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    When you have discrete data and are calculating an upper bound, after you've done value + accuracy/2 do you then need to minus one? As you can't count in part numbers? Thanks
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    (Original post by mayjb)
    When you have discrete data and are calculating an upper bound, after you've done value + accuracy/2 do you then need to minus one? As you can't count in part numbers? Thanks
    Could you post an example? I'm pretty sure even if you have discrete data, the bounds should be the same even if they are "part" numbers because well, that's the whole point of a bound.
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    (Original post by Zacken)
    Could you post an example? I'm pretty sure even if you have discrete data, the bounds should be the same even if they are "part" numbers because well, that's the whole point of a bound.
    In my notes I have written for continuous data the upper bound of 150 to the nearest ten accuracy would be 155 as it's basically the same as 149.99999 etc so you just use it as the upper bound even though it would round to 160, but for discrete the upper bound would be 154 as you can't get part numbers as discrete data is all about counting. Is this correct???
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    (Original post by mayjb)
    In my notes I have written for continuous data the upper bound of 150 to the nearest ten accuracy would be 155 as it's basically the same as 149.99999 etc so you just use it as the upper bound even though it would round to 160, but for discrete the upper bound would be 154 as you can't get part numbers as discrete data is all about counting. Is this correct???
    No, you'd still put it as 150.
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    (Original post by Zacken)
    No, you'd still put it as 150.
    do you mean 155??
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    (Original post by mayjb)
    do you mean 155??
    Yes I do, whoops. Good catch.
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    (Original post by Zacken)
    Yes I do, whoops. Good catch.
    ok thanks for the help
 
 
 
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