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Math Help - I need help with normal distributions...

  1. #1
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    I need help with normal distributions...

    Hello! :]

    I've had no trouble with normal distribution questions until I got to this one...

    Q. The mean of normal distribution is 50 and the standard deviation is 5. What percentage of the data:

    a. lies between 38 and 62

    (b+c+d are pretty much the same question, but with different numbers.)

    e. Is less than 38.

    I haven't worked with numbers that haven't fallen into any of the standard deviations. Could I have a little help please? I have no idea what to do. :[
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  2. #2
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    Quote Originally Posted by Hikaru View Post
    Hello! :]

    I've had no trouble with normal distribution questions until I got to this one...

    Q. The mean of normal distribution is 50 and the standard deviation is 5. What percentage of the data:

    a. lies between 38 and 62

    (b+c+d are pretty much the same question, but with different numbers.)

    e. Is less than 38.

    I haven't worked with numbers that haven't fallen into any of the standard deviations. Could I have a little help please? I have no idea what to do. :[
    Have you been taught how to get the z-value? And have you been taught how to use tables of values of the standard normal distribution?
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  3. #3
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    I have, but I've completely forgotten about it... I've been in and out of hospital recently, so I might have missed some vital notes here and there.

    I guess I'll ask my teacher tomorrow on how to use Z-values, thanks =]

    (But if it follows an easy formula, could you please post it anyway?)

    Uh... I don't think we've used any tables... O_o

    Quote Originally Posted by mr fantastic View Post
    Have you been taught how to get the z-value? And have you been taught how to use tables of values of the standard normal distribution?
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  4. #4
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    Hello,

    The principle of using a z-table is as following :

    If a random variable X is normally distributed, with mean m and standard deviation s, then the variable Z=(X-m)/s will follow a standard normal distribution, which representation will be the red curve here :


    Note that P(X<x)=P(Z<(x-m)/s).

    And there exists a table (http://www.science.mcmaster.ca/psych...e/z-table2.jpg) that gives you P(Z<z).

    After that, it's all juggling with these values. See the table so that you may be able to understand a little more efficiently.
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  5. #5
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    Quote Originally Posted by Hikaru View Post
    I have, but I've completely forgotten about it... I've been in and out of hospital recently, so I might have missed some vital notes here and there.

    I guess I'll ask my teacher tomorrow on how to use Z-values, thanks =]

    (But if it follows an easy formula, could you please post it anyway?)

    Uh... I don't think we've used any tables... O_o
    Z = (X - mean)/sd.

    X = 38 => Z = (38 - 50)/5 = -12/5 = -2.4

    X = 62 => Z = (62 - 50)/5 = 2.4.

    Therefore:

    Pr(38 < X < 62) = Pr(-2.4 < Z < 2.4) = Pr(Z < 2.4) - Pr(Z < -2.4)

    = Pr(Z < 2.4) - Pr(Z > 2.4)

    using symmetry of the normal distribution

    = Pr(Z < 2.4) - (1 - Pr(Z < 2.4))

    = 2 Pr(Z < 2.4) - 1.

    And you get Pr(Z < 2.4) using the standard normal distribution tables (unless you use technology).

    Looking at your previous posted questions (and the replies), I'm surprised that you're having trouble here .....?
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