Refer to Exercise 3.145. Use the uniqueness of moment-generating functions to give the distribution of a random variable with moment-generating function m (t) = (.6et +.4) 3.

The random variable with m (t) = (0.6e^t + 0.4) ^3

follows binomial random variable with parameters p = 0.6 and n = 3

Step-by-step explanation:

Given that p + q = 1

q = 1 - p

The general form of moment generating function, MGF, m (t) is given as

m (t) = [pe^t + q]^n for a binomial distribution

Comparing this to the moment-generating function to [0.6e^t + 0.4]^3

These m (t) functions are exactly the same with

p = 0.6,

q = 1 - 0.6

q = 0.4

n = 3.

Thus, the random variable with m (t) = (0.6e^t + 0.4) ^3

follows binomial random variable with parameters p = 0.6 and n = 3

Answer: it follows a binomial random variable with parameters p = 0.6 and n=3

Step-by-step explanation:

Let's look at the moment - generating function of a binomial distribution which is given as,

m (t) = [pet+q]n. If we look closely, these moment - generating functions are exactly the same or we can say they are identical to each other, where

p = 0.6, q = 0.4 and n = 3. Thus, the random variable with moment-generating function m (t) = (0.6et + 0.4) 3 follows binomial random variable with parameters p = 0.6 and n = 3