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Chapter 6: Problem 10

Consider the random variable \(y=\) the number of broken eggs in a randomly selected carton of one dozen eggs. Suppose the probability distribution of \(y\) is as follows: \(\begin{array}{cccccc}y & 0 & 1 & 2 & 3 & 4 \\ p(y) & 0.65 & 0.20 & 0.10 & 0.04 & ?\end{array}\) 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 \(p(1)=0.20 ?\) c. Calculate \(P(y \leq 2)\), the probability that the carton contains at most two broken eggs, and interpret this probability. 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)? e. What is the probability that the carton contains exactly 10 unbroken eggs? f. What is the probability that at least 10 eggs are unbroken?

Short Answer

Expert verified
a. \(p(4)=0.01\) b. \(p(1)=0.20\) means there is a 20% chance of having exactly 1 broken egg in a randomly selected carton. c. \(P(y \leq 2)= 0.95\), which suggests a 95% chance of having at most 2 broken eggs in a randomly chosen carton. d. \(P(y < 2)=0.85\), which is smaller than \(P(y \leq 2)=0.95\) because it doesn't include cartons with exactly 2 broken eggs. e. \(P(\text{exactly 10 unbroken eggs}) = 0.10\) f. \(P(\text{at least 10 unbroken eggs}) = 0.95\)

Step by step solution

01

a. Find \(p(4)\)

Since we only have the \(y\) values of \(0\), \(1\), \(2\), \(3\), and \(4\), and the probabilities must add up to 1, we can determine \(p(4)\) using the following equation: \[1= p(0)+p(1)+p(2)+p(3)+p(4)\] \[p(4)=1-0.65-0.20-0.10-0.04=0.01\] The probability of having 4 broken eggs in the carton is \(0.01\). b. Interpret the probability of having exactly 1 broken egg in the carton
02

b. Interpretation of \(p(1)=0.20\)

The probability \(p(1)=0.20\) means that if we randomly select a carton of one dozen eggs, there is a 20% chance that the carton contains exactly one broken egg. c. Calculate and interpret the probability of having at most 2 broken eggs in the carton.
03

c. Calculate and interpret \(P(y \leq 2)\)

To calculate \(P(y \leq 2)\), we need to sum the probabilities for \(y=0\), \(y=1\), and \(y=2\): \[P(y \leq 2) = p(0) + p(1) + p(2) = 0.65 + 0.20 + 0.10 = 0.95\] The probability is 0.95, which suggests that 95% of the time, when choosing a carton of one dozen eggs randomly, we can expect it to have at most 2 broken eggs inside. d. Calculate and compare the probability of having fewer than 2 broken eggs in the carton.
04

d. Calculate and compare \(P(y < 2)\)

To calculate \(P(y < 2)\), we need to sum the probabilities for \(y=0\) and \(y=1\): \[P(y < 2) = p(0) + p(1) = 0.65 + 0.20 = 0.85\] \(P(y < 2)=0.85\) is smaller than the probability \(P(y \leq 2)=0.95\) because \(P(y<2)\) only considers cartons with \(0\) or \(1\) broken eggs, whereas \(P(y \leq 2)\) also includes cartons having exactly 2 broken eggs. e. Find the probability of having exactly 10 unbroken eggs in the carton.
05

e. Calculate \(P(\text{exactly 10 unbroken eggs})\)

If a carton contains exactly 10 unbroken eggs, it means that there are 2 broken eggs in the carton. So we have: \[P(\text{exactly 10 unbroken eggs}) = p(2) = 0.10\] There is a 10% probability that a randomly chosen carton has exactly 10 unbroken eggs. f. Determine the probability of having at least 10 unbroken eggs in the carton.
06

f. Calculate \(P(\text{at least 10 unbroken eggs})\)

If a carton has at least 10 unbroken eggs, it means that the carton can contain 0, 1, or 2 broken eggs. Thus, we need to sum the probabilities for \(y=0\), \(y=1\), and \(y=2\): \[P(\text{at least 10 unbroken eggs}) = P(y \leq 2) = 0.65 + 0.20 + 0.10 = 0.95\] There is a 95% probability that a randomly chosen carton has at least 10 unbroken eggs.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Random Variable
A random variable is a numerical description of the outcome of a statistical experiment. It's a variable that can take on different values, each associated with a probability. When we say a random variable 鈥榅鈥� represents the number of heads in a coin toss, we鈥檙e defining a function that assigns numbers to each possible outcome of the coin tosses. In the case of our exercise, the random variable 'y' represents the number of broken eggs in a carton.

A random variable can be discrete or continuous. Discrete random variables have countable outcomes (like the number of eggs), while continuous random variables have infinite outcomes within a range (like the height of students in a class). The provided example of 'y' is a discrete random variable because the count of broken eggs can only be whole numbers within a range of 0 to, presumably, 12 for a dozen.
Cumulative Probability
The concept of cumulative probability refers to the probability of a random variable falling within a certain range. It's the sum of the probabilities of the variable taking on values less than or equal to a specific value. The notation P(y ≤ k) is often used, where 'y' is our random variable and 鈥榢鈥� is the value up to which probabilities are being added together.

In the context of our egg carton problem, when calculating P(y ≤ 2), we added the probabilities of y = 0, y = 1, and y = 2. This cumulative probability tells us about the likelihood of getting up to 2 broken eggs in our statistical experiment. Cumulative distributions are very useful in understanding the overall probability trends for certain ranges of our random variable's possible values.
Probability Interpretation
The probability interpretation helps us understand what the probability value signifies in practical terms. It鈥檚 a narrative of what the probability means in real life. For instance, if we have p(1) = 0.20 for our egg-carton problem, it means that there's a 20% chance that a randomly chosen carton of eggs will contain exactly one broken egg.

Understanding these interpretations aids in decisions based on probability. For example, if a grocery store manager observes that P(y ≤ 2) = 0.95, they can infer that there鈥檚 a high likelihood the cartons will have 0 to 2 broken eggs, which might be acceptable for quality standards. Conversely, a low probability of finding cartons with no broken eggs might indicate an issue in handling or packaging that needs addressing.

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Most popular questions from this chapter

A business has six customer service telephone lines. Consider the random variable \(x=\) number of lines in use at a randomly selected time. Suppose that the probability distribution of \(x\) is as follows: \(\begin{array}{lccccccc}x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ p(x) & 0.10 & 0.15 & 0.20 & 0.25 & 0.20 & 0.06 & 0.04\end{array}\) a. Calculate the mean value and standard deviation of \(x\). b. What is the probability that the number of lines in use is farther than 3 standard deviations from the mean value?

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The paper referenced in Example 6.21 suggested that a normal distribution with mean 3500 grams and standard deviation 550 grams is a reasonable model for birth weights of babies born in Canada. a. One common medical definition of a large baby is any baby that weighs more than 4000 grams at birth. What is the probability that a randomly selected Canadian baby is a large baby? b. What is the probability that a randomly selected Canadian baby weighs either less than 2000 grams or more than 4000 grams at birth? c. What birth weights describe the \(10 \%\) of Canadian babies with the greatest birth weights? 6.52 The time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated by a normal distribution with a mean of 120 seconds and a standard deviation of 20 seconds. The fastest \(10 \%\) are to be given advanced training. What task times qualify individuals for such training?

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