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    If someone doesn't mind spending a little time, can they explain a step missed in the mark scheme for June 2001 S4 Q3 c)

    Paper: http://www.aqa.org.uk/qual/gceasa/qp...W-QP-Jun01.pdf
    Mark Scheme: http://www.aqa.org.uk/qual/gceasa/qp...W-MS-Jun01.pdf

    In the mark scheme, on their normal distribution diagram, where does the value 0.727 come from? I'm fine with the standardising bit, (finding the z value), but have a mindblock on where to go from there!

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    hi, basically with you look up, on the normal distribution tables, the probability which corresponds to 0.603 (your z value), which is 0.727. then since you want the probability of exceeding 1.3, you just calculate (1-0.727). Hope this makes sense, namrata x
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    yeah pretty much what NIKSIP said:

    1) you standardise the x variable (1.3 in this case)

    2) That give 0.603 which you look up in the Z-tables BUT as you can't look up to that accuracy you just look up 0.60 which is 0.72575.

    3) The question asks for the probablity that it EXCEEDS 1.3. So therefore you do 1 - 0.72575 = 0.274
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    (Original post by xskater)
    2) That give 0.603 which you look up in the Z-tables BUT as you can't look up to that accuracy you just look up 0.60 which is 0.72575.
    Ah that was the bit that got me, I was looking for the exact value
 
 
 

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