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    P.docx
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    any clue ?
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    Can't make sense of what you've done.

    PS: It would be better if you put your question/working into your post, rather than a Word document. Just copy & paste would do it.
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    my question simply was for the Z value 0.1 . If i convert this to 0.5398 rather than -1.2816 , would my solution then be correct ? because for that value you can choose probability on either Table ( Z < z) or ( Z > z)
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    Hi Charly2012

    I'm also studying A level Stats at the moment and the Normal Distribution is also causing me a problem or two, so was drawn to your predicament.

    Anyhow, as I'm self tutoring, I can't be certain that my solution(s) is/are correct, however, for waht it's worth, I've had a bash and this is what I've come up with:

    Mean= 55

    Standard Deviation: to be determined (and required to answer second part of question).

    We're looking to find a z value which can be used to calculate the unknown Standard Deviation.

    So, we're told that 10 per cent of the cups contain less than the 50 ml stated on machine, so we're looking to turn this 10 percent into a 'z' value. If you were to sketch the normal Distribution curve above an axis (asymptotes to the axis at both ends don't forget and bell shaped in the middle where the mean od zero is situated) the z value we want (exceeds 10 per cent) would be found at the left hand side of the distribution i.e. on the negative side so the 'z' must be negative.
    As the table doesn't contain values for percentages below 50 per cent, we use symmetry to convert the 10 per cent value into a value we can use therefore, (100 - 10) per cent = 90 per cent. From the tables, this gives a value of 1.2816, however, remember ours was at the left hand end below zero so we require the negative value.

    Now, using z = x - mean/ standard deviation and a little transposition, we find that the standard deviation = x - mean/ z

    This is: (50 - 55)/ -1.2816 = 3.9 (to 2 s.f.)

    Therefore Standard Deviation = 3.9

    Now part two of the question: what percentage contain more than 61 mls, I think it said.

    So, we want a probability that X>61 and to do this we need to convert this x value to a 'z' value using

    z = (x - mean)/ standard deviation

    = (61 - 55)/ 3.9

    = 1.54 to 2 d.p.

    So, in z terms, we want a probability that Z>1.54

    Again from a sketch of the normal distribution curve, we find 1.54 appears on the right, and we shade the area immediate;y to the right of this value to indicate an area greater than 1.54.

    To calculate this, we can see that this is equal to 1- (P Z< 1.54)

    = 1 - 0.93822 (gleaned from the tables)

    = 0.062 to 2 S.F

    or, in other words, approximately 6 per cent of the cups in the machine contain more than 61 ml of whatever it is they're supposed to contain.

    I hope and trust you can follow this, and sincerest apologies if I'm wide of the mark.

    Regards,

    Jonty
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    Oh sorry so we need the Z value , cuz i changed the Z value to P which is other way around isnt it ?

    i gotta change Probability valur into Z which is gonna be on small table 0.1000 >>> 1.2816 and since this value is for table (Z>z ) we gonna write -1.2816 for (Z < z )

    am i going about it in the right way ?


    thank you
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    Oh sorry so we need the Z value , cuz i changed the Z value to P which is other way around isnt it ?

    i gotta change Probability valur into Z which is gonna be on small table 0.1000 >>> 1.2816 and since this value is for table (Z>z ) we gonna write -1.2816 for (Z < z )

    am i going about it in the right way ? ?


    thank you
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    Hi Charly2012.

    I think the penny has dropped and you've successfully identified your error. What you seem to have done is confused your Probability and z values and how to go about determining one from the other in the process. Not that you need me to tell you

    Remember:

    Tables exist to:

    a) determine a probability from a given z value - Norm. Dist. Func. table, and

    b) determine a z value from a given percentage using the Percentage Points table.

    This question involves process b) and from what I can gather, you seem to have confused your percentage points with probability and then lost your way with the z values in the process.

    Here's what (I think) you've done: you've looked for a percentage value for 10% in the percentage points table - which only goes down as far as 50% - and being unable to find one, you've looked across to the z value tables in desperation looking for a 10% value - or anything vaguely resembling it - and looking at the z values in the left hand column have found salvation in the 0.1 value - which is afterall what 10% looks like as a decimal
    You've then selected the 0.53983 value alongside it, at which point you've totally lost yourself: and me too, I think!!!!! In the meantime, I think it's dawned on you that the 0.1 is NOT 10% but the z value 0.1 - which is definitely NOT the z value we want!

    Hopefully, my previous post shows how I went about finding z, although if this is still unclear, just let me know. Hopefully my mutterings about sketches of Norm. Dist. 'bell shaped, curves will ring a bell (sorry) but if not, again let me know and I'll try and find a way of attaching a suitable sketch. The Heinemann books - if you can get your hands on one - have plenty of e.g.s and are my sources of reference.

    Incidentally, I'm doing the AQA syllabus.

    Hope this helps,

    Jonty.
 
 
 
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