1. Let Y denote the number of broken eggs in a randomly selected carton of one-d

Question

1. Let Y denote the number of broken eggs in a randomly selected carton of one-dozen “store brand” eggs at a local supermarket. Suppose that the probability distribution of Y is as follows: Y 0 1 2 3 4 P (Y) 0.78 0.11 0.07 0.03 (a) Find the probability of having 4 broken eggs in a dozen. (b) What is the probability that at least 10 eggs in a randomly selected carton are unbroken? (c) Calculate the expected (average) number of broken eggs in a carton. (d) Calculate the standard deviation of broken eggs in a carton. (e) What is the probability that the number of broken eggs in the carton will be within 1.5 standard deviations from the mean?

Explanation / Answer

Given Y : 0 1 2 3 4

P(Y=y): 0.78 0.11 0.07 0.03 0.01

a) the probability of having 4 broken eggs in a dozen is P(X = 4 ) = 0.01

Since sum of all probabilties is equal to 1

b) The probability that at least 10 eggs in a randomly selected carton are unbroken

i.e. P(at most 2 eggs in randomly selected carton are broken) =

i.e. P(X = 2) + P(X=1) + P(X=0) = 0.78+0.11+0.07 = 0.96

c)
From the given data

E(Y) = 0.38

d) Var(Y) = 0.82--0.382 = 0.6756

SD(Y) = sqrt(0.6756) = 0.82195

e) P(0..38 - 1.5*0.82195 < Y < 0.38+1.5*0.82195) = P(-0.8523 <Y < 1.613) = P(Y=0) + P(Y=1) = 0.78+0.11 = 0.89

Y P(Y=y) yP(Y=y) y^2P(Y=y) 0 0.78 0 0 1 0.11 0.11 0.11 2 0.07 0.14 0.28 3 0.03 0.09 0.27 4 0.01 0.04 0.16 Total: 0.38 0.82

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