Z-Score and probability

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July 14th 2011, 01:52 PM
sfspitfire23

Z-Score and probability

Fellas,

Say I have a list of numbers (normally distributed) with a mean of -200 and a SD of 50. Now, say I would like to find the probability of getting a less than -100 and greater than +100. For the -100 I get Z=2 and for +100 I get z=6. Now I integrate against a standard normal distribution from (-infty to 2) for the -100 and from (6 to infty) for the 100 one.

Something seems wrong with this method...Basically, I'm trying to figure out how to calculate the probability that the Dow Jones makes a 500 point swing in either direction given that its returns are normally distributed.

Thanks!
July 14th 2011, 02:14 PM
Siron

Re: Z-Score and probability

To find the probability after calculated the z-score you can also use 'the table of the normal distribution'.
July 14th 2011, 02:25 PM
pickslides

Re: Z-Score and probability

Quote:

Originally Posted by sfspitfire23

Basically, I'm trying to figure out how to calculate the probability that the Dow Jones makes a 500 point swing in either direction given that its returns are normally distributed.

Thanks!

Then you are looking for $\displaystyle P(-200-500<X<-200+500) = P(-700<X<300)$

$\displaystyle = P\left(\frac{-700-(-200)}{50}<Z<\frac{300-(-200)}{50}\right)=\dots$
July 14th 2011, 03:02 PM
sfspitfire23

Re: Z-Score and probability

Sorry pickslides, I should have been more clear. The numbers I used in the problem are not from the Dow Jones. I just used that as an example. What I understand you saying is this:

If I want to find the probability of a 500 point swing, I should do this: P(mean-500<X<mean+500)=P(t<X<y)=P(t-mean/sigma<Z<y-mean/sigma).

Then if I wanted the prob of just a 500 point decline it would be P(mean-500<)=P(t<X)=P(t-mean/sigma<Z) and integrate from -infty to Z.

Right?

Thx
July 14th 2011, 03:12 PM
pickslides

Re: Z-Score and probability

Quote:

Originally Posted by sfspitfire23

If I want to find the probability of a 500 point swing, I should do this: P(mean-500<X<mean+500)=P(t<X<y)=P(t-mean/sigma<Z<y-mean/sigma).

Right?

Thx

This seems ok, but I would just use a table as suggested in post #2.
July 14th 2011, 10:07 PM
CaptainBlack

Re: Z-Score and probability

Quote:

Originally Posted by sfspitfire23

Fellas,

Say I have a list of numbers (normally distributed) with a mean of -200 and a SD of 50. Now, say I would like to find the probability of getting a less than -100 and greater than +100. For the -100 I get Z=2 and for +100 I get z=6. Now I integrate against a standard normal distribution from (-infty to 2) for the -100 and from (6 to infty) for the 100 one.

Something seems wrong with this method...Basically, I'm trying to figure out how to calculate the probability that the Dow Jones makes a 500 point swing in either direction given that its returns are normally distributed.

Thanks!

What evidence do you have that they are normally distributed (or stationary)?

CB