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    (Original post by notanooblol)
    No worries. I'm glad it went well for us both
    Guys for iv sorry but the mean should be -5 as it was per 12middle and per final
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    (Original post by Rodneykimster)
    Guys for iv sorry but the mean should be -5 as it was per 12middle and per final
    There is nothing to be sorry about. We are perfectly correct. You obviously don't have the intelligence to realise that our answers will be the same using numerous methods, one of which is our method that appears on the mark scheme every single time. Your method with using the mean as -5 is unnecessary and easier to slip up on. Next time, think before you argue with what the mean SHOULD be with some superior mathematicians.
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    (Original post by notanooblol)
    There is nothing to be sorry about. We are perfectly correct. You obviously don't have the intelligence to realise that our answers will be the same using numerous methods, one of which is our method that appears on the mark scheme every single time. Your method with using the mean as -5 is unnecessary and easier to slip up on. Next time, think before you argue with what the mean SHOULD be with some superior mathematicians.
    Very funny mate.
    It is not really the same method as with my method you get 0.43792 as the probability
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    (Original post by Rodneykimster)
    Very funny mate.
    It is not really the same method as with my method you get 0.43792 as the probability
    Then you conducted your method incorrectly
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    (Original post by notanooblol)
    Then you conducted your method incorrectly
    We dont know yet^^
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    (Original post by Rodneykimster)
    We dont know yet^^
    I'm going to give you the information that we were given and then explain to you why an answer below 0.5 is not possible here. In addition I'd like to point out that you may lose marks for overspecification, you cannot find the fifth decimal place from the tables and it is not a valid point saying "my calculator can do it though".

    We were given that the mean for the middle questions was 220. We were given also that the mean for the end questions was 250. I do not have to even look at the standard deviations to prove you wrong. The questions asked the probability of the mean for the middle questions being at least 25 less than the end questions. We can see by doing 250-220 to find that the mean=30. 30 is greater than 25 and so the answer will be at least 0.5.

    You need to accept failure.
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    (Original post by notanooblol)
    I'm going to give you the information that we were given and then explain to you why an answer below 0.5 is not possible here. In addition I'd like to point out that you may lose marks for overspecification, you cannot find the fifth decimal place from the tables and it is not a valid point saying "my calculator can do it though".

    We were given that the mean for the middle questions was 220. We were given also that the mean for the end questions was 250. I do not have to even look at the standard deviations to prove you wrong. The questions asked the probability of the mean for the middle questions being at least 25 less than the end questions. We can see by doing 250-220 to find that the mean=30. 30 is greater than 25 and so the answer will be at least 0.5.

    You need to accept failure.
    chill out
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    (Original post by rubycstoddart)
    chill out
    I'm perfectly calm.
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    Well i dont know whether you are clever enough to know about the interpolation to get the 5th decimal place from the table firstly. Secondly look at the question precisely... it is said p (m>f-25) not the weird stuff that you are on about.
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    (Original post by Rodneykimster)
    Well i dont know whether you are clever enough to know about the interpolation to get the 5th decimal place from the table firstly. Secondly look at the question precisely... it is said p (m>f-25) not the weird stuff that you are on about.
    Knowing about it wouldn't make one clever. Knowing about it would make one knowledgeable. I know that a 5th decimal place does not appear on the mark schemes since interpolating to the 5th decimal place is considered to be too inaccurate. I have looked at the question precisely, I've done all the past papers, achieving A*s in every single one. p(m<f-25) would be correct.
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    (Original post by notanooblol)
    Knowing about it wouldn't make one clever. Knowing about it would make one knowledgeable. I know that a 5th decimal place does not appear on the mark schemes since interpolating to the 5th decimal place is considered to be too inaccurate. I have looked at the question precisely, I've done all the past papers, achieving A*s in every single one. p(m<f-25) would be correct.
    Yu still clearly havent looked at it correctly and you just changed your whole argument. You will clearly loose your a* for this paper^^. It is siad m is at least f-25 which means m>f-25. Stop being cocky and get refference from old mark schemes which is not the mark scheme for this paper buddy.
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    (Original post by Rodneykimster)
    Yu still clearly havent looked at it correctly and you just changed your whole argument. You will clearly loose your a* for this paper^^. It is siad m is at least f-25 which means m>f-25. Stop being cocky and get refference from old mark schemes which is not the mark scheme for this paper buddy.

    I haven't altered my argument at all, I have merely corrected yours. I'm sorry but you are not correct for this question. You are clearly too stubborn to accept this.
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    Guys just wondering, for the first question when they asked you about mean times, would you do (for first times) 4 times mean and then 4 times var, and then 1/4 times mean and 1/16 times var???
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    Yep that's exactly what you do for the first part 😊
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    does anyone happen to have an official markscheme for this, perhaps they might have been given one by teachers?
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    (Original post by bob6969)
    does anyone happen to have an official markscheme for this, perhaps they might have been given one by teachers?
    Hi please see attached. Hope this helps.
    Attached Images
  1. File Type: pdf 2016 STATISTICS 3 OFFICIAL MARK SCHEME.pdf (417.4 KB, 31 views)
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    (Original post by Mathathonica)
    Hi please see attached. Hope this helps.
    Thankyou!!
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    guys I'm confused with normal distribution values. usually the ms givers standardised values to 4dp and then finds it but even if i use a graphical calculator to find the normal value its always slightly different to the MS. are there any rules on what they want us to do with this?
 
 
 
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