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Help/Advice re 'The Normal Distribution' - 'A' Level Maths/Stats 1 Watch

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    Hi everybody.

    This is my first time here so please be gentle

    First, a bit of background about myself, If I may, just so you know you're dealing with no brain of Britain!! - plenty of patience required, I fear.

    I'm an external candidate who, at the grand old age of 52 and in his infinite wisdom, has decided to try and self tutor himself at 'A' Level Maths - probably need to get a life - Respect btw to all my fellow candidates who are sixth formers preparing for University; how do you do it all?.

    Anyway, am with AQA Exam Board and using the Heinemann books as sources of reference. To date, I've completed and sat exams in Cores 1, 2, 3 & 4, and am now attempting Stats. 1(B); which will (hopefully) be completed with Mechanics 1 if I don't decide to call it all a day in the meantime. That's the plan anyway.

    Have attempted all of the past papers on the AQA site and not done too badly - pleased that those I couldn't attempt at the time I later went back and managed to solve after several cups of tea and biccies.

    Anyway, I'll come to the point. Have had a bash at an Exam Practice Paper at the rear of the Heinemann AQA Advancing Maths 2nd edition - Stats - but got a little stuck on one part of a question on the topic regarding the Normal Distribution. At the risk of making this thread a little too lengthy, I'll post the question in whole so that I don't overlook any info that might prove essential to any kind soul trying to assist me and also to show you my 'thought processes', just in case I'm barking up the wrong tree. It's actually part c) which is the problem. So here goes:

    Rahul works for a bank. He can travel to the branch where he works by car, bus or bike. He does not wish to arrive too early as he cannot enter the building before it is unlocked at 8.30 am. He does not wish to arrive after his starting time which is 9.00 am.

    a) When he travels by bus his journey time may be modelled by a normally distributed random variable with mean 40 minutes and standard deviation 10 minutes. Given that he leaves home at 8.00 am and travels by bus, find the probability that he will arrive at the branch:

    i) before 9.00 am

    I needed an x value to convert to a 'z' value to determine the required probability using the percentage points of normal distribution table.
    I reckoned that leaving home at eight to get there for nine took 60 minutes (mean and standard deviation are in minutes so that's where I needed to be) so considered the probability needed to be x<60 mins.

    So to find 'Z' = (60 - 40) / 10
    = 2
    therefore, probability of z< 2 which according to the table was 0.977 (to 3 s.f.). which was correct according to the answers.

    ii) between 8.30 am and 9.00 am:

    which, converted to 'z' values are between 2 and -1. Using a normal dist. curve sketch I determined that the probability was: p(z<2) - (1 - P(z<1)) = 0.97725 - (1 - 0.84134) = 0.819 (to 3 s.f.) which again was correct according to the answers at the back of the book.

    b) When he travels by bike, his journey time may be modelled by a normally distributed random variable with mean 35 minutes and standard deviation 2 minutes. What time should he leave home when he travels by bike if he wishes to have a probability of 0.99 of arriving before 9.00 am?

    Using another sketch and tables, I determined a (positive) value of 'z' of 2.3263 that has a probability greater than 99 per cent.

    This was converted to x using x=mean (+/-) 'z'(Stn. Dev.) which gave a value for x of 40 minutes.
    Therefore, if it takes him 40 mins. to travel, he needs to leave home by 08.20 am to arrive at 9.00 am. I subtracted 1 minute from this to give an answer of 08.19 am so that he would arrive 1 minute before as 08.20 am is not arriving before but at 9.00 am, and the answer asked 'before' 9.00 am. The answer in the book gave 8.20 am so perhaps my 8.19am was wrong, or perhaps I'd have still have gained a mark if I'd included my reasoning? Dunno.

    And now for the stinker!!!!

    c) When he travels by car, his journey time may be modelled by a normally distributed random variable with mean 28 minutes and standard deviation 8 minutes. What time should he leave home when he travels by car, in order to maximise his chance of arriving between 8.30am and 9.00am?

    The answer is given as 8.17 am. Almost smack bang between arriving at 8.30 and 9.00.

    Is it something to do with the majority of values (99.9 percent) lying within the mean plus 3 X Standard deviations which would give a value in this case of 28 plus 3 X 8 = 52 mins?

    What flummoxes me most is that no time of leaving from home is provided.

    Can anyone help, please?

    Regards,

    A hairless Jonty.
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    I can't help you answer the question, but there is a website called exams solution, if you type it up in google it should come up, so of you ever get stuck on a certain topic it should be able to help you out
    Sorry I can't answer your question, I'm doing edexcel...


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    (Original post by Tplox)
    I can't help you answer the question, but there is a website called exams solution, if you type it up in google it should come up, so of you ever get stuck on a certain topic it should be able to help you out
    Sorry I can't answer your question, I'm doing edexcel...


    Posted from TSR Mobile
    Hi, Tplox, and thanks for your prompt response. Very kind.

    Anything that might help is a bonus so thank you for that.

    Good luck with exam(s)

    Jonty
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    (Original post by jonty)
    Hi everybody.

    This is my first time here so please be gentle

    First, a bit of background about myself, If I may, just so you know you're dealing with no brain of Britain!! - plenty of patience required, I fear.

    I'm an external candidate who, at the grand old age of 52 and in his infinite wisdom, has decided to try and self tutor himself at 'A' Level Maths - probably need to get a life - Respect btw to all my fellow candidates who are sixth formers preparing for University; how do you do it all?.

    Anyway, am with AQA Exam Board and using the Heinemann books as sources of reference. To date, I've completed and sat exams in Cores 1, 2, 3 & 4, and am now attempting Stats. 1(B); which will (hopefully) be completed with Mechanics 1 if I don't decide to call it all a day in the meantime. That's the plan anyway.

    Have attempted all of the past papers on the AQA site and not done too badly - pleased that those I couldn't attempt at the time I later went back and managed to solve after several cups of tea and biccies.

    Anyway, I'll come to the point. Have had a bash at an Exam Practice Paper at the rear of the Heinemann AQA Advancing Maths 2nd edition - Stats - but got a little stuck on one part of a question on the topic regarding the Normal Distribution. At the risk of making this thread a little too lengthy, I'll post the question in whole so that I don't overlook any info that might prove essential to any kind soul trying to assist me and also to show you my 'thought processes', just in case I'm barking up the wrong tree. It's actually part c) which is the problem. So here goes:

    Rahul works for a bank. He can travel to the branch where he works by car, bus or bike. He does not wish to arrive too early as he cannot enter the building before it is unlocked at 8.30 am. He does not wish to arrive after his starting time which is 9.00 am.

    a) When he travels by bus his journey time may be modelled by a normally distributed random variable with mean 40 minutes and standard deviation 10 minutes. Given that he leaves home at 8.00 am and travels by bus, find the probability that he will arrive at the branch:

    i) before 9.00 am

    I needed an x value to convert to a 'z' value to determine the required probability using the percentage points of normal distribution table.
    I reckoned that leaving home at eight to get there for nine took 60 minutes (mean and standard deviation are in minutes so that's where I needed to be) so considered the probability needed to be x<60 mins.

    So to find 'Z' = (60 - 40) / 10
    = 2
    therefore, probability of z< 2 which according to the table was 0.977 (to 3 s.f.). which was correct according to the answers.

    ii) between 8.30 am and 9.00 am:

    which, converted to 'z' values are between 2 and -1. Using a normal dist. curve sketch I determined that the probability was: p(z<2) - (1 - P(z<1)) = 0.97725 - (1 - 0.84134) = 0.819 (to 3 s.f.) which again was correct according to the answers at the back of the book.

    b) When he travels by bike, his journey time may be modelled by a normally distributed random variable with mean 35 minutes and standard deviation 2 minutes. What time should he leave home when he travels by bike if he wishes to have a probability of 0.99 of arriving before 9.00 am?

    Using another sketch and tables, I determined a (positive) value of 'z' of 2.3263 that has a probability greater than 99 per cent.

    This was converted to x using x=mean (+/-) 'z'(Stn. Dev.) which gave a value for x of 40 minutes.
    Therefore, if it takes him 40 mins. to travel, he needs to leave home by 08.20 am to arrive at 9.00 am. I subtracted 1 minute from this to give an answer of 08.19 am so that he would arrive 1 minute before as 08.20 am is not arriving before but at 9.00 am, and the answer asked 'before' 9.00 am. The answer in the book gave 8.20 am so perhaps my 8.19am was wrong, or perhaps I'd have still have gained a mark if I'd included my reasoning? Dunno.

    And now for the stinker!!!!

    c) When he travels by car, his journey time may be modelled by a normally distributed random variable with mean 28 minutes and standard deviation 8 minutes. What time should he leave home when he travels by car, in order to maximise his chance of arriving between 8.30am and 9.00am?

    The answer is given as 8.17 am. Almost smack bang between arriving at 8.30 and 9.00.

    Is it something to do with the majority of values (99.9 percent) lying within the mean plus 3 X Standard deviations which would give a value in this case of 28 plus 3 X 8 = 52 mins?

    What flummoxes me most is that no time of leaving from home is provided.

    Can anyone help, please?

    Regards,

    A hairless Jonty.


    It's okay all; have managed to solve it.

    For anyone who's interested, between 8.30 am and 9.00 am, makes it 8.45 am. If you treat this time as the centre of the bell curve, then 8.45 am minus the average time taken to travel i.e. 8.45 - 28 minutes = 08.17 am, which is the latest you can leave to maximise your chances of arriving within the 08.30 and 9.00 window.

    Hope this helps.

    Jonty.
 
 
 
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