Chapter 7.2, Problem 43E

To determine

To find: the probability that 75% or more women in the sample have been on a diet.

Answer to Problem 43E

  $P\left(\widehat{p}\ge 75\%\right)=0.03754$

Explanation of Solution

Given:

  $\begin{array}{l}p=70\%=0.70\\ \widehat{p}=75\%=0.75\\ n=267\end{array}$

Formula used:

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

  $z=\frac{x-\mu }{\sigma }$

Calculation:

  ${\mu }_{\widehat{p}}=p=0.70$

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.70\left(1-0.70\right)}{267}}\\ =0.02804\end{array}$

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{0.75-0.70}{0.02804}\\ =1.78\end{array}$

Required probability

  $\begin{array}{c}=P\left(\widehat{p}\ge 75\%\right)\\ =P\left(z\ge 1.78\right)\\ =1-P\left(z\le 1.78\right)\\ =1-0.96246\\ =0.03754\end{array}$