Chapter 18, Problem 21E

To determine

To find out what is the probability the newspaper’s sample will lead them to predict defeat and also verify that the assumptions and conditions for your analysis are met.

Answer to Problem 21E

  $0.212$ .

Explanation of Solution

It is given in the question that a local newspaper polls $400$ voters in an attempt to predict whether the budget will pass. And suppose that the budget actually has the support of $52\%$ of the voters. Thus, let us first verify the assumptions and conditions for the analysis as:

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

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

Success/failure condition: as $np=208,n(1-p)=192$ 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:

  $\begin{array}{l}\mu =p=0.52\\ \sigma =\sqrt{\frac{pq}{n}}\\ =\sqrt{\frac{0.52(0.48)}{400}}\\ =0.025\end{array}$

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

  $\begin{array}{l}z=\frac{0.50-0.52}{0.025}\\ =-0.80\end{array}$

So, the probability the newspaper’s sample will lead them to predict defeat is calculated as:

  $\begin{array}{l}P(z<-0.80)=normalcdf(-E99,-0.80,0,1)\\ =0.212\end{array}$

Therefore, the probability the newspaper’s sample will lead them to predict defeat is $0.212$ .