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    Could someone possibly help with the second and third parts?
    I've got 0.3108 for the first part
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    For the first part of the question I did:

     \frac {0.4-0}{1} =0.4

\phi (0.4) = 0.6554

1-0.6554=0.3446

so, 1-(2 \times 0.3446) = 0.3108
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    (Original post by Mutleybm1996)
    For the first part of the question I did:

     \frac {0.4-0}{1} =0.4

\phi (0.4) = 0.6554

1-0.6554=0.3446

so, 1-(2 \times 0.3446) = 0.3108
    Looks good so far.

    So, we have E\sim N(0,0.1^2)

    What then is the distribution of the sample mean for 9 observations?

    The mean will be 0, but what about the variance?
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    (Original post by ghostwalker)
    Looks good so far.

    So, we have E\sim N(0,0.1^2)

    What then is the distribution of the sample mean for 9 observations?

    The mean will be 0, but what about the variance?
    X9? I think....


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    (Original post by Mutleybm1996)
    X9? I think....


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    Needs to be divided by 9, so, 0.1^2/9

    You should be familiar with this - it's fairly basic for sampling.
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    (Original post by ghostwalker)
    Needs to be divided by 9, so, 0.1^2/9

    You should be familiar with this - it's fairly basic for sampling.
    Wait, sorry, I read it as 9 different butterflies! In this case would it be x9?


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    (Original post by Mutleybm1996)
    Wait, sorry, I read it as 9 different butterflies! In this case would it be x9?


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    If it was 9 different butterflies, then they'd have different weights. The error on the mean weight would have the same distribution I think, but I'm not 100% sure.
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    (Original post by ghostwalker)
    Looks good so far.

    So, we have E\sim N(0,0.1^2)

    What then is the distribution of the sample mean for 9 observations?

    The mean will be 0, but what about the variance?

    So, we have E\sim N(0, \frac{0.1^2}{9})

    Then do I do the same again as for the first question?

    What about the third part?




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    (Original post by Mutleybm1996)
    So, we have E\sim N(0, \frac{0.1^2}{9})

    Then do I do the same again as for the first question?
    Yes.

    \bar{E}\sim N(0, \frac{0.1^2}{9})


    What about the third part?
    Standard, constructing a confidence interval.
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    (Original post by ghostwalker)
    Yes.

    \bar{E}\sim N(0, \frac{0.1^2}{9})



    Standard, constructing a confidence interval.
    Doing the same again gives z=45? I think. For part b


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    Sorted it! Thanks


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