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..

Printable View

- Aug 5th 2013, 06:19 AMTrefoil2727Normal 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.. - Aug 5th 2013, 07:07 PMchiroRe: 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 PMTrefoil2727Re: Normal distribution, again
- Aug 5th 2013, 08:41 PMchiroRe: 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 AMHallsofIvyRe: 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 AMTrefoil2727Re: 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 AMHallsofIvyRe: 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.