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Math Help - Mean/Standard Deviation question

  1. #1
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    Mean/Standard Deviation question

    I need help on this question, I don't know what equation to use to find the solution.

    Suppose that the wrapper of a certain candy bar lists its weight as 2.13 ounces. Naturally, the weights of individual bars vary somewhat. Suppose that the weights of these candy bars vary according to a normal distribution with a mean 2.2 ounces and standard deviation 0.04 ounces.

    Question 1) If the manufacturer wants to adjust the production process so that only 1 candy in 1000 weights less than the advertised weight, what should the mean of the actual weights be(assuming that the standard deviation of the weights remain 0.04 ounces)?
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  2. #2
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    Quote Originally Posted by chobozor View Post
    I need help on this question, I don't know what equation to use to find the solution.

    Suppose that the wrapper of a certain candy bar lists its weight as 2.13 ounces. Naturally, the weights of individual bars vary somewhat. Suppose that the weights of these candy bars vary according to a normal distribution with a mean 2.2 ounces and standard deviation 0.04 ounces.

    Question 1) If the manufacturer wants to adjust the production process so that only 1 candy in 1000 weights less than the advertised weight, what should the mean of the actual weights be(assuming that the standard deviation of the weights remain 0.04 ounces)?
    The z-score such that P(Z<z)=0.999 is z=3.091, so the required mean is:

    m=2.13+3.091*0.04

    CB
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  3. #3
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    why is it 0.999 instead of using 0.001? i though we were looking for 1 out of 1000 isnt 0.999 out of 1 means 999 out of 1000?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by chobozor View Post
    why is it 0.999 instead of using 0.001? i though we were looking for 1 out of 1000 isnt 0.999 out of 1 means 999 out of 1000?

    P(Z>z)=1-P(Z<z)

    so if P(Z<z)=0.999, then P(Z>z)=0.001

    CB
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