Chapter 18, Problem 23E

To determine

To find out what is the probability that the shipment will be accepted anyway.

Answer to Problem 23E

  $0.088$ .

Explanation of Solution

It is given in the question that when a truckload of apples arrives at a packing plant a random sample of $150$ is selected and examined. The whole truckload will be rejected if more than $5\%$ of the sample is unsatisfactory and also suppose that in fact $8\%$ of the apples on the truck do not meet the desired standard. Thus, let us first verify the assumptions and conditions for the analysis as:

Random condition: It is satisfied as we assume that the apples were randomly selected.

  $10\%$ condition: The sample size less than $10\%$ of the population size.

Success/failure condition: as $np=12,n(1-p)=138$ and both are greater than ten then this is also true.

Thus, all the conditions and assumptions are met. Thus, now find the mean and the standard deviation for the following as:

  $\begin{array}{l}\mu =p=0.08\\ \sigma =\sqrt{\frac{pq}{n}}\\ =\sqrt{\frac{0.08(0.92)}{150}}\\ =0.022\end{array}$

Thus, let us find the z -score to standardized as:

  $\begin{array}{l}z=\frac{0.05-0.08}{0.022}\\ =-1.354\end{array}$

So, the probability that the shipment will be accepted anyway is calculated as:

  $\begin{array}{l}P(z>1.399)=normalcdf(-E99,-1.354,0,1)\\ =0.088\end{array}$

Therefore, the probability that the shipment will be accepted anyway is $0.088$ .