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Math Help - Normal Distribution

  1. #1
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    Normal Distribution

    The average height for players on a certain basketball team is 73 inches, with a standard deviation of 6 inches. What percent of players' hgeights are:

    a.) between 70 and 73 inches?

    I did:

    z1 = 70 - 73/6
    z1 = -3/6
    z1 = -0.5
    z1 = 0.3085


    z2 = 73 - 73/6
    z2 = 0/6
    z2 = 0
    z2 = 0.5000

    0.3085 + 0.5000 = 0.8135
    1 - 0.8135 = 0.1915 which is 19.15%

    So 19.15% of the people are between 70 and 73 inches. But my question is there a faster simpler way of solving this?
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  2. #2
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    Quote Originally Posted by lax600 View Post
    The average height for players on a certain basketball team is 73 inches, with a standard deviation of 6 inches. What percent of players' hgeights are:

    a.) between 70 and 73 inches?

    I did:

    z1 = 70 - 73/6
    z1 = -3/6
    z1 = -0.5
    z1 = 0.3085


    z2 = 73 - 73/6 this work is redundant
    z2 = 0/6
    z2 = 0
    z2 = 0.5000

    0.3085 + 0.5000 = 0.8135
    1 - 0.8135 = 0.1915 which is 19.15%

    So 19.15% of the people are between 70 and 73 inches. But my question is there a faster simpler way of solving this?

    By definition for a normal distribution the mean is the mode and median value, so there was no need to calculate the probability of someone being less then 73 inches.

    Bobak
    Last edited by bobak; June 7th 2008 at 04:08 PM.
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  3. #3
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    i need to find those who were in between 70 and 73 inches
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  4. #4
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    Quote Originally Posted by lax600 View Post
    The average height for players on a certain basketball team is 73 inches, with a standard deviation of 6 inches. What percent of players' hgeights are:

    a.) between 70 and 73 inches?

    I did:

    z1 = 70 - 73/6
    z1 = -3/6
    z1 = -0.5
    z1 = 0.3085


    z2 = 73 - 73/6
    z2 = 0/6
    z2 = 0
    z2 = 0.5000

    0.3085 + 0.5000 = 0.8135
    1 - 0.8135 = 0.1915 which is 19.15%

    So 19.15% of the people are between 70 and 73 inches. But my question is there a faster simpler way of solving this?
    Yes. As bobak already mentioned, 73 is the mean so you know that z2 = 0 without having to do the calculation you've done to get it. Then the required probability is 0.5 - 0.3085 = 0.1915 etc.
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