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Math Help - Normal distribution, again

  1. #1
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    Normal distribution, again

    If X is normally distributed with a mean of 5 and a variance of 4, find the value of k so that P(-k<X<k)=0.80

    P(-k<X<k)=0.80
    let (k-5)/2=z
    1-2P(Z>z)=0.8
    P(Z>z)=0.1
    (k-5)/2=1.282
    k=7.564

    and the answer is 2.564..
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  2. #2
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    Re: Normal distribution, again

    Hey Trefoil2727.

    I think the problem is that you have P(Z>z) instead of P(Z<z). Since you are looking at P(-k < X < k) then it means that you are looking at 1 - 2*P(X < k) due to symmetry of the normal distribution which means that 1 - 2*P(Z < z) = 0.8 where z = (k - 5)/2

    If you use this you get using R:

    > (qnorm(0.1,0,1)*2+5)
    [1] 2.436897

    Since I used a computer for this, if the answer is based on using tables then it will probably lose precision since the answer of 2.43 is in the tails of the distribution.
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  3. #3
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    Re: Normal distribution, again

    Quote Originally Posted by chiro View Post
    Hey Trefoil2727.

    I think the problem is that you have P(Z>z) instead of P(Z<z). Since you are looking at P(-k < X < k) then it means that you are looking at 1 - 2*P(X < k) due to symmetry of the normal distribution which means that 1 - 2*P(Z < z) = 0.8 where z = (k - 5)/2

    If you use this you get using R:

    > (qnorm(0.1,0,1)*2+5)
    [1] 2.436897

    Since I used a computer for this, if the answer is based on using tables then it will probably lose precision since the answer of 2.43 is in the tails of the distribution.
    huh, how can this be done?
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  4. #4
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    Re: Normal distribution, again

    The R package allows you to do a lot of statistical computations very easily.

    qnorm(0.1,0,1) returns the value of z where P(Z < z) = 0.1.

    You can do the same thing by getting a statistical normal table and finding the value of z where P(Z < z).

    The difference is that qnorm uses a computational algorithm where-as a table just gives values (calculating by the same kind of computer algorithm) and you look up those values on paper rather than with a computer.
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  5. #5
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    Re: Normal distribution, again

    You seem to have used .08 rather than .8. If you look at the Normal Distribution table here, you will see that "z= .1" corresponds to P(z)= .04, NOT .40. (Since this app is giving P(z> 0), and your problem has -.8< z< .8, you divide by 2 to get 0< z< .4.) P(z< .4) corresponds to z= 1.28, not .1.
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  6. #6
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    Re: Normal distribution, again

    sorry I still couldn't get it, why there's -0.8<z<0.8? i'm suppose to find the k right?
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  7. #7
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    Re: Normal distribution, again

    Oh, blast. I was reading the problem backwards! P(-k< z< k)= .80 is the same as P(0< z< k)= .40 and I get z= 1.28 for that.
    So (z- 5)/2= 1.28, z- 5= 2(1.28)= 2.56, z= 7.56 which is the answer you got.
    Thanks from Trefoil2727
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