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    The question is from Jan 20120

    Josephine accurately measures the widths of A4 sheets of paper and then rounds the widths to the nearest 0.1 cm. The rounding error, X centimetres, follows a rectangular distribution.
    A randomly selected A4 sheet of paper is measured to be 21.1 cm in width.

    (c) Calculate P(-0.01<=X<=0.03)

    I get that its just the area of the rectangle between these two x values so you would have 0.04 x k. where k is the constant - the problem is I don't see how you get K, can anyone help me please.

    Thank you.
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    Are you sure you have provided all the given info ?
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    (Original post by Bric)
    Are you sure you have provided all the given info ?
    Its from an exam paper, its part c of the first question, you can find the paper here: http://filestore.aqa.org.uk/subjects...B-QP-JAN12.PDF.
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    The rounding ERROR follows a rectangular distribution (sorry I didn't read carefully enough). So you now know the width of the rectangle (confirm from part a), so you know the height (k).
    So yes, your calculation for the probability of 0.04 x k is correct.
    Bric
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    Thanks ,I realised the width is equal to 0.1 and of course the area underneath is equal to 1 so k is just 1/0.1. Can't believe I didn't see that.
 
 
 
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