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Math Help - Please Help with this Stat Q

  1. #1
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    Talking Please Help with this Stat Q

    Hi Everybody,

    I am stuck on these stat questions and I would very much appreciate any help that I get.

    The introductory salaries for medical billing clerks are normally distributed with a mean of $24,800 and a standard deviation of $2850.

    a. What percent of introductory salaries are less than $23,000?
    b. What percent of introductory salaries are between $21,250 and $23,750?

    -and-

    The amount of annual snowfall in a certain mountain range is normally distributed with a mean of 109 inches, and a standard deviation of 10 inches.

    a. What is the probability that the mean annual snowfall during 40 randomly picked years will exceed 111.8 inches?
    b. What is the probability that the mean annual snowfall during 40 randomly picked years will be between 105 inches and 112 inches?


    Thanks.
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  2. #2
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    Quote Originally Posted by al_pacino View Post
    Hi Everybody,

    I am stuck on these stat questions and I would very much appreciate any help that I get.

    The introductory salaries for medical billing clerks are normally distributed with a mean of $24,800 and a standard deviation of $2850.

    a. What percent of introductory salaries are less than $23,000?
    b. What percent of introductory salaries are between $21,250 and $23,750?

    [snip]
    Let X be the random variable introductory salaries ($).

    a. Calculate Pr(X < 23000) and multiply the answer by 100.

    I get approximately 26.38%.

    b. Calculate Pr(21250 < X < 23750) and multiply the answer by 100.

    I get approximately 24.98%.

    Without knowing how you've been taught to calculate such probabilities (using tables, technology such as TI-84 or TI-89 calculators, numerical integration of the pdf etc.) it's impossible to provide further relevant help.

    Quote Originally Posted by al_pacino View Post
    [snip]
    The amount of annual snowfall in a certain mountain range is normally distributed with a mean of 109 inches, and a standard deviation of 10 inches.

    a. What is the probability that the mean annual snowfall during 40 randomly picked years will exceed 111.8 inches?
    b. What is the probability that the mean annual snowfall during 40 randomly picked years will be between 105 inches and 112 inches?


    Thanks.
    Let X be the random variable annual snowfall (inches).

    a. Calculate Pr(X > 111.8)

    I get 0.3897 (correct to four decimal places).

    Let Y be the random variable number of years out of 40 randomly picked years that X > 111.8.

    Then Y ~ Binomial(n = 40, p = 0.3897).

    You need to calculate Pr(Y = 40) ......

    b. Proceed similar to part a. Calculate Pr(105 < X < 112) etc.
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  3. #3
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    Quote Originally Posted by al_pacino View Post
    Hi Everybody,

    I am stuck on these stat questions and I would very much appreciate any help that I get.

    The introductory salaries for medical billing clerks are normally distributed with a mean of $24,800 and a standard deviation of $2850.

    a. What percent of introductory salaries are less than $23,000?
    b. What percent of introductory salaries are between $21,250 and $23,750?
    Al,

    To see the details of how Mr. Fantastic came up with Question a, go here:

    Sampling Distribution of the Mean

    \overline{x} = 23000
    \mu = 24800
    \sigma = 2850
    n = 1

    This will show you the math behind Fantastic's answer. Like he said, you want P(Z >= z). Know that \mu is the "sample" mean.
    Last edited by mathceleb; August 13th 2009 at 12:48 AM.
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  4. #4
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    Ok, I've built the other lesson as well. For Fantastics 2nd answer, go here again:

    Sampling Distribution of the Mean

    All your entries from above will remain the same except for \overline{X}.

    For that, you have 2 of them now, so enter them as 21250,23750. All other inputs stay the same. Press the P(a < Z < b) button and you will get Fantastic's answer within 0.0002.

    Let me know if you have questions.
    Last edited by mathceleb; August 13th 2009 at 12:48 AM.
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