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Math Help - Standard deviation help!

  1. #1
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    Standard deviation help!

    Hiya guys/girls. I have a self study stats assignment to do which I have completed apart from one question where I have no idea what to do. Any help would be amazing.

    " a batch of 800 components produced by a machine is expected to be normally distributed. A sample was taken and measured and found to have a mean of 15mm with a standard deviation of 0.02. Calculate how many of the batch will have a length:
    a) between 15.00mm and 15.06mm
    b) between 14.99mm and 15.05mm
    c) greater than 15.04mm
    d) less than 14.94mm"

    once again thanks in advance for any help/advice/solutions

    Shaun
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  2. #2
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    Re: Standard deviation help!

    Are you aware of the 68, 95, 99.7 rule?
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  3. #3
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    Re: Standard deviation help!

    I am not sorry. However as its a self taught assignment if that's the way to go ill read up on that!
    Shaun
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    Re: Standard deviation help!

    Yes definitely. You need to know that 68% of the data lies within one standard deviation from the mean, 95% lies within two standard deviations from the mean, and 99.7% lies within three standard deviations from the mean.
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  5. #5
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    Re: Standard deviation help!

    hi again ive read up on the 68 95 99.7 rule and it helped me find the answer for a) c) and d) however for b) im struggling as the numbers all between the deviation parameter so to speak. there any advice how I can get round this
    Thanks
    Shaun
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  6. #6
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    Re: Standard deviation help!

    If you can't use the 68, 95, 99.7 rule, then you have to do a few things:

    1) Draw a diagram of your normal distribution and shade in the area you are trying to find, i.e. between say \displaystyle \begin{align*} x_1 \end{align*} and \displaystyle \begin{align*} x_2 \end{align*}. We can represent this area as \displaystyle \begin{align*} \textrm{Pr}\,{ \left( x_1 \leq X \leq x_2 \right) }  \end{align*}.

    2) Standardise your normal distribution, so that you get the corresponding standard normal values (say \displaystyle \begin{align*} z_1 \end{align*} and \displaystyle \begin{align*} z_2 \end{align*}) using \displaystyle \begin{align*} z = \frac{x - \bar{x}}{s} \end{align*}. This means that you will be making use of the fact that \displaystyle \begin{align*} \textrm{Pr}\,{\left( x_1 \leq X \leq x_2 \right) } = \textrm{Pr}\,{ \left( z_1 \leq Z \leq z_2 \right) } \end{align*}

    3) If you want to find \displaystyle \begin{align*} \textrm{Pr}\, \left( z_1 \leq Z \leq z_2 \right) \end{align*} then you need to realise that this is equal to \displaystyle \begin{align*} \textrm{Pr}\,{ \left( Z \leq z_2 \right) } - \textrm{Pr}\,{ \left( Z \leq z_1 \right) } \end{align*}. If you have a diagram, can you see why this is? These two values can be read from a standard normal distribution table, though you may have to apply some symmetry as well.
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