Let r denote the number of broken eggs in a randomly selected carton of one doze
ID: 3199999 • Letter: L
Question
Let r denote the number of broken eggs in a randomly selected carton of one dozen eggs. suppose that the probability distribution of is as follows. Only y values of 0, 1, 2, 3, and 4 have positive probabilities. What is pr4)? How would you interpret p(1) 020? The proportion of egg that will be broken in each carton mom this populations 0.20. If you check large number of cartons, the proportion that will have at most one broken egg equal to. The probability of one randomly chosen carton having broken eggs in it In the long run the proportion of cartons that have exactly one broken egg will equal 0.20. Calculate P(y lessthanorequalto 2), the probability that the carton contains at two broken eggs, Interpret this probability. The proportion of eggs that will be broken in any two cartons from this population is 0.9s. The probability two randomly chosen cartons having broken them If you unlock a large number of cartoons, the processor that will have at meant two broken egg will equal to 0.05. In this long run, the probability of cartoons that have exactly two broken egg will equal to 0.095.Explanation / Answer
Part a
We know P(Y) = 1
So, P(4) = 1 – (0.64+0.20+0.11+0.03) = 1 – 0.98 = 0.02
Required probability = 0.02
Part b
How would you interpret P(1) = 0.20
Answer:
In the long run, the proportion of carton that have exactly one broken egg will equal 0.20.
Part c
P(Y2) = P(Y=0) + P(Y=1) + P(Y=2)
P(Y2) = 0.64 + 0.20 + 0.11
P(Y2) = 0.95
Required probability = 0.95
Interpretation:
If you check a large number of cartons, the proportion that will have at most two broken eggs will equal 0.95.
Part d
P(Y<2) = P(Y=0) + P(Y=1)
P(Y<2) = 0.64 + 0.20
P(Y<2) = 0.84
Required probability = 0.84
The probability is less than the probability in part c because the event y = 2 is not included.
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