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    (Original post by reubenkinara)
    What help do you need? I'll try to, if it's within the range of my abilities. Unfortunately, my answers may be delayed as I'm doing some other stuff . I've also edited your post in my QUOTE .
    LOL anyway i should stop posting on this thread after wht ive done
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    (Original post by Robbie242)
    So would the denominator be P(T\geq 4)?
    Yeah.

    Your numerator isn't quite right though. Think what P( A \cap B ) would be if A was T < 6 and B was T \geq 4.
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    (Original post by Boy_wonder_95)
    Linear interpolation? Median's gonna be 1/2n so you look in your table what value's the median going to be between, then you subtract, divide, multiply, then add... but I guess I was taught a different method to you.
    I don't know what linear interpolation is.
    My teacher only gave the formula, I guess it basically does the same thing. But she didn't tell we have to know it, so I assumed I will never be asked to find a median in this way. Then I was doing the Jan '13 paper.
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    (Original post by Revisionbug)
    I've flopped the easier maths units. As a result, I'm forced to aim for 100 UMS in both C3 and C4 to gain the grade I'm hoping to come out with. In retrospect, Id've done things differently this week if I could start it all over. Just a sidenote: the reason for my poor use of spelling, punctuation and grammar is that I'm currently using a poor cellular device to type my posts.
    What grade are you aiming for?
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    (Original post by justinawe)
    Yeah.

    Your numerator isn't quite right though. Think what P( A \cap B ) would be if A was T < 6 and B was T \geq 4.
    I'm thinking around 4 or 5, being less than 6 those two values fit so I'm thinking perhaps...

    P(4\leq T\leq 5)?
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    (Original post by purplemind)
    I don't know what linear interpolation is.
    My teacher only gave the formula, I guess it basically does the same thing. But she didn't tell we have to know it, so I assumed I will never be asked to find a median in this way. Then I was doing the Jan '13 paper.
    Really? . It;s just a way of obtaining quartlies or percentiles using the cumulative frequency graph and the equation I've given.
    That's where it's derived anyway but many teachers have a simpler way of teaching it using a line-method.
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    (Original post by Robbie242)
    I'm thinking around 4 or 5, being less than 6 those two values fit so I'm thinking perhaps...

    P(4\leq T\leq 5)?
    No, T isn't discrete.

    It's a lot simpler than you think. What is the area where T \geq 4 and T < 6?
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    (Original post by justinawe)
    No, T isn't discrete.

    It's a lot simpler than you think. What is the area where T \geq 4 and T < 6?
    From 4 to 6? so P(4\leq T<6) ?
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    (Original post by reubenkinara)
    Really? . It;s just a way of obtaining quartlies or percentiles using the cumulative frequency graph and the equation I've given.
    That's where it's derived anyway but many teachers have a simpler way of teaching it using a line-method.
    No, I have never seen this.
    Well, if I were to find a percentile using the cumulative frequency, eg. frequency goes up to 800 and I want top 10%, I would be looking at the value where frequency is 720.
    Line-method? I have no idea what it is either.
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    (Original post by Robbie242)
    P(A+B) wouldn't it be intersect and surely if it was P(A)+P(B) that would go to 10 ? I was thinking this perhaps?

    P(P(T<6) | P(T\leq 4)=\dfrac{P(T<6)}{P(T\leq 4)} ? A bit lost tbh, cheers dj for any help
    My apologies for the notation - I'm writing P(A and B) as P(A+B) as I don't know how to get the correct symbol for it. In this case, the probability of A and B is the probability that she spends more than 4 minutes and less than 6, which is easily worked out from the uniform distribution. You can also comfortably work out the probability that she spends more than 4 minutes, which is the probability of B.

    (Also, why are you and justin using a strict inequality for one and not for the other? For a continuous distribution they're notationally equivalent.)
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    (Original post by DJMayes)
    My apologies for the notation - I'm writing P(A and B) as P(A+B) as I don't know how to get the correct symbol for it. In this case, the probability of A and B is the probability that she spends more than 4 minutes and less than 6, which is easily worked out from the uniform distribution. You can also comfortably work out the probability that she spends more than 4 minutes, which is the probability of B.

    (Also, why are you and justin using a strict inequality for one and not for the other? For a continuous distribution they're notationally equivalent.)
    This helps in the bold, I didn't know if there was a strict difference, so would my probability I gave Justin be right?
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    (Original post by reubenkinara)
    Really? . It;s just a way of obtaining quartlies or percentiles using the cumulative frequency graph and the equation I've given.
    That's where it's derived anyway but many teachers have a simpler way of teaching it using a line-method.
    I remember I used to use linear interpolation to check my percentile / median answers from graphs in IGCSE maths in year 10, and I had no idea what it was called or that it was a well-documented method :lol:
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    (Original post by reubenkinara)
    What grade are you aiming for?
    A* but i need an A for my offer but the way things are now im either gettin a B or an A* as i need to clean up c2-4 and 480ums and 90+ in c3 and c4 is A* the thingbis im a gap year student i cant do another year its do or die from now on
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    (Original post by DJMayes)
    My apologies for the notation - I'm writing P(A and B) as P(A+B) as I don't know how to get the correct symbol for it. In this case, the probability of A and B is the probability that she spends more than 4 minutes and less than 6, which is easily worked out from the uniform distribution. You can also comfortably work out the probability that she spends more than 4 minutes, which is the probability of B.

    (Also, why are you and justin using a strict inequality for one and not for the other? For a continuous distribution they're notationally equivalent.)
    because I feel like it, I don't want to say T > 4 because technically she could finish immediately :mmm: even if the probability is the same for a continuous distribution
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    (Original post by Robbie242)
    From 4 to 6? so P(4\leq T<6) ?
    Yes.
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    (Original post by justinawe)
    Yes.
    Ah it was so obvious, thanks Justin for the help
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    (Original post by Robbie242)
    Ah it was so obvious, thanks Justin for the help
    No problem, glad I could help
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    (Original post by purplemind)
    No, I have never seen this.
    Well, if I were to find a percentile using the cumulative frequency, eg. frequency goes up to 800 and I want top 10%, I would be looking at the value where frequency is 720.
    Line-method? I have no idea what it is either.
    Name:  interpolation.PNG
Views: 91
Size:  234.3 KBHere's a solution scan from my textbook (Edexcel AS and A level Modular Mathematics) on interpolation. Hopefully the yellow boxes and the text within the example are sufficient without the actual question and its data!
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    (Original post by Robbie242)
    This helps in the bold, I didn't know if there was a strict difference, so would my probability I gave Justin be right?
    Yes. You're looking for  \dfrac{P(4<T<6)}{P(T>4)}

    There is another way to look at it - you know more than four minutes have been spent at the till, and that the probability of her stopping at any given time is equally likely. Using this, you can write the amount of time extra she spends as another uniform distribution - in this case, as you know there are 5 minutes left (She spends at most 5 minutes) and she is equally likely to stop at any given time, so you can write the amount of extra time she spends as a uniform distribution between 0 and 5 and solve this way.
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    (Original post by Revisionbug)
    x
    I see. Good luck then
 
 
 
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