for a standard normal random variable

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November 5th 2008, 05:48 PM
stewardt

for a standard normal random variable

a) find P(-.12<Z<.54)

b)find P(z=0)

c) find the z-score that seperates the bottom 40% frin the top 60%

D) find the z-scores for which 95 % of the distribution is between -z and z

If you are able to help me I want to thank you in advance
If you can help me more on a one on one basis then tell me so
let me know how much you would charge to help me
November 6th 2008, 12:15 AM
mr fantastic

Quote:

Originally Posted by stewardt

a) find P(-.12<Z<.54)

Mr F says: = Pr(Z < 0.54) - Pr(Z < - 0.12) = Pr(Z < 0.54) - Pr(Z > 0.12) = Pr(Z < 0.54) - [1 - Pr(Z < 0.12)] = Pr(Z < 0.54) + Pr(Z < 0.12) - 1.

Now use your tables.

b)find P(z=0) Mr F says: You should know that since Z is a continuous random variable, Pr(Z = a) = 0.

c) find the z-score that seperates the bottom 40% frin the top 60%

Mr F says: Pr(Z < - z*) = 0.4 => Pr(Z > z*) = 0.4 => Pr(Z < z*) = 0.6. Get z* by using your tables in a reverse way. Hence get -z*.

D) find the z-scores for which 95 % of the distribution is between -z and z

Mr F says: Pr(Z > z) = 0.025 => Pr(Z < z) = 0.975. Get z by using your tables in a reverse way.

If you are able to help me I want to thank you in advance
If you can help me more on a one on one basis then tell me so
let me know how much you would charge to help me

..
November 6th 2008, 01:47 AM
stewardt

im still confused

Mr F says: = Pr(Z < 0.54) - Pr(Z < - 0.12) = Pr(Z < 0.54) - Pr(Z > 0.12) = Pr(Z < 0.54) - [1 - Pr(Z < 0.12)] = Pr(Z < 0.54) + Pr(Z < 0.12) - 1.
do I look at the table where -1 is
Mr F says: Pr(Z < - z*) = 0.4 => Pr(Z > z*) = 0.4 => Pr(Z < z*) = 0.6. Get z* by using your tables in a reverse way. Hence get -z*.
are you saying here 1-.xxxx
Mr F says: Pr(Z > z) = 0.025 => Pr(Z < z) = 0.975. Get z by using your tables in a reverse way.
I have no idea what to do
November 6th 2008, 04:26 AM
mr fantastic

Quote:

Originally Posted by stewardt

im still confused

Mr F says: = Pr(Z < 0.54) - Pr(Z < - 0.12) = Pr(Z < 0.54) - Pr(Z > 0.12) = Pr(Z < 0.54) - [1 - Pr(Z < 0.12)] = Pr(Z < 0.54) + Pr(Z < 0.12) - 1.
do I look at the table where -1 is

Mr F says: No. You use your tables to look up Pr(Z < 0.54) and Pr(Z < 0.12).

Mr F says: Pr(Z < - z*) = 0.4 => Pr(Z > z*) = 0.4 => Pr(Z < z*) = 0.6. Get z* by using your tables in a reverse way. Hence get -z*.
are you saying here 1-.xxxx

Mr F says: No. I'm saying go to the main body of your tables and find 0.6. Then work backwards from it to find what the value of z is.

Mr F says: Pr(Z > z) = 0.025 => Pr(Z < z) = 0.975. Get z by using your tables in a reverse way.
I have no idea what to do

Mr F says: If that's true then you're strongly advised to seek urgent help from your teacher.

Go to the main body of your tables and find 0.975. Then work backwards from it to find what the value of z is.

..
November 6th 2008, 05:07 AM
stewardt

Quote:

Originally Posted by mr fantastic

..

D) find the z-scores for which 95 % of the distribution is between -z and z

Mr F says: Pr(Z > z) = 0.025 => Pr(Z < z) = 0.975. Get z by using your tables in a reverse way.

the answer to this question is 1.96 then
November 6th 2008, 05:09 AM
mr fantastic

Quote:

Originally Posted by stewardt

D) find the z-scores for which 95 % of the distribution is between -z and z

Mr F says: Pr(Z > z) = 0.025 => Pr(Z < z) = 0.975. Get z by using your tables in a reverse way.

the answer to this question is 1.96 then

Yes, z = 1.96 (correct to two decimal places)
November 6th 2008, 05:32 AM
stewardt

Quote:

Originally Posted by mr fantastic

Yes, z = 1.96 (correct to two decimal places)

find the z-score that corresponds to the third quartile Q3
is it .7486
November 6th 2008, 05:37 AM
mr fantastic

Quote:

Originally Posted by stewardt

find the z-score that corresponds to the third quartile Q3
is it .7486

No. Pr(Z > z*) = 0.25 => Pr(Z < z*) = 0.75 => z* = 0.6745 (correct to four decimal places).

And new questions should be asked in a new thread.