Normal Distribution

In a normal distribution with u=200 and o=30, find the percentage of values that are between 200 and 225

z= (200-200)/30
=0/30 = 0

=25/30= .83


Is this right?


New Member
what is u and o in this case? :confused:
Usually to find this you'd need to integrate from 200 to 225 the normal prob. density f-n with the parameters given.
If u is the mean, o is variance, you'd want to integrate 1/sqrt(2Pi)*exp{-(x-200)^2/30} from 200 to 225.


TS Contributor
Your answer is correct. You don't need to do integration here - it's just a problem of finding the areas under the normal curve, using tables.