S2 Normal dist but with what seems discrete bus times

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acomber
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This is a question in S2 which is at the start of learning Normal distribution topic BEFORE we have gone onto modelling discrete situations.

The question is about catching a bus which leaves at set times. How do I work these out.

Here is the question.

Each weekday a man goes to work by bus. His arrival time at the bus stop is Normally distributed with standard deviation 3 minutes. His mean arrival time is 8:30am. Buses leave promptly every five minutes at 8:21am, 8:26am, etc. Find the probabilities that he catches the bus at 8:26am.

Because there are sort of time slots I wasn't sure how to calculate this.

My attempt was:

Normalise time to minutes. Eg 8:26am = 506 minutes.

Code:
z = (506 - 510) / 3 = -4/3 = -1.3 recurring
use +ve value so phi(1.333) = 0.9087

So P(8:26am) = 1 - 0.9087 = 0.0913

Book answer is 0.0900. Yes I know its close but sure I am not calculating correctly. So how do I calculate in this situation???
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ghostwalker
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(Original post by acomber)
Book answer is 0.0900. Yes I know its close but sure I am not calculating correctly. So how do I calculate in this situation???
Your general methodology is fine, however you're not understanding the situation correctly.

You've assumed that if he arrives befroe 8:26, then he will catch the 8:26 bus.

But if he arrives before 8:21, he will catch the 8:21 bus.
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atsruser
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(Original post by ghostwalker)
Your general methodology is fine, however you're not understanding the situation correctly.

You've assumed that if he arrives befroe 8:26, then he will catch the 8:26 bus.

But if he arrives before 8:21, he will catch the 8:21 bus.
Or to spell it out even more: he will catch the 8.26 bus if and only if he arrives in the time period (8.21, 8.26).
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acomber
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(Original post by atsruser)
Or to spell it out even more: he will catch the 8.26 bus if and only if he arrives in the time period (8.21, 8.26).
Yes I went out and was thinking about it in the car and came to this conclusion. Thanks.
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