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14 January, 09:51

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.

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  1. 14 January, 10:09
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    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
  2. 14 January, 11:09
    0
    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
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