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    (Original post by ii-mohsin-ii)
    Roger is an active retired lecturer. Each day after breakfast, he decides whether the
    weather for that day is going to be fine (F), dull (D) or wet (W). He then decides on
    only one of four activities for the day: cycling (C), gardening (G), shopping (S) or
    relaxing (R). His decisions from day to day may be assumed to be independent.
    The table shows Roger’s probabilities for each combination of weather and activity.
    Weather
    Fine (F) Dull (D) Wet (W)
    Activity
    Cycling (C) 0.30 0.10 0
    Gardening (G) 0.25 0.05 0
    Shopping (S) 0 0.10 0.05
    Relaxing (R) 0 0.05 0.10

    (iii) to go cycling, given that he had decided that it was going to be fine;
    (iv) not to relax, given that he had decided that it was going to be dull;
    (v) that it was going to be fine, given that he did not go cycling. (9 marks)
    (b) Calculate the probability that, on a particular Saturday and Sunday, Roger decided
    that it was going to be fine and decided on the same activity for both days.
    (3 marks)

    HELPPPPPP PLEASE
    (iii)

    P(C | F) : 0.30/(0.30+0.25)

    (iv)

    P(R' | D) : (0.10 + 0.05 + 0.10)/(0.10 + 0.05 + 0.10 + 0.05)

    (v)


    I got this one wrong, and I can't figure it out, sorry.

    (b)

    P(F and C)² + P(F and G)² :

    (0.30 * 0.30) + (0.25 * 0.25)

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    Mr Stickler processes 32 applications each day. His records for the previous 10 days show that the numbers of photographs that he deemed unusable were
    8 6 10 7 9 7 8 9 6 7


    Can someone please explain why the above is a population deviation and not a ...

    Sample standard deviation ??

    Thanks
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    Please can someone tell me when a question says " Construct a 99% confidence interval " : are the values the ones on page 25 of the AQA formula book ?? Secondly how do you tell which value of p it is under from the top row of the table " ...

    Because I know for 99% confidence interval it is 0.99 ( left side of the table on page 25h , but how to tell which value from the top of the table ?

    Thank You
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    (Original post by Tha Realest)
    Mr Stickler processes 32 applications each day. His records for the previous 10 days show that the numbers of photographs that he deemed unusable were
    8 6 10 7 9 7 8 9 6 7


    Can someone please explain why the above is a population deviation and not a ...

    Sample standard deviation ??

    Thanks
    How is that a population deviation?

    It's sample deviation :-)


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    (Original post by Tha Realest)
    Please can someone tell me when a question says " Construct a 99% confidence interval " : are the values the ones on page 25 of the AQA formula book ?? Secondly how do you tell which value of p it is under from the top row of the table " ...

    Because I know for 99% confidence interval it is 0.99 ( left side of the table on page 25h , but how to tell which value from the top of the table ?

    Thank You
    99% confidence interval ==> z=2.5758

    You look at 0.995 on the tables in pg 25

    P(z<Z<z)=0.99

    P(Z<z)=0.99+ 0.05

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    (Original post by Son234)
    How is that a population deviation?

    It's sample deviation :-)


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    Lol ... What's the difference ? Please could you tell me if a question came up : how would I tell if its a sample or population and if possible can you give me an example of an actual population deviation thanks
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    P(Z<z)=0.99+0.005 sorry mistake


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    (Original post by Son234)
    99% confidence interval ==> z=2.5758

    You look at 0.995 on the tables in pg 25


    This was posted from The Student Room's iPhone/iPad App
    Thanks but I still don't get why are we looking at 0.995 ?? Because I thought 99% = 0.99 not 0.995 ??

    I have no confidence in these confidence intervals lol
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    (Original post by Tha Realest)
    Lol ... What's the difference ? Please could you tell me if a question came up : how would I tell if its a sample or population and if possible can you give me an example of an actual population deviation thanks
    It really doesn't matter you get marks for both

    But the sake of your question you assume it's always the sample unless it states population

    Also my teacher said that too (-:


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    (Original post by Tha Realest)
    Thanks but I still don't get why are we looking at 0.995 ?? Because I thought 99% = 0.99 not 0.995 ??

    I have no confidence in these confidence intervals lol
    Its the fact that P(z<Z<z)=0.99

    So P(Z<z)=0.99 + 0.005= 0.995

    It's so much easier to explain it using the diagram! Lol




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    When do I use N+1/2 for the median and when do I use n/2?

    If they give me a list of 11 numbers, 1-11 then the median would be 6? I can get this from doing n+1/2 or n/2 and then rounding up?
    What happens if the list of numbers was from 1-10 and then I used n/2? I would get the 5th value but apparently you round to 5.5th value so half way inbetween 5 and 6? I don't understand this?
    Wouldn't it juts be easier to use n+1/2 in this case?

    I am really confused. So when you do n+1/2 you don't round. You just stick with the value you got?
    But if you do n/2 then you have to round up if its a decimal and if you get a whole number you look between that number and the next one up?
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    (Original post by Tha Realest)
    Thanks but I still don't get why are we looking at 0.995 ?? Because I thought 99% = 0.99 not 0.995 ??

    I have no confidence in these confidence intervals lol
    Think of it this way

    95% CI you use 0.975

    96% CI you use 0.98

    97% CI you use 0.985

    98% CI you use 0.99

    99% CI you use 0.995




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    (Original post by Son234)
    Think of it this way

    95% CI you use 0.975

    96% CI you use 0.98

    97% CI you use 0.985

    98% CI you use 0.99

    99% CI you use 0.995




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    Thanks Alot ... Thanks very much :dance:
    • Thread Starter
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    (Original post by Son234)
    Think of it this way

    95% CI you use 0.975

    96% CI you use 0.98

    97% CI you use 0.985

    98% CI you use 0.99

    99% CI you use 0.995




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    Or your confidence interval plus 100 then divide by 2 :-P
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    (Original post by Son234)
    Its the fact that P(z<Z<z)=0.99

    So P(Z<z)=0.99 + 0.005= 0.995

    It's so much easier to explain it using the diagram! Lol




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    Cheers :bigsmile:
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    (Original post by ollyhal)
    Or your confidence interval plus 100 then divide by 2 :-P
    That's convenient ... Thanks :excited:
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    Need some help with January 2011 question 2:

    http://filestore.aqa.org.uk/subjects/AQA-MS-SS1B-W-QP-JAN11.PDF


    I can do all of it apart from part c. I can get to 0.026, bet then I don't understand why it has to be multiplied by 6.

    Thanks.
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    (Original post by Son234)
    How's revision going for everyone?

    There is always a question where you have to comment or suggest and that's where I seem to lose my marks

    I have gone through all the past papers so I have learnt the mark scheme answers which I understand but I cant really do anything more

    So can anyone throw me some of the worst or difficult questions they can think off?





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    This is EXACTLY the same as what happens to me. The Maths seems to be fine for me its just the questions where you have to comment in context or explain why.
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    (Original post by andyt8765)
    Need some help with January 2011 question 2:

    http://filestore.aqa.org.uk/subjects/AQA-MS-SS1B-W-QP-JAN11.PDF


    I can do all of it apart from part c. I can get to 0.026, bet then I don't understand why it has to be multiplied by 6.

    Thanks.
    Because of the number of different combinations, I think
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    (Original post by andyt8765)
    Need some help with January 2011 question 2:

    http://filestore.aqa.org.uk/subjects/AQA-MS-SS1B-W-QP-JAN11.PDF


    I can do all of it apart from part c. I can get to 0.026, bet then I don't understand why it has to be multiplied by 6.

    Thanks.
    It's really simple

    At first you have 3 choices either labour, conservative, lib dem + other

    At the Second you have 2 choices either conservative or lib dem+other

    At the final you have 1 choice

    3X2X1=6

    So you multiply your probability by 6

    :-)



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