Chapter 7.2, Problem 44E

To determine

To find: the probability that contains 20% or more than that.

Answer to Problem 44E

  $P\left(\widehat{p}\ge 0.20\right)=0.00008$

Explanation of Solution

Given:

  $\begin{array}{c}p=14\%=0.14\\ n=500\end{array}$

Formula used:

  ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$

  $z=\frac{x-\mu }{\sigma }$

Calculation:

  ${\mu }_{\widehat{p}}=p=0.14$

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.14\left(1-0.14\right)}{500}}\\ =0.0155\\ z=\frac{x-\mu }{\sigma }\\ =\frac{0.20-0.14}{0.0155}\\ =3.87\end{array}$

Required probability

  $\begin{array}{c}=P\left(\widehat{p}\ge 0.20\right)\\ =P\left(z\ge 3.78\right)\\ =1-P\left(z\le 3.78\right)\\ =1-0.99992\\ =0.00008\end{array}$