Let y denote the number of broken eggs in a randomly selected carton of one doze
ID: 3261873 • Letter: L
Question
Let y denote the number of broken eggs in a randomly selected carton of one dozen "store brand" eggs at a certain market. Suppose that the probability distribution of y is as follows.
(a) Only y values of 0, 1, 2, 3, and 4 have positive probabilities. What is p(4)?
(b) How would you interpret p(1) = .19?
The proportion of eggs that will be broken in each carton from this population is .19.If you check a large number of cartons, the proportion that will have at most one broken egg will equal .19. In the long run, the proportion of cartons that have exactly one broken egg will equal .19.The probability of one randomly chosen carton having broken eggs in it is .19.
(c) Calculate P(y 2), the probability that the carton contains at most two broken eggs.
Interpret this probability.
The proportion of eggs that will be broken in any two cartons from this population is .0.95.The probability of two randomly chosen cartons having broken eggs in them is .0.95. If you check a large number of cartons, the proportion that will have at most two broken eggs will equal .0.95.In the long run, the proportion of cartons that have exactly two broken eggs will equal .0.95.
(d) Calculate P(y < 2), the probability that the carton contains fewer than two broken eggs.
Why is this smaller than the probability in part (c)?
This probability is less than the probability in part (c) because in probability distributions, P(y k) is always greater than P(y < k).This probability is less than the probability in part (c) because the proportion of eggs with any exact number of broken eggs is negligible. This probability is not less than the probability in part (c) because the two probabilities are the same for this distribution.This probability is less than the probability in part (c) because the event y = 2 is now not included.
(e) What is the probability that the carton contains exactly 10 unbroken eggs? (Hint: What is the corresponding value of y?)
(f) What is the probability that at least 10 eggs are unbroken?
Explanation / Answer
a)
p(4) = 1 - (0.64+0.19+0.12+0.03) = 1 - 0.98 = 0.02
b)
p(1) = 0.19
About 19% of the cartons have exactly one broken egg.
That is in the long run, the proportion of cartons that have exactly one broken egg will equal to 19.
c)
p(y<=2) = p(y=0) + p(y=1) + p(y=2)
= 0.64 +0.19 + 0.12
= 0.95
If you check a large number of cartons, the proportion that will have at most two broken eggs will equal 0.95
d)
p(y<2) = p(y=0) + p(y=1)
= 0.64 + 0.19
= 0.83
This probability is less than the probability in part (c) because the event y = 2 is now not included.
e)
In one dozen cartons, 10 unbroken egg mean 2 broken eggs.
p(y=2) = 0.12
That is probability that the carton contains exactly 10 unbroken eggs is 0.12.
f)
At least 10 unbroken egg means 10,11,12 unbroken this means 2,1,0 broken eggs.
p(0,1,2) = 0.64+0.19+0.12 = 0.95
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