Question

Most major airlines allow passengers to carry two pieces of luggage (of a certain maximum size) onto the plane. However, their studies show that the more carry-on baggage passengers have, the longer it takes to unload and load passengers. One regional airline is considering changing its policy to allow only one carry-on per passenger. Before doing so, it decided to collect some data. Specifically, a random sample of 1,000 passengers was selected. The passengers were observed, and the number of bags carried on the plane was noted. Out of the 1,000 passengers, 345 had more than one bag.
The domestic version of Boeing's 747 has a capacity for 568 passengers. Determine an interval estimate of the number of passengers that you would expect to carry more than one piece of luggage on the plane. Assume the plane is at its passenger capacity.
a) (171.651, 216.214)
b) (181.514, 208.313)
c) (174.412, 217.218)
d) (179.20, 212.716)

Answer 1

Answer:

The correct option is

d) (179.20, 212.716)

Step-by-step explanation:

We have out of a random sample of 1000, 345 carried more than a bag

Therefore in the question our X = 345 passengers carry more than a piece of luggage

n = 1000

Therefore the probability of a passenger carrying more than one luggage = 345/1000 = 0.345

The confidence interval estimator of p is given by

$p'+/-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }$ where p' = 0.345 = probability of desired outcome

n = 1000 = Population size

z at 95 %   = Confidence interval estimate, the value is sought from the distribution table as z value is 1.96 at 95 % confidence level

We have  $0.345+/-1.96\sqrt{\frac{0.345*(0.655)}{1000} }$ which gives

0.3745 or 0.3155

Which gives the range of confidential interval estimate as

212.695 to 179.224 which is equivalent to

d) (179.20, 212.716).