# Normal distribution, again

• Aug 5th 2013, 06:19 AM
Trefoil2727
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

• Aug 5th 2013, 07:07 PM
chiro
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.
• Aug 5th 2013, 08:24 PM
Trefoil2727
Re: Normal distribution, again
Quote:

Originally Posted by chiro
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?
• Aug 5th 2013, 08:41 PM
chiro
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.
• Aug 6th 2013, 06:10 AM
HallsofIvy
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.
• Aug 7th 2013, 05:07 AM
Trefoil2727
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?
• Aug 7th 2013, 07:00 AM
HallsofIvy
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.