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Normal distribution probability question

  1. Jan 17, 2006 #1
    Got a question I need a little bit of help.

    Assume the scores on an aptitude are normally distributed with mean=500 and standard deviation=100

    What is the middle 40%?

    My workings


    => p(z1≤z≤z2)=p(z≤z2)-p(z≥z1)=p(z≤z2)-[1-p(z≤z1)]

    p(z≤z2)=0.7 p(z≤z1)=0.3
    from statistical tables
    => z2= -0.525 z1= 0.525

    z1=(x-500)/100=0.525 => x=500+52.5=552.5

    z2=(x-500)/100=-0.525 => x=500-52.5=447.5

    therefore the middle 40% is between 447.5 and 552.5.

    Now my question. Is that the correct method and approach, it is a bonus if I got it right. I only want to check the method really. Thanks.
  2. jcsd
  3. Jan 17, 2006 #2
    I think it's correct.

    Normal distribution is given [tex]\frac 1 {\sigma \sqrt{\pi}} exp(-(x-\mu)^2/\sigma^2)[/tex], where [tex]\mu[/tex] is the mean and [tex]\sigma[/tex] is the standard deviation.

    I looked at table and center 20% probability is (standard deviation)*(between 0.52 and 0.53)

    So your calculation looks fine. :smile:
    Last edited: Jan 17, 2006
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