Question

A normal distribution has a mean of 0.40 and standard deviation of 0.028. What percentage of observations will lie between 0.372 and 0.428?

Answer 1

Solution:
To evaluate for P(0.372<x<0.428) we shall proceed as follows:
z-score is given by:
z=(x-μ)/σ
thus when x=0.372:
z=(0.372-0.4)/0.028
z=-1
thus
P(x<0.372)=P(z<-1)=0.1587

when x=0.428
z=(0.428-0.4)/(0.028)=1
P(x<0.428)=P(z<1)=0.8413
thus
P(0.372<x<0.428) =P(z<1)-P(z<-1)=0.8413-0.1587=0.6826