Normal Distribution

#1
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

z=(225-200)/30
=25/30= .83
=.2967

=.2967-0
=30%

Is this right?
 

heD

New Member
#2
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.
 

JohnM

TS Contributor
#4
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.