Normal distribution

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 8.30 am. Buses leave promptly every 5 minutes at 8.21 am, 8.26 am, etc. Find the probability that he catches the bus at 8.26 am assuming he gets on the first bus to arrive.

Does anyone know how I can start this? I've never tried a normal distribution question like this, its normally more than, less than or between 2 limits.

Reply 1

Rubgish

This is between two limits, he needs to arrive between 8:21;01s and 8:25;59 seconds. Find out how many standard deviations these are away are solve like normal. That should get the answer (depending on how they worked the times)

Reply 2

JBKProductions

Rubgish

Oh ok thanks. :smile:

Reply 3

JBKProductions

Rubgish

I tried using that method and it didnt work, it didnt tell how the times work but i assume you dont include seconds so 8.21 < Z < 8.26. but i got 0.0068 and the answer is 0.0900. Unless I've done something wrong. For my working out i put P(Z < -0.03) = 1-0.5120 and P(Z > -0.013) = 0.5052) and did P(-0.030<Z<-0.013) = 1-(0.4880+0.5052). Can you please start me off on the question if this is wrong. Thanks.

Reply 4

JBKProductions

Anyone? please :frown:

Reply 5

technicolour

I think part of the problem is that you used 8:21 < Z < 8:26, you can't do this as the numbers here are times, not decimal numbers. So, for example, 8:59 + 0:01 = 9:00 and not 8:60, and this would screw up the calculation somewhat.

Reply 6

danny111

dont you love normality? wonderful isnt it!

Reply 7

JBKProductions

technicolour

You know how i can get around this? and sorry i was supposed to write P(-0.03<Z<-0.013) and not P(8.21<Z<8.26)

Reply 8

technicolour

JBKProductions

You know how i can get around this? and sorry i was supposed to write P(-0.03<Z<-0.013) and not P(8.21<Z<8.26)

I would discard the times themselves and set 8:30 as zero, meaning you would be finding the probability that he arrives between "-9 minutes" and "-4 minutes". Using this method I appear to get the correct answer.

Reply 9

JBKProductions

technicolour

oh ok thanks :biggrin:

Reply 10

MARYCONTRARYMARY

Hi guys, I hope someone could help me with the final part of this question-let me just remind u of 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 5mins.Q4iv)The man is late for work if he catches a bus after 8.31am.What mean arrival time would ensure that ,on average ,he is not late for work more than one day in five?(Assuming standard deviation can't be changed and give answer to nearest 10seconds).I have been stuck on this question for an hour could someone help me pleeeeease !

Reply 11

ghostwalker

Original post by MARYCONTRARYMARY

What have you got so far?

Reply 12

User1255194

Yeah you need to change it into a standard normal distribution of Z by using (x - mean)/SD, then work from there.

Standard Normal Dist, Z ~ N(0,1^2)

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