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    a study shows attendance at football games revealed that the distribution of attendance is normally distributed with a mean of 10,000 and a standard deviation of 2,000
    What is the probability a game has an attendance of 13,500 or more?

    What percent of the games have an attendance between 8,000 and 11,500?

    Ten percent of the games have an attendance of how many or less?
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  2. #2
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    Quote Originally Posted by numnuts View Post
    a study shows attendance at football games revealed that the distribution of attendance is normally distributed with a mean of 10,000 and a standard deviation of 2,000
    What is the probability a game has an attendance of 13,500 or more?

    What percent of the games have an attendance between 8,000 and 11,500?

    Ten percent of the games have an attendance of how many or less?
    I'm not an expert in statistics, so I don't remember what the pdf is for normal distributions, but it seems to me that all of these questions could be answered by integrating the pdf. The first question you integrate from 13,500 to infinity, the second you integrate from 8,000 to 11,500, and the third you solve for x in the equation \int _{-\infty}^x p(t) dt = 0.1, where p(t) is the pdf of your distribution.
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    huh

    I don't understand at all
    That went right over this girls head
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    The normal distribution has what is known as a probability density function (pdf). If you integrate it for certain ranges of values, it tells you what the probability is of a value being in that range. Alternatively, you may have tables that tell you percentages based on how far the values are in terms of the standard deviation from the mean. I can tell you this: the probability that a value will fall within one standard deviation from the mean is about two-thirds. I know that doesn't directly answer any of the questions you have, but your book probably has a table of probabilities based on the number of standard deviations from the mean the endpoints of your target range are, and you could use this table to answer these questions as well.
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    Quote Originally Posted by numnuts View Post
    a study shows attendance at football games revealed that the distribution of attendance is normally distributed with a mean of 10,000 and a standard deviation of 2,000
    What is the probability a game has an attendance of 13,500 or more?

    What percent of the games have an attendance between 8,000 and 11,500?

    Ten percent of the games have an attendance of how many or less?
    I'm always a sucker for a damsel in distress:


    1. Z = \frac{X - \mu}{\sigma} = \frac{13,500 - 10,000}{2,000} = 1.75.

    Therefore \Pr(X > 13,500) = \Pr(Z > 1.75).

    Use your Standard Normal Distribution tables to get this probability.

    Note: Depending on your tables, you might need to do the following:

    \Pr(Z > 1.75) = 1 - \Pr(Z < 1.75).

    Get \Pr(Z < 1.75) from your tables.

    ------------------------------------------------------------------------------------------------------

    2. Z = \frac{X - \mu}{\sigma} = \frac{8,000 - 10,000}{2,000} = -1.

    Z = \frac{X - \mu}{\sigma} = \frac{11,500 - 10,000}{2,000} = 0.75.

    Therefore \Pr(8,000 < X < 11,500) = \Pr(-1 < Z < 0.75) = \Pr(Z < 0.75) - \Pr(Z < -1).

    Use your Standard Normal Distribution tables to get these probabilities. I cheated and used my TI-89 and got 0.6147 as the answer.

    So the percentage is (100)(0.6147) = 61.47%.

    Note: To get \Pr(Z < -1) from your tables you might need to do the following gymnastics:

    \Pr(Z < -1) = \Pr(Z > 1)

    by symmetry of the normal distribution

    = 1 - \Pr(Z < 1).

    Get \Pr(Z < 1) from your tables.

    ----------------------------------------------------------------------------------------------------------------

    3. You want the value of x* such that \Pr(X < x*) = 0.1.

    To get it you first need to find the value of z* such that \Pr(z < z*) = 0.1:

    \Pr(z < z*) = 0.1 \Rightarrow \Pr( z > -z*) = 0.1

    by symmetry of the normal distribution

    \Rightarrow \Pr(z < -z*) = 0.9 \Rightarrow -z* = 1.2816

    using the tables in the 'inverse way'

    \Rightarrow z* = -1.2816.

    Z = \frac{X - \mu}{\sigma} \Rightarrow z* = \frac{x* - 10,000}{2,000} \Rightarrow -1.2816 = \frac{x* - 10,000}{2,000} \Rightarrow x* = \, .....

    and that's the answer.
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