A director of reservations believes that 8% of the ticketed passengers are no-shows. If the director is right, what is the probability that the proportion of no-shows in a sample of 596 ticketed passengers would differ from the population proportion by more than 3%?

Answer :

Given :

Population proportion $(p)=0.08$

Sample size $(n)=596$

Solution :

→ The probability that the proportion of no-show would differ from the population proportion by more than 3% :

$$\text{P}(|\hat{p}–p|>0.03)=\text{P}(\hat{p}–p<-0.03\text{or}\hat{p}–p>0.03)$$ $$=\text{P}(\hat{p}<p–0.03\text{or}\hat{p}>p+0.03)$$ $$=\text{P}(\hat{p}<0.08–0.03\text{or}\hat{p}>0.08+0.03)$$ $$=\text{P}(\hat{p}<0.05\text{or}\hat{p}>0.11)$$ $$=1–\text{P}(0.05<\hat{p}<0.11)$$ $$=1–\text{P}(\frac{0.05–p}{\sqrt{\frac{p(1-p)}{n}}}<z<\frac{0.11–p}{\sqrt{\frac{p(1-p)}{n}}})$$ $$=1–\text{P}(\frac{0.05–0.08}{\sqrt{\frac{0.08(1-0.08)}{596}}}<z<\frac{0.11–0.08}{\sqrt{\frac{0.08(1-0.08)}{596}}})$$ $$=1–\text{P}(-2.7<z<2.7)$$ $$=1–(\text{P}(z<2.7)–\text{P}(z<-2.7))$$ $$=1–(0.9965–0.0035)$$ $$=1–0.9930$$ $$=0.0070$$