Question

Let y denote the number of broken eggs in a randomly selected carton of one dozen eggs. ply) 0.60 0.20 0.15 0.04 0.01 (a) Cal

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Answer #1

Answer)

A)

Mean = sum of y*y(x)

= 0*0.6 + 1*0.2 + 2*0.15 + 3*0.04 + 4*0.01

= 0.66

B)

P(x<mean) = p(x<0.66) = p(0) = 0.6

Yes as it is large

C)

This computation of mean is incorrect because it does not take into account that the probabilities with which the number broken eggs need to be weighted

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