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Math Help - The normal distribution

  1. #1
    Senior Member chella182's Avatar
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    The normal distribution

    Okay, so this may seem a bit simple for here, and I wasn't sure, but this is from my University past paper. The question reads:

    A chocolate company produce boxes of chocolates with mean 1kg. It is decided that all boxes weighing less than 990g or more than 1020g will be re-packed. If the weight of the boxes is Normally distributed with standard deviation 20g, what percentage of boxes will be repacked?

    So I got from this that...

    X\sim N(1,0.02^{2}) and I was trying to find 1-\left(P(0.99\leq X\leq 1.02)\right) (because the stuff in the P would be the boxes that were the right weight, so 1 minus that would be boxes to be repacked). However, after standardising and using the tables I'm given and what not, with this I get a percentage that, to me, seems a bit high something like 43%.

    Can anyone help? Have I made a misunderstanding somewhere?
    Last edited by chella182; August 12th 2009 at 01:34 PM. Reason: Forgot to subscribe
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by chella182 View Post
    Okay, so this may seem a bit simple for here, and I wasn't sure, but this is from my University past paper. The question reads:

    A chocolate company produce boxes of chocolates with mean 1kg. It is decided that all boxes weighing less than 990g or more than 1020g will be re-packed. If the weight of the boxes is Normally distributed with standard deviation 20g, what percentage of boxes will be repacked?

    So I got from this that...

    X\sim N(1,0.02^{2}) and I was trying to find 1-\left(P(0.99\leq X\leq 1.02)\right) (because the stuff in the P would be the boxes that were the right weight, so 1 minus that would be boxes to be repacked). However, after standardising and using the tables I'm given and what not, with this I get a percentage that, to me, seems a bit high something like 43%.

    Can anyone help? Have I made a misunderstanding somewhere?
    \Pr (0.99 \leq X \leq 1.02) = 0.5328 (I used technology, using tables will probably give a slightly different answer). So your answer looks OK.

    In fact, this large a probability is hardly surprising given that the interval (0.99, 1.02) is equivalent to \left( \mu - \frac{\sigma}{2}, \, \mu + \sigma \right) ....
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  3. #3
    Senior Member chella182's Avatar
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    Quote Originally Posted by mr fantastic View Post
    \Pr (0.99 \leq X \leq 1.02) = 0.5328 (I used technology, using tables will probably give a slightly different answer). So your answer looks OK.

    In fact, this large a probability is hardly surprising given that the interval (0.99, 1.02) is equivalent to \left( \mu - \frac{\sigma}{2}, \, \mu + \sigma \right) ....
    Cheers. I guess I just thought that (realistically rather than statistically) repacking 43% of the boxes seems a lot of extra work is all. Thanks again.
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