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    I just don't even know where to start with this question, help me please

    A firm makes spacer bars that are fitted between the arms of a metal yoke. The bars are cut from long lengths of square-sectioned steel and must be between1.200m and 1.205m in length, Any spacer bar longer than 1.205m can be machined down to size at a cost of 60p per bar but any bar less than 1.200m is only fit for scrap, valued at 80p. The bars cost £2.10 each to make and satisfactory bars sell for £2.90. The automatic cutting process is believed to work to a mean length of 1.203m, Normally distributed with standard deviation 2mm.

    What percentage of bars will finally be saleable?
    What is the expected profit per bar?
    How would the situation change if the mean cut were increased to 1.204m?

    S1 is easily my hardest module, I just can't grasp it. Any tips I would appreciate.
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    (Original post by xyz9856)
    I just don't even know where to start with this question, help me please

    A firm makes spacer bars that are fitted between the arms of a metal yoke. The bars are cut from long lengths of square-sectioned steel and must be between1.200m and 1.205m in length, Any spacer bar longer than 1.205m can be machined down to size at a cost of 60p per bar but any bar less than 1.200m is only fit for scrap, valued at 80p. The bars cost £2.10 each to make and satisfactory bars sell for £2.90. The automatic cutting process is believed to work to a mean length of 1.203m, Normally distributed with standard deviation 2mm.

    What percentage of bars will finally be saleable?
    What is the expected profit per bar?
    How would the situation change if the mean cut were increased to 1.204m?

    S1 is easily my hardest module, I just can't grasp it. Any tips I would appreciate.
    So, X \sim N(1.203, 2^2) and you want \mathbb{P}(1.200 < X < 1.205) = \mathbb{P}(X < 1.205) - \mathbb{P}(X < 1.200) = \cdots
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    (Original post by Zacken)
    So, X \sim N(1.203, 2^2) and you want \mathbb{P}(1.200 < X < 1.205) = \mathbb{P}(X < 1.205) - \mathbb{P}(X < 1.200) = \cdots
    the sd is in mm and the mean is in m, do i have to convert. Thanks for the response
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    (Original post by xyz9856)
    the sd is in mm and the mean is in m, do i have to convert. Thanks for the response
    Yes, you do! Either convert everything to metres or everything to millimetres.
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    (Original post by Zacken)
    Yes, you do! Either convert everything to metres or everything to millimetres.
    Ok will do, will tag you again if thats ok (if I get stuck). Would rep you but i've already repped you too much haha. Thanks again
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    (Original post by xyz9856)
    Ok will do, will tag you again if thats ok (if I get stuck). Would rep you but i've already repped you too much haha. Thanks again
    No problem, I'm off to do some stuff so won't be able to reply straight away if you do tag me (which is okay, yes. ).
 
 
 
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