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A manuscript is sent to a typing firm consisting of typists A, B and C. If it is typed by A, then the number of errors made is a Poisson random variable with mean 2.6; if typed by B then the number of errors made is a Poisson random variable with mean 3; and if typed by C then it is a Poisson random variable with mean 3.4. Let X denote the number of errors in the manuscript. Assume that each typist is equally likely to do the work. (a) Find E(X) (b) Find Var(X)

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Answer:

(a) E(X) = 3

(b) Var(X) = 12.1067

Explanation:

(a) E[X]

E[X]T = E[X]T=A + E[X]T=B + E[X]T=C

         = (2.6 + 3 + 3.4)/3

         = 2.6 (1/3) + 3(1/3) + 3.4(1/3)

         = 2.6/3 + 1 + 3.4/3

         = 3

(b) Var (X) = E[X²]−(E[X])²

Recall that if Y ∼ Pois(λ), then E[Y 2] = λ+λ2. This implies that

E[X²] = [(2.6 + 2.6²) + (3 + 3²) + (3.4 + 3.4²)]/3

         = (9.36 + 12 + 14.96)/3

         = 36.32/3

         = 12.1067

Var(X) = E[X²]−(E[X])²

          = 12 - 3²

          = 12.1067 - 9

          = 3.1067

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