Normal distribution + calculation of Z values

[JamesGoh]

For a normal distribution with E[x]=0 and Var(X)=1, how do we determine the Z-value of a particular percentage ?

i.e. if the percentage is 5%, how do we know that Z(5%)= 1.645 ?

is there a calculation involved or do we get it from observing the x-axis of the normal distribution ?

 

[mathman]

There are tables of the error function (integral of normal) for mean 0 and variance 1. These have been constructed by numerical integration. For a particular value, just look it up.

Google "normal distribution table".

 

[Bacle2]

A nice rule to remember too, is the "1-2-3 rule" aka 68-95-99.7 rule:

In a normal distribution, 68% of the data is within 1σ of the mean ,

(so that, by symmetry, 34% is right of the mean and 34% is left- of the mean)

95% of the data is within 2σ, and 99.7% of all data is within 3σ of μ.

Also, using the fact that the normal distribution is symmetric also simplifies

a lot of other calculations.

Notice an approximation for your 5% question: you know that the percentile for

the mean ; z(μ)=0 , is 50-percentile. Then, by symmetry, the value σ=1 gives

you the 84th percentile. Now, z=2 would give you the 97.5th percentile--

too far. So 95th percentile is somewhere between z=1 and z=2 . More

advanced tricks will allow you to zone-in more carefully, but this is a nice

rule- of- thumb.

 

[ashimgiyanani]

To add to this question itself. I have a CDF so a column of 19 values. [0.05, 0.1, 0.15...0.95] and i have the corresponding x values [779, 784, 793...877 ]...again 19 values

When i plot graphically each other, it gives a smooth CDF following a normal curve however i am not sure if its normal, how do u derive if its normal since i do not have the random numbers.

Also I made a PDF formula for these values with d(CDF)/dx which means...(cdf2-cdf1)/(x2-x1) as coming from various textbooks. Is it right?

How do i generate Standard deviation from such a CDF?? Currently thinking that as 95% data is under 4σ area...[x(95) - x(5)]/4 will approximately give me the Standard deviation...Can someone suggest me the right way here