Chapter 7, Problem T7.3SPT

To determine

To find: the value for which p would it be safe to use the normal approximation to the sampling distribution of $\widehat{p}$ .

Answer to Problem T7.3SPT

0.85

Explanation of Solution

Given:

  $n=100$

Calculation:

The sampling distribution of $\widehat{p}$ is about to normal if both $np$ and $n\left(1-p\right)$ are greater than $10$ .

  $\begin{array}{l}\left(a\right)np=100\times 0.01=1n\left(1-p\right)=100\times \left(1-0.01\right)=99\\ \left(b\right)np=100\times \frac{1}{11}=9.0909n\left(1-p\right)=100\times \left(1-\frac{1}{11}\right)=90.9091\\ \left(c\right)np=100\times 0.85=85n\left(1-p\right)=100\times \left(1-0.85\right)=15\\ \left(d\right)np=100\times 0.975=97.5n\left(1-p\right)=100\times \left(1-0.975\right)=2.5\\ \left(e\right)np=100\times 0.999=99.9n\left(1-p\right)=100\times \left(1-0.999\right)=0.1\end{array}$

Hence, the correct option is (c)