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Thread: Computing pobabilities for any normal probability Dist.

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    Computing pobabilities for any normal probability Dist.




    Greetings all,

    I am a math challenged individual in the middle of a stats class and (you've heard it before) seriously struggling with story problems. Here is one:

    Sleep foundation states that the average nights sleep is 6.8 hrs. Assume the standard deviation is .6 hrs. and that the probability distribution is normal.

    a. What is the probability that a randomly selected person sleeps more than 8 hrs.?
    b. " " 6 hrs. or less
    c. Doctors suggest getting between 7 and 9 hrs. of sleep each night. What percentage of the pop. gets this much sleep?

    I'm not so much looking for someone to just give me the answers, but to show me HOW,( in a manner that is understandable to a complete stats novice) to get the answers. I need to know how so I can pass a final. (BTW,This question is not on the final).

    Thanks for your help!

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    Re: Computing pobabilities for any normal probability Dist.

    Let X be the amount oh night sleep (in hour).

    First of all, from the question you should know X \sim \mathcal{N}(6.8, 0.6^2)

    Once you know a random variable follows a normal distribution, you can compute the related probabilities by either:

    1) Using statistical software like R;
    2) Looking at the statistical table like http://www.normaltable.com/ in the link above

    to obtain an approximate answer. So it really depends on which tool you use. Once you get use to the method, the question is straight forward - it can be just 1 command in the software, or just cost you several seconds to pick out the values in the table (in this way of course you need to convert to standard normal first).

    Of course another important part is that you know what probability the question is asking for. For example a) part ask you about \Pr\{X > 8\}.

    You may have a try first and ask follow-up questions if any.

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    Re: Computing pobabilities for any normal probability Dist.


    Ok,

    So for (a). I used excel and entered this formula:=NORMDIST(8,6.8,.6TRUE) and got .97725 then subtracted 1-.97725 which = .02275 which is the probability that a randomly selected person sleeps more than 8 hrs. Correct?

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