Answer: 0.5
Step-by-step explanation:
Given : Delta Airlines quotes a flight time of 2 hours, 5 minutes for its flights from Cincinnati to Tampa.
i.e. Flight time = 2(60) +5= 125 minutes [∵ 1 hour = 60 minutes]
Actual flight times are uniformly distributed between 2 hours and 2 hours, 20 minutes.
i.e. In minutes the flight times are between 120 minutes and 140 minutes.
Let x be a uniformly distributed variable in [120 minutes, 140 minutes] that represents the flight time.
Since the probability density function for x uniformly distributed in [a,b] is ![f(x)=\dfrac{1}{b-a}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cdfrac%7B1%7D%7Bb-a%7D)
⇒ Probability density function for flight time : ![\dfrac{1}{140-120}=\dfrac{1}{20}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B140-120%7D%3D%5Cdfrac%7B1%7D%7B20%7D)
5 minutes late than usual time = Flight time+ 5 = 125+5 = 130 minutes
Now , the probability that the flight will be no more than 5 minutes late will be :-
![\int^{130}_{120} \dfrac{1}{20}\ dx\\\\=\dfrac{1}{20}[x]^{130}_{120}\\\\= \dfrac{130-120}{20}\\\\=\dfrac{1}{2}=0.5](https://tex.z-dn.net/?f=%5Cint%5E%7B130%7D_%7B120%7D%20%5Cdfrac%7B1%7D%7B20%7D%5C%20dx%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B20%7D%5Bx%5D%5E%7B130%7D_%7B120%7D%5C%5C%5C%5C%3D%20%5Cdfrac%7B130-120%7D%7B20%7D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B2%7D%3D0.5)
Hence, the probability that the flight will be no more than 5 minutes late is 0.5.