Question

Students on diets Suppose that $70\%$of college women have been on a diet

within the past $12$months. A sample survey interviews an SRS of $267$college women. What is the probability that $75\%$or more of the women in the sample have been on a diet?

Short answer

The probability is$P(\hat{p}\ge 75\%)=0.03754$

Step-by-step solution

Given Information

It is given that $p=70\%=0.70$

$\hat{p}=75\%=0.75$

$n=267$

Simplification

Here ${\mu }_{\hat{p}}=p=0.70$

We know that ${\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$

$=\sqrt{\frac{0.70(1-0.70)}{267}}=0.02804$

Also,

$z=\frac{x-\mu }{\sigma }$

$=\frac{0.75-0.70}{0.02804}$

$=1.78$

Required probability is $P(\hat{p}\ge 75\%)$

$=P(z\ge 1.78)=1-P(z\le 1.78)$

$=1-0.96246=0.03754$

Required probability is$0.03754$