I'm trying to prep myself for entrance exams and struggle with the following problem: Mr. X uses t

cazinskup3

cazinskup3

Answered question

2022-06-16

I'm trying to prep myself for entrance exams and struggle with the following problem:
Mr. X uses two buses daily on his journey to work. Bus A departs 6.00 and Mr. X always catches that. The journey time follows normal distribution (24,42). Bus B departs at 6.20, 6.30 and 6.40, journey time following normal distribution (20,52), independent of bus A. We assume the transfer from bus A to B won't take time and Mr. X always takes the first bus B available. Calculate probability for Mr. X being at work before 6.55.
So I am aware that the sum of two normal distribution variables follows also normal distribution, but don't know how two proceed with the given departure times of bus B (6.20, 6.30, 6.40) as clearly I can't just calculate the probability of A+B being under 55 minutes.. Any tips how to start with this?

Answer & Explanation

Zayden Wiley

Zayden Wiley

Beginner2022-06-17Added 21 answers

Some hints:

Let's A and B denote the times bus A and B take. So, we have A N ( 24 , 16 ) and B N ( 20 , 25 )

There are three disjoint cases, how a time below 55min can be reached:

1. A 20 B < 35
2. 20 < A 30 B < 25
3. 30 < A 40 B < 15
Now, use addition rule for probabilities and independence of A and B.
Ayanna Trujillo

Ayanna Trujillo

Beginner2022-06-18Added 13 answers

P ( at work before 6:55 ) = k { A , B , C } P ( bus k ) P ( on time | bus k ) .

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