Answer: 0.75
Step-by-step explanation:
Given : Interval for uniform distribution : [0 minute, 5 minutes]
The probability density function will be :-
The probability that a given class period runs between 50.75 and 51.25 minutes is given by :-
![P(x>1.25)=\int^{5}_{1.25}f(x)\ dx\\\\=(0.2)[x]^{5}_{1.25}\\\\=(0.2)(5-1.25)=0.75](https://tex.z-dn.net/?f=P%28x%3E1.25%29%3D%5Cint%5E%7B5%7D_%7B1.25%7Df%28x%29%5C%20dx%5C%5C%5C%5C%3D%280.2%29%5Bx%5D%5E%7B5%7D_%7B1.25%7D%5C%5C%5C%5C%3D%280.2%29%285-1.25%29%3D0.75)
Hence, the probability that a randomly selected passenger has a waiting time greater than 1.25 minutes = 0.75
The pattern is n-3, so we will start from the beginning, from 1.
7, 4, 1, -2, -5, -8, -11, -14, -17, -20, -23, -26, -29, -32, -35, -38, -41, -44.
-44 is the 18th term in the sequence.
Answer:
more likely
Step-by-step explanation:
the probability of 0.05 suggests there is a chance of that event happening. A probability of 0 says that such event would never happen. so despite being small the 0.05 chance is higher than the probability of 0
