Chapter 18, Problem 22E

To determine

To find out what is the probability that more than $\text{95\% of the 160}$ seeds in the packet will generate and also discuss the assumptions and conditions that support your model.

Answer to Problem 22E

  $0.081$ .

Explanation of Solution

It is given in the question that information on a pocket of seeds that the germination rate is $92\%$ . Thus, let us first verify the assumptions and conditions for the analysis as:

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

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

Success/failure condition: as $np=147.2,n(1-p)=12.8$ 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:

  $\begin{array}{l}\mu =p=0.92\\ \sigma =\sqrt{\frac{pq}{n}}\\ =\sqrt{\frac{0.92(0.08)}{160}}\\ =0.021\end{array}$

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

  $\begin{array}{l}z=\frac{0.95-0.92}{0.021}\\ =1.399\end{array}$

So, the probability that more than $\text{95\% of the 160}$ seeds in the packet will generate is calculated as:

  $\begin{array}{l}P(z>1.399)=normalcdf(1.399,E99,0,1)\\ =0.081\end{array}$

Therefore, the probability that more than $\text{95\% of the 160}$ seeds in the packet will generate is $0.081$ .