Let y denote the number of broken eggs in a randomly selected carton of one dozen eggs. Suppose that the probability distribution of y is as follows. y 0 1 2 3 4 p(y) 0.63 0.21 0.11 0.03 ? (a) Only y values of 0, 1, 2, 3, and 4 have positive probabilities. What is p(4)? (Hint: Consider the properties of a discrete probability distribution.) p(4) = (b) How would you interpret p(1) = 0.21? In the long run, the proportion that will have at most one broken egg will equal 0.21.In the long run, the proportion of cartons that have exactly one broken egg will equal 0.21. The probability of one randomly chosen carton having broken eggs in it is 0.21.The proportion of eggs that will be broken in each carton from this population is 0.21. (c) Calculate P(y ≤ 2), the probability that the carton contains at most two broken eggs. P(y ≤ 2) = Interpret this probability. The probability of two randomly chosen cartons having broken eggs in them is this probability.In the long run, the proportion that will have at most two broken eggs will equal this probability. In the long run, the proportion of cartons that have exactly two broken eggs will equal this probability.The proportion of eggs that will be broken in any two cartons from this population is this probability. (d) Calculate P(y < 2), the probability that the carton contains fewer than two broken eggs. P(y < 2) = 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 event y = 2 is now not included. 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.

Let y denote the number of broken eggs in a randomly selected carton of one dozen eggs. Suppose that the probability distribution of y is as follows. y 0 1 2 3 4 p(y) 0.63 0.21 0.11 0.03 ? (a) Only y values of 0, 1, 2, 3, and 4 have positive probabilities. What is p(4)? (Hint: Consider the properties of a discrete probability distribution.) p(4) = (b) How would you interpret p(1) = 0.21? In the long run, the proportion that will have at most one broken egg will equal 0.21.In the long run, the proportion of cartons that have exactly one broken egg will equal 0.21. The probability of one randomly chosen carton having broken eggs in it is 0.21.The proportion of eggs that will be broken in each carton from this population is 0.21. (c) Calculate P(y ≤ 2), the probability that the carton contains at most two broken eggs. P(y ≤ 2) = Interpret this probability. The probability of two randomly chosen cartons having broken eggs in them is this probability.In the long run, the proportion that will have at most two broken eggs will equal this probability. In the long run, the proportion of cartons that have exactly two broken eggs will equal this probability.The proportion of eggs that will be broken in any two cartons from this population is this probability. (d) Calculate P(y < 2), the probability that the carton contains fewer than two broken eggs. P(y < 2) = 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 event y = 2 is now not included. 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.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

Let y denote the number of broken eggs in a randomly selected carton of one dozen eggs. Suppose that the probability distribution of y is as follows.

y	0	1	2	3	4
p(y)	0.63	0.21	0.11	0.03	?

(a)

Only y values of 0, 1, 2, 3, and 4 have positive probabilities. What is p(4)? (Hint: Consider the properties of a discrete probability distribution.)

p(4) =

(b)

How would you interpret

p(1) = 0.21?

In the long run, the proportion that will have at most one broken egg will equal 0.21.In the long run, the proportion of cartons that have exactly one broken egg will equal 0.21. The probability of one randomly chosen carton having broken eggs in it is 0.21.The proportion of eggs that will be broken in each carton from this population is 0.21.

(c)

Calculate

P(y ≤ 2),

the probability that the carton contains at most two broken eggs.

P(y ≤ 2) =

Interpret this probability.

The probability of two randomly chosen cartons having broken eggs in them is this probability.In the long run, the proportion that will have at most two broken eggs will equal this probability. In the long run, the proportion of cartons that have exactly two broken eggs will equal this probability.The proportion of eggs that will be broken in any two cartons from this population is this probability.

(d)

Calculate P(y < 2), the probability that the carton contains fewer than two broken eggs.

P(y < 2) =

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 event y = 2 is now not included. 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.

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