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Math Help - Help with probability question

  1. #1
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    Help with probability question

    Hi all I'm currently taking Statistics 1 and I'm stuck on a homework problem that I've been trying to figure out for a while...hopefully one of you guys can enlighten me on how to do it..heres the question:

    The weight of corn chips dispensed into a 10-ounce bag by the dispensing machine has been identified as possessing a normal distribution with a mean of 10.5 ounces and a standard deviation of .2 ounce. Suppose 100 bags of chips are randomly selected. Find the probability that the mean weight of these 100 bags exceeds 10.45 ounces.

    Any help will be greatly appreciated. Thanks
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  2. #2
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    Quote Originally Posted by lilyungn View Post
    Hi all I'm currently taking Statistics 1 and I'm stuck on a homework problem that I've been trying to figure out for a while...hopefully one of you guys can enlighten me on how to do it..heres the question:

    The weight of corn chips dispensed into a 10-ounce bag by the dispensing machine has been identified as possessing a normal distribution with a mean of 10.5 ounces and a standard deviation of .2 ounce. Suppose 100 bags of chips are randomly selected. Find the probability that the mean weight of these 100 bags exceeds 10.45 ounces.

    Any help will be greatly appreciated. Thanks
    The mean of a sample follows a normal distribution ..... Read this: Sampling Distribution of the Mean

    Use this distribution to calculate \Pr(\overline{X} > 10.45).
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  3. #3
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    I got this 0.40129367 i'm not sure if i did it right...
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    Quote Originally Posted by lilyungn View Post
    I got this 0.40129367 i'm not sure if i did it right...
    Show all your working that led to this answer and someone will review it.
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  5. #5
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    actually I got 0.598706 since I dont have my TI-83 I used this site to fill in the required fields...Z table - Normal Distribution
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    Quote Originally Posted by lilyungn View Post
    actually I got 0.598706 since I dont have my TI-83 I used this site to fill in the required fields...Z table - Normal Distribution
    Both this and your other answer are wrong. If you had shown your work like I asked you to do, I could have reviewed it and pointed out your mistake(s).
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    sorry im stuck the professor let's us use the TI-83 to solve the problems so I'm not to sure how to solve it by hand...can you help me out
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  8. #8
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    I gave it a shot and this is what I got...Since 10.5 - 10.45 = 0.05 which is (1/4) of a standard deviation you're to look for the z-score of 0.25 and add it to 0.5.

    0.5 + 0.0943 = 0.50943 is that right?
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  9. #9
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    Quote Originally Posted by lilyungn View Post
    sorry im stuck the professor let's us use the TI-83 to solve the problems so I'm not to sure how to solve it by hand...can you help me out
    By working I mean saying what mean and what standard deviation you're using.

    And if you're using a TI-83 you don't need to convert to a z-score (I suggest you carefully review how to use the TI-83 to solve normal distribution problems).
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  10. #10
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    Is this right?

    Mean(100 bags) = 10.5 ounces
    Std deviation(100 bags) = 0.2 / sqrt(100) = 0.02 ounces

    P(x-bar > 10.45) = P(Z > -0.05/0.02) = P(Z>-2.5) ~= 99.38%
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  11. #11
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    Quote Originally Posted by lilyungn View Post
    Is this right?

    Mean(100 bags) = 10.5 ounces
    Std deviation(100 bags) = 0.2 / sqrt(100) = 0.02 ounces

    P(x-bar > 10.45) = P(Z > -0.05/0.02) = P(Z>-2.5) ~= 99.38%
    Yes, but I would write it as 0.9938 (probability is a number that lies between 0 and 1 inclusive).
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