Question

Exercise \(7.9\) introduced the following probability distribution for \(y=\) the number of broken eggs in a carton: $$ \begin{array}{lrrrrr} y & 0 & 1 & 2 & 3 & 4 \\ p(y) & .65 & .20 & .10 & .04 & .01 \end{array} $$ a. Calculate and interpret \(\mu_{y}\). 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. Why doesn't \(\mu_{y}=(0+1+2+3+4) / 5=2.0\). Explain.

Short answer

\(\mu_{y} = 1.21\), 85% of the time, the breakage is less than the mean. A simple average does not take probabilities into account, therefore, the mean isn't 2.

Step-by-step solution

Calculate Mean (\(\mu_{y}\))

Start by calculating the mean of the distribution. The formula for the mean of a probability distribution is \(\Sigma xp(x)\). So, multiply each value of \(y\) with its corresponding probability and sum up these products: \(\mu_{y} = 0*.65 + 1*.20 + 2*.10 + 3*.04 + 4*.01 = 0.65 + 0.20 + 0.20 + 0.12 + 0.04 = 1.21.\

Calculate Percentage

Next, calculate the percentage of cartons with broken egg count less than \(\mu_{y}\) (1.21). This would be a summation of probabilities for \(y\) values less than 1.21, which are just 0 and 1. Hence, the required percentage is \( p(0) + p(1) = 0.65 + 0.20 = 0.85 or 85%.

Explain Mean Disparity

The expected value or mean of a distribution depends on the associated probabilities of the values. Simply taking an average of the values does not take into account the likelihood of the occurrence of these values. In this case, the lower values of \(y\) (0,1) have a much higher probability of occurrence as compared to the higher ones (2,3,4), which skews the mean towards a lower value. Thus, \(\mu_{y} = 2.0\) only when every value of \(y\) has an equal chance of occurrence which is not the case here.

Interpret Results

It can be observed that 85% of the time, the number of broken eggs in a carton is less than the mean value (1.21). This does not surprise as the probabilities for 0 and 1 broken eggs are quite high, which skews the distribution towards the lower values. The mean isn't 2 because the probabilities of different \(y\) values are not evenly distributed.