What is the mean number of smokers in the sample?what is standard deviation?

A) EX+3Y = 44 Var X+3Y = 10 EX-3Y = -4 Var X-3Y = 10 B)7C4 .64^4 x.36^3 = 0.274 C) N = 18 P = .28 Binomial to Normal transformations. Mean = NP Var = NP(1-P) Mean = 5.04 Var = 3.6288 SD = 1.905 So Mean = 5.04 SD = 1.905.

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