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Math Help - Standard Deviation and Normal Distribution

  1. #1
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    Standard Deviation and Normal Distribution

    Hi I know how to calculate standard deviation but am having trouble with questions the in the image. Please help.. Either 8 or 10 worked out would be a huge help.
    Attached Thumbnails Attached Thumbnails Standard Deviation and Normal Distribution-math-problems.jpg  
    Last edited by Arbitur; January 25th 2006 at 10:24 AM.
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    Since 173 is the mean and standard deviation is 7.7 cm. Thus, One standard deviation to the left and right takes 68% (a rule) thus, 165.3-180.7 take 68% of all the students. Thus 32% below 165.3 or above 180.7 cm. Now the normal distribution is symettric thus, 16% below 165.3 and 16% aboce 180.7 cm. Look at the problem it say the taller 15% thus, approximately the taller 16% thus the range is approximately 180.7 cm.
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    Look at the hint to look for the lower part of normal distribution. This problem unlike the first requires you to move 2 standard deviations. This will take on 95% (a rule) thus, between .9-4.1 years are included in these 95%. Thus, 5% below .9 or above 4.1 years. But because the normal distribution is symettric thus, 2.5% below .9 years or above 4.1 years. The problem asks for the 2.5% thus the range is .9 years.
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    thanks... I know what your saying but could you just show how you calculated just one of those if possible? Thanks.
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    Let P(a\leq x\leq b) be the precentage (or probability) of all values between a and b.

    There is a theorem (theorem-1) in statistics that if something is normal distributed and \bar x is the mean, and \sigma is the standard deviation then,
    P(\bar x-\sigma\leq x\leq \bar x+\sigma)=.68
    P(\bar x-2\sigma\leq x\leq \bar x+2\sigma)=.95.
    Also because (theorem-2) the normal distribution has symettrism we have that If P(\bar x-a\leq x\leq \bar x+a)=p Then, P(x\leq \bar x-a)=P(\bar x+a\leq x)=\frac{1-p}{2}. The reason being, because 1-p excludes the first one and then diving by 2 is because of symettrism.


    Now to do problem 2, the mean \bar x=2.5 and the standard deviation is \sigma=.8. By the above explanations and theorem-1 P(.9\leq x\leq 4.1)=.95 (because 2 standard deviations). Next, use theorem-2 thus, P(x\leq .9)=\frac{1-.95}{2}=.025=2.5% which you where trying to find.
    Thus the answer is x\leq .9.
    Q.E.D.
    Last edited by ThePerfectHacker; January 25th 2006 at 03:11 PM.
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    Arr.. thanks for trying but I still don't get how you got to the answer. I'll try a tutor somewhere for 1 session or something. (Taking this course online.) Thanks anyways.
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    Heh.. feel stupid.. cant even figure out how you did it. Thanks neways guess im just not smart enough

    Shouldnt it be 17% instead of 16%?
    Last edited by Arbitur; January 25th 2006 at 03:38 PM.
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    Quote Originally Posted by Arbitur
    Heh.. feel stupid.. cant even figure out how you did it. Thanks neways guess im just not smart enough

    Shouldnt it be 17% instead of 16%?
    How? 100-68=32 and 32/2=16
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    Quote Originally Posted by Arbitur
    Hi I know how to calculate standard deviation but am having trouble with questions the in the image. Please help.. Either 8 or 10 worked out would be a huge help.
    8. If the school has a normal distribution of height h, with mean 173 cm and
    standard deviation 7.7 cm, then:

    z=\frac{h-173}{7.7},

    has a standard normal distribution. The height of the beams is such that
    15 \% will hit their heads, call this height h_c.

    Call the corresponding z score z_c, then:

    z_c=\frac{h_c-173}{7.7},

    So 15 \% of z scores exceed z_c, or 85 \% of z scores are less than z_c.

    Now look up the value of z_c which gives p(z<z_c)=0.85 in a table
    of the cumulative normal distribution, doing this gives z_c \approx 1.036.

    The corresponding height (which will be the height of the beams) will be
    the solution of:

    z_c = \frac{h_c - 173}{7.7},

    which when we plug in the value of z_c gives:

    h_c=180.98.

    Thus all students taller than 180.98 cm. will be affected.

    RonL
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