Consider the random variable y = Number of broken eggs in a randomly selected carton of one dozen eggs.
Suppose the probability distribution of y is as follows.
y 0 1 2 3 4
p(y) 0.64 0.19 0.11 0.05 ?
(a) Only y values of 0, 1, 2, 3, and 4 have probabilities greater than 0. What is p(4)?
(b) How would you interpret p(1) = 0.19?
The proportion of eggs that will be broken in each carton from this population is 0.19.
In the long run, the proportion of cartons that have exactly one broken egg will equal
0.19.
If you check a large number of cartons, the proportion that will have at most one broken egg
will equal 0.19.
The probability of one randomly chosen carton having broken eggs in it is 0.19.
(c) Calculate P(y ≤ 2), the probability that the carton contains at most two broken eggs.
Probability = ______
If you check a large number of cartons, the proportion that will have at most two broken
eggs will equal this value.
The proportion of eggs that will be broken in any two cartons from this population is this
value.
The probability of two randomly chosen cartons having broken eggs in them is this value.
In the long run, the proportion of cartons that have exactly two broken eggs will equal this
value.
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For (a), since the sum of all probabilities must equal 1, we can subtract the probabilities of y=0, y=1, y=2, and y=3 from 1 to find the probability p(4).
p(4) = 1 - 0.64 - 0.19 - 0.11 - 0.05 = 0.01
So, p(4) is equal to 0.01.
For (c), to calculate P(y ≤ 2), we need to sum the probabilities of y=0, y=1, and y=2.
P(y ≤ 2) = 0.64 + 0.19 + 0.11 = 0.94
So, the probability that the carton contains at most two broken eggs is 0.94.