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Suppose \(y=\) the number of broken eggs in a randomly selected carton of one dozen eggs. The probability distribution of \(y\) is as follows: $$ \begin{array}{lccccc} y & 0 & 1 & 2 & 3 & 4 \\ p(y) & 0.65 & 0.20 & 0.10 & 0.04 & 0.01 \end{array} $$ a. Calculate and interpret \(\mu_{y}\). (Hint: See Example 6.13) b. In the long run, for what percentage of cartons is the number of broken eggs less than \(\mu_{y} ?\) Does this surprise you? c. Explain why \(\mu_{y}\) is not equal to \(\frac{0+1+2+3+4}{5}=2.0\).

Short Answer

Expert verified
\(\mu_{y} = 0.56\), indicating on average, there are expected to be 0.56 broken eggs per carton. 65% of cartons have a number of broken eggs less than \(\mu_{y}\). The expected value (\(\mu_{y}\)) is not a straightforward average of all possible outcomes, but a weighted average that accounts for the probability of each outcome. Hence, it is not equal to 2.0.

Step by step solution

01

Calculate \(\mu_{y}\)

Given the probability distribution function, \(y\) and \(p(y)\), the mean or expected value, \(\mu_{y}\), can be calculated as the sum of the product of each probability and its corresponding value. Thus we get, \(\mu_{y} = (0*0.65)+(1*0.20)+(2*0.10)+(3*0.04)+(4*0.01)=0+0.2+0.2+0.12+0.04=0.56.\)
02

Answer Part b: Interpret \(\mu_{y}\) and its relation with broken eggs

As per the calculation, \(\mu_{y} = 0.56\) which represents the expected number of broken eggs in a carton. This implies that if we were to randomly select many cartons, the average number of broken eggs per carton would be approximately 0.56. To find the percentage of cartons that have broken eggs less than \(\mu_{y}=0.56\), we need to sum up the probabilities of getting 0 broken eggs. Here, that is \(p(0)=0.65\) or 65%. This might be surprising because although the mean is 0.56, 65% of the time, the carton has no broken eggs which is less than the mean.
03

Explain why \(\mu_{y}\) is not simply an average of the outcomes

Expected value in a probability distribution is not simply the average of all outcomes, it is a weighted average where each outcome is weighted by its probability of occurrence. This takes into account the fact that some outcomes may be more likely than others. Therefore, in this case, even though the simple average of 0, 1, 2, 3, and 4 is 2.0, the expected number of broken eggs \(\mu_{y}=0.56\) is less than 2 because of the higher probabilities associated with fewer broken eggs.

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Key Concepts

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

Expected Value
When working with probability distributions, the expected value is a fundamental concept that helps us understand long-term averages. In simple terms, it represents the mean of a random variable across many trials.
To find the expected value, also denoted as \( \mu \), we multiply each possible outcome of a random variable by its probability, then sum all these values. For example, if we have a random variable \( y \) with a certain probability distribution as shown above, the expected value \( \mu_y \) is calculated as follows:
  • Multiply each outcome \( y \) by its probability \( p(y) \).
  • Add these products together.
This computation results in the average number of broken eggs you would expect per carton in the long run. Unlike a simple arithmetic average, it accounts for the probabilities of each result, providing a weighted average that reflects real-world scenarios more accurately. That’s why, in our example, even though the simple average of broken eggs is 2.0, the expected value is actually 0.56.
Random Variables
In probability theory, a random variable is a variable that can take on different values based on the outcome of a random event. Its value is not certain but can be described in terms of probabilities.
Random variables can be discrete, like in our egg carton example, where values are distinct and countable such as 0, 1, 2, 3, or 4 broken eggs. Each value corresponds to a probability that tells us how likely each outcome is to occur.
  • The random variable \( y \) is defined as the number of broken eggs.
  • Its possible values are integers corresponding to the number of broken eggs in a carton.
  • The probability \( p(y) \) expresses how frequently each scenario happens.
Understanding random variables is crucial for calculating expected values and interpreting the results in a meaningful way. It provides the foundational basis for constructing probability distributions.
Probability Weights
Probability weights play a critical role in determining the expected value of a random variable. These weights are essentially the probabilities assigned to each possible outcome and indicate the relative frequency or likelihood of each occurring.
In the context of our example, the following steps explain why probability weights matter:
  • Each outcome of broken eggs, like 0 or 1, is multiplied by its probability (weight).
  • These weighted values are combined to compute the expected value.
The reason \( \mu_y \) is not just the average of the numbers 0, 1, 2, 3, and 4 is because the probabilities, or weights, differ and are not equal. Most weights are skewed towards fewer broken eggs, reflecting that it’s more common to have 0 or 1 broken egg than 3 or 4. This impact of unequal weights results in an expected value of 0.56 rather than the simple average of 2.0, providing a more accurate measure that considers both value and frequency.

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Most popular questions from this chapter

A chemical supply company currently has in stock 100 pounds of a certain chemical, which it sells to customers in 5 -pound lots. Let \(x=\) the number of lots ordered by a randomly chosen customer. The probability distribution of \(x\) is as follows: $$ \begin{array}{lcccc} x & 1 & 2 & 3 & 4 \\ p(x) & 0.2 & 0.4 & 0.3 & 0.1 \end{array} $$ a. Calculate and interpret the mean value of \(x\). b. Calculate and interpret the variance and standard deviation of \(x\).

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The Los Angeles Times (December 13,1992 ) reported that what \(80 \%\) of airline passengers like to do most on long flights is rest or sleep. Suppose that the actual percentage is exactly \(80 \%,\) and consider randomly selecting six passengers. Then \(x=\) the number among the selected six who prefer to rest or sleep is a binomial random variable with \(n=6\) and \(p=0.8\) a. Calculate \(p(4)\), and interpret this probability. b. Calculate \(p(6),\) the probability that all six selected passengers prefer to rest or sleep. c. Calculate \(P(x \geq 4)\).

Consider the variable \(x=\) time required for a college student to complete a standardized exam. Suppose that for the population of students at a particular university, the distribution of \(x\) is well approximated by a normal curve with mean 45 minutes and standard deviation 5 minutes. a. If 50 minutes is allowed for the exam, what proportion of students at this university would be unable to finish in the allotted time? b. How much time should be allowed for the exam if you wanted \(90 \%\) of the students taking the test to be able to finish in the allotted time? c. How much time is required for the fastest \(25 \%\) of all tudents to complete the exam

Consider the population of all one-gallon cans of dusty rose paint manufactured by a particular paint company. Suppose that a normal distribution with mean \(\mu=5 \mathrm{ml}\) and standard deviation \(\sigma=0.2 \mathrm{ml}\) is a reasonable model for the distribution of the variable \(x=\) amount of red dye in the paint mixture Use the normal distribution to calculate the following probabilities. (Hint: See Example 6.21\()\) a. \(P(x < 5.0)\) b. \(P(x < 5.4)\) c. \(P(x \leq 5.4)\) d. \(P(4.6 < x < 5.2)\) e. \(P(x > 4.5)\) f. \(P(x > 4.0)\)

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