The distribution of scores is normal with a ì = 100 and ó =15. What proportion of the population have scores A) Above 130? B) Below 90? C) Above 110?
Given μ=100; σ=15.
Standardize the distribution by
z(x,μ,σ*sup2;)=(x-μ)/σ
For example,
(a)
z(130,100,15^2)=(130-100)/15=2
Use a normal distribution table to find
P(z<2)=0.9772
P(z>2)=1-0.9772=0.0228
P(x>130)=P(z>2)=0.0228
(b)
z=(90-100)/15=-2/3
P(x<90)=
=P(z<-2/3)
=P(z>2/3)
=1-P(z<2/3)
=1-0.7475
=0.2525
(c)
P(x>110)
can be found similar to (a)
Many normal distribution tables are available online, such as:
http://www.math.unb.ca/~knight/utility/NormTble.htm