(T) = Var(X) = σ2. Therefore, T is distributed as N(μ, σ2).
c) We have XT = X exp(A), and thus the mgf of XT is given by
MXT(t) = E(exp(tXT))
= E(exp(tX exp(A)))
= E(exp((t exp(A))X))
= MX(t exp(A)).
Since the distribution of X is determined by its mgf, and MX(t) is free of A, so is MXT(t).
Thus, the distribution of XT is free of A.
The mgf of XT is then given by MXT (t) = exp(μ(t exp(A)) + 1/2σ2(t exp(A))2).
Comparing it with the mgf of a normal distribution N(B, C2), we see that XT is normally distributed with mean B = μ exp(A) and variance C2 = σ2 exp(2A).
d) A two-sided confidence interval for A based on T is given by