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Thread: probability with mean and standard deviation

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    probability with mean and standard deviation




    The local hardware store sells screws in 1lb bags. Because the screws aren't identical, the number of screws per bag, although distributed normally,varies with a μ=115, and σ=6.

    A carpenter needs a total of 590 screws for a particular project. What is the probability that he will have enough screws if he buys five bags?

    --

    I don't know why I have trouble calculating even the simplest probability problems, but they always confuse me. I've been trying to figure out how to do it but nothing that I get makes sense! I'm frustrated and I need to get the correct answer by tonight.

    If anyone could point me in the right direction, I would be eternally grateful!

    Thanks.

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    Quote Originally Posted by i need a brain View Post
    The local hardware store sells screws in 1lb bags. Because the screws aren't identical, the number of screws per bag, although distributed normally,varies with a μ=115, and σ=6.

    A carpenter needs a total of 590 screws for a particular project. What is the probability that he will have enough screws if he buys five bags?

    --

    I don't know why I have trouble calculating even the simplest probability problems, but they always confuse me. I've been trying to figure out how to do it but nothing that I get makes sense! I'm frustrated and I need to get the correct answer by tonight.

    If anyone could point me in the right direction, I would be eternally grateful!

    Thanks.

    The key is that you have to add the population means and variances. That is, if the carpenter buys 5 bags then distriubtion will be normally distributed with Mu=575 and Sigma^2 = 180.

    As such, use the usual transformation to find the probability i.e.

    Z = (590 - 575) / Sqrt [ 180 ].

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    Thanks, Dragan!

    This is what I got:
    Z = (590 - 575) / Sqrt [ 180 ]= 1.11

    Looked up the z-score of 1.11 in the Unit Normal Table in the back of my text book and the proportion that I got that corresponds to that z-score(i used the proportion in the tail) is .1335.

    So, the probability that he will have enough screws if he buys five bags=.1335

    Is this correct?

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    Quote Originally Posted by i need a brain View Post
    Thanks, Dragan!

    This is what I got:
    Z = (590 - 575) / Sqrt [ 180 ]= 1.11

    Looked up the z-score of 1.11 in the Unit Normal Table in the back of my text book and the proportion that I got that corresponds to that z-score(i used the proportion in the tail) is .1335.

    So, the probability that he will have enough screws if he buys five bags=.1335

    Is this correct?
    Yes, on second thought, you're interpretation is the correct one.
    Last edited by Dragan; 06-04-2008 at 06:52 PM.

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    Re: probability with mean and standard deviation


    Please why did you use the variance to calculate the standard deviation and not standard deviation divided by square root of 5. Again when I checked 1.11 against the z table answer is 0.8665 whereas -1.11 is 0.1335 but, the result you got is not minus 1.11.
    Please could you clarify. Thank u.


    Quote Originally Posted by i need a brain View Post
    Thanks, Dragan!

    This is what I got:
    Z = (590 - 575) / Sqrt [ 180 ]= 1.11

    Looked up the z-score of 1.11 in the Unit Normal Table in the back of my text book and the proportion that I got that corresponds to that z-score(i used the proportion in the tail) is .1335.

    So, the probability that he will have enough screws if he buys five bags=.1335

    Is this correct?

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