THE NORMAL
DISTRIBUTION
Lesson 2
Starter: Find P (Z<1.63)
To do this we begin with a
sketch of the normal
distribution.
We then mark a line to
represent Z=1.63
P(Z<1.63) is the area under
the curve to the left of a.
We now use the table to
look up this probability
1.63
P(Z<1.63) = 0.9484
Remember this from last lesson!
Note this result..
P(Z>a) = 1-P(Z<a)
And we found this really useful result
P(Z<-a)
=
1 - P(Z<a)
Objectives
• Finding the value of Z from a given probability.
Points to note
• P(Z<a) is greater than 0.5
•a>0
• P(Z<a) is less than 0.5
• a <0
Points to note
• P(Z>a) is less than 0.5
•a>0
• P(Z>a) is greater than 0.5
•a<0
Ex 1 Find the value of a such that
P(Z<a)=0.7852
Ex 1 Find the value of a such that
P(Z<a)=0.7852
Use the table
P(Z<a) = 0.7852
a = 0.79
a
Ex 2 Find the value of a such that
P(Z>a)=0.01
P(Z>a) = 0.01
P(Z<a) = 0.990
Look up this result is in the
main table.
It’s not there! Instead look at
the table of Percentage Points
to see if p=0.01 is listed.
a
P(Z>a) = 0.01
a = 2.3263
Ex 3 Find the value of a such that
P(Z>a)=0.0314
P(Z>a) = 0.0314
P(Z<a) = 1 – P(Z>a)
= 1-0.0314
= 0.9686
So from the main table
a = 1.86
a
a
Ex 4 Find the value of a such that
P(Z<a) = 0.0197
The table only lists values
for z>0 so we need to
reflect the problem in the
vertical axis…
a
p=0.0197
P(Z<z) = 1 – 0.0197
= 0.9803
From the table
z = 2.06
Therefore a = -2.06
z
p=0.9830
Further Learning:
• Read through Example 2 on P180
• Do Ex 9B on P181