Question

There are three options for the meal choice on a flight on a particular air route: beef, fish and vegetarian. Over many flights it has been observed that the probability that a passenger chooses beef is 0.5917. For a sample of 322 passengers on one flight what Z value would you use to find the probability that more than 55.26% of passengers request beef? Give your result rounded to four decimal places

Answer 1

$\sigma=\sqrt{(p(1-p))/n}=$ $\sqrt{0.5917*(0.4083)/322}=\sqrt{ 0.00075}=0.027$

$P(p>0.5526)=P((p-P)/\sigma>(0.5526-P)/\sigma)=P(z>-1.448)$

z-value that we use equals to -1.448