Posted by **chante** on Tuesday, March 20, 2012 at 3:09pm.

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?

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**MathMate**, Tuesday, March 20, 2012 at 5:25pm
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

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