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**MathRules!** (i)Given the name of the distribution of X; (ii)find the values of ” and σ^2.

(a)M(t) = (0.3 + 0.7e^t)^5

(b)M(t) = 0.45 + 0.55e^t

(c)M(t) = 0.3e^t + 0.4e^2t + 0.2e^3t + 0.1e^4t

For (a) I said that it was a binomial distribution for part (i). Then for part (ii),

M'(t) = 5(0.3 + 0.7e^t)^4(0.7e^t)

M'(0) = 5(0.3 + 0.7e^0)^4(0.7e^0) = 5 (0.7) = 3.5

therefore ” = 3.5

M''(t) = 20(0.3 + 0.7e^t)^3(0.7e^t)^2 + 5(0.3 + 0.7e^t)^4(0.7e^t)

M''(0) = 20(0.3 + 0.7e^0)^3(0.7e^0)^2 + 5(0.3 + 0.7e^0)^4(0.7e^0) = 13.3

therefore σ^2 = M''(0) - [M'(0)]^2 = 13.3 - 3.5^2 = 1.05

For (b) I said that it was a Bernoulli distribution for part (i). Then for part (ii),

” = np = 1(0.55) = 0.55

σ^2 = np (1-p) = 0.55 ( 1- 0.55) = 0.2475

For (c) I said that it was not a usual distribution

” = M'(0) = 2.1

σ^2 = M''(0) - [M'(0)]^2 = 3.2 - 2.1 = 1.1 <---- I am very not sure here

Is my logic correct?