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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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Most popular questions from this chapter

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