A Travel Weekly International Air Transport Association survey asked business travelers about the purpose for
their most recent business trip. 19% responded that it was for an internal company visit. Suppose 950 business travelers are randomly selected.
a. What is the probability that more than 20% of the business travelers say that the reason for their most recent business trip was an internal company visit?
b. What is the probability that between 18% and 20% of the business
a) P = 0.19 z(0.20) = (0.20-0.19)/sqrt[0.19*0.81/950] = 0.786 p-value to the right of z = 0.786 is : P(p... View the full answer