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
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 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).
