Final answer:
The question asks for the probability that more than 100 of the 109 ticket holders will show up for a flight, given an 89% show rate. The approach involves using a normal approximation to the binomial distribution and finding the z-score for 100 passengers, then subtracting the corresponding probability from 1.
Step-by-step explanation:
The student is asking for the probability that more than 100 ticket holders will show up for a flight, given an airline's estimate that 89% of ticket holders actually show up, and 109 tickets are sold. To find this probability, we essentially need to find the probability of the complementary event, which is 100 or fewer ticket holders showing up, and subtract it from 1.
Since we are dealing with a large number of tickets, we can model the number of passengers who show up as a binomial distribution, where the number of trials is 109 (the number of tickets sold), and the probability of success (a passenger showing up) is 0.89. However, for practical purposes and easier computation, we can use the normal approximation to the binomial distribution since the sample size is large.
Calculating the mean (μ) and standard deviation (σ) of the distribution:
- μ = n * p = 109 * 0.89
- σ = sqrt(n * p * (1 - p))
Using these parameters, we would then find the z-score for 100 passengers showing up:
Z = (X - μ) / σ
Where X is 100. Then we find the probability corresponding to that z-score using a standard normal distribution table, or a computational tool like a statistics calculator or software. Finally, we subtract this probability from 1 to obtain the probability of more than 100 passengers showing up.