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Math Help - probability and random samples

  1. #1
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    probability and random samples

    Hi I have some uni problems in my exam revision i can't sort out.
    m=215mg
    sd=15mg
    normal distribution
    215mg of nicotine in a standard cigarette

    a) find the probability a single cigarette has a nicotine content more than 220mg

    b) if a random sample of 25 cigarettes is selected, what is the probability that the mean is more than 220mg

    I tried using Z=y-u/sd but the z score from the table was 0.629 which can be correct right?
    PLEASE help i dont know what model i should be using!!

    Also, i have no idea about part b! HELP!
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  2. #2
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    Quote Originally Posted by kate204 View Post
    Hi I have some uni problems in my exam revision i can't sort out.
    m=215mg
    sd=15mg
    normal distribution
    215mg of nicotine in a standard cigarette

    a) find the probability a single cigarette has a nicotine content more than 220mg

    b) if a random sample of 25 cigarettes is selected, what is the probability that the mean is more than 220mg

    I tried using Z=y-u/sd but the z score from the table was 0.629 which can be correct right?

    Mr F says: Obviously NOT since {\color{red}220 > \mu = 215}.

    PLEASE help i dont know what model i should be using!!

    Also, i have no idea about part b! HELP!
    The model you're using is plainly the following:

    X ~ Normal( \mu = 215, ~ \sigma = 15) where X is the random variable nicotine content of a single cigarette.

    (a) \Pr(X > 220) = \Pr \left( Z > \frac{220 - 215}{15} \right).

    (b) You need to know the distribution of the sample mean:

    \overline{X} ~ Normal \left( \mu = 215, ~ \sigma = \frac{15}{\sqrt{25}}\right)

    assuming an 'infinite' population of cigarettes. You must have a theorem in your textbook or class notes that shows where this has come from.

    Now find \Pr (\overline{X} > 220) = \Pr(Z > .......).
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  3. #3
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    Quote Originally Posted by kate204 View Post
    Hi I have some uni problems in my exam revision i can't sort out.
    m=215mg
    sd=15mg
    normal distribution
    215mg of nicotine in a standard cigarette

    a) find the probability a single cigarette has a nicotine content more than 220mg

    b) if a random sample of 25 cigarettes is selected, what is the probability that the mean is more than 220mg

    I tried using Z=y-u/sd but the z score from the table was 0.629 which can be correct right?
    PLEASE help i dont know what model i should be using!!

    Also, i have no idea about part b! HELP!

    (a) z=\frac{220-215}{15}=0.3333..

    Now look this up to find P(Z<0.3333)\approx 0.631, so to within the limits of accuracy of your table your answer looks OK.

    (b) for the second part you need to use the result that the SD of a sample mean is the SD of an individual divided by the square root of the sample size.

    RonL
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    (a) z=\frac{220-215}{15}=0.3333..

    Now look this up to find P(Z<0.3333)\approx 0.631, so to within the limits of accuracy of your table your answer looks OK.
    [snip]
    Sorry to be disagreeable but the answer is not OK. Not as a final answer anyway, which is how it was presented ....

    The question asks "find the probability a single cigarette has a nicotine content MORE than 220mg" so the answer is 1 - 0.631 ......
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  5. #5
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    sorry im still a little confused,
    why is the final answer 1-0.631 instead of 0.631?
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  6. #6
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    Quote Originally Posted by kate204 View Post
    sorry im still a little confused,
    why is the final answer 1-0.631 instead of 0.631?
    \Pr (Z < 0.3333) \approx 0.631. But the question is not asking this.

    It should be clear (and my posts make this abundantly clear) that the question wants \Pr(Z > 0.3333) .....

    As I said at the start, 220 > 215 (the mean) ...... it must be obvious then that Pr(X > 220) cannot possibly be larger than 0.5 .....
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