An alternative proof of Theorem 2 may be based on the fact that if X1, X2, ..., and Xn are independent ran-dom variables having the same Bernoulli distribution with the parameter θ, then Y = X1 + X2 +···+ Xn isa random variable having the binomial distribution withthe parameters n and θ.Verify directly (that is, without making use of the factthat the Bernoulli distribution is a special case of thebinomial distribution) that the mean and the variance ofthe Bernoulli distribution are μ = θ and σ2 = θ (1 − θ

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.8: Probability
Problem 32E
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An alternative proof of Theorem 2 may be based on
the fact that if X1, X2, ..., and Xn are independent ran-
dom variables having the same Bernoulli distribution
with the parameter θ, then Y = X1 + X2 +···+ Xn is
a random variable having the binomial distribution with
the parameters n and θ.
Verify directly (that is, without making use of the fact
that the Bernoulli distribution is a special case of the
binomial distribution) that the mean and the variance of
the Bernoulli distribution are μ = θ and σ2 = θ (1 − θ
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How do you show that if X1, X2,...Xn are independent random variables having the same Bernoulli distribution with the parameter θ and Y =X1 + X2 + ... + Xn, then E(Y) = nθ and var(Y) = nθ(1-θ)

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