Normal Distribution 2
To
be able to transform a normal distribution into Z
and use tables
To be able to use normal tables to find  and 
To use the normal distribution to answer questions
in context
Standard Normal Distribution
X ̴ N(, ²) can be transformed
into a Z score where Z ̴ N(0, 1²)
•z=x-

X ̴ N(50,
4²) so =50 and =4
Find P(x<53)
z = 53 – 50 = 0.75
4
Find P(z<0.75)
P(Z<0.75) = 0.7734
X ̴ N(50,
4²) so =50 and =4
Find P(x≤45)
z = 45 – 50 = -1.25
4
Find P(z<-1.25)
1-0.8944
P(Z<-1.25) = 0.1056
Y ̴ N(20,
9) so =20 and =3
P(Y>b)=0.0668
z = b – 20
3
0.0668
1- 0.0668=0.9332
From tables Z=1.5
1.5 = b – 20
3
4.5 = b – 20
b = 24.5
The random variable X ̴ N(,3²)
Given P(X>20)=0.20, find the value of 
P z = 20 –  = 0.20
3
P=0.20
From tables Z=0.8416
0.8416 = 20 – 
3
2.5248 =20–
 = 17.4752
The random variable X ̴ N(50,  ²)
Given P(X<46)=0.2119, find the value of 
P z = 46 – 50 = 0.2119

P=0.2119
P=0.7881
From tables Z=0.8 so on graph Z=-0.8
-0.8 = 46 –50

 = -4/-0.8
=5
The heights of a large group of women are normally
distributed with a mean of 165cm and a standard
deviation of 3.5cm. A woman is selected at random from
this group.
A) Find the probability that she is shorter than 160cm.
Steven is looking for a woman whose height is between
168cm and 174cm for a part in his next film.
B) Find the proportion of women from this group who met
Steven’s criteria
H=heights of women
H ̴ N(165, 3.5²)
A) Find the probability that she is shorter than 160cm.
Find P(x<160)
z = 160 – 165 = 0.75
3.5
P(z <1.43) = 0.9236
P(z < -1.43 = 1-0.923
= 0.1895
P(z < -1.43)
H=heights of women
H ̴ N(165, 3.5²)
B) Find P(168<x<174)
z = 168 – 165 = 0.86
3.5
z = 174 – 165 = 2.55
3.5
P(z >0.86) = 0.8051
P(z < 2.55) = 0.9946
P(0.86 <z < 2.55) = 0.9946-0.8051 = 0.1895
Boxes of chocolates are produced with a mean weight of
510g. Quality control checks show that 1% of boxes are
rejected because their weight is less than 485g.
A) Find the standard deviation of the weight of a box of
chocolates
b) Hence find the proportion of boxes that weigh more
than 525g.
W=weights of box of chocolates
W ̴ N(510,  ²)
P(W<485) = 0.01
P z <485 – 510

485 – 510

 = 10.7
= 0.01
= -2.3263
b) Hence find the proportion of boxes that weigh more
than 525g.
P(W > 525)
z = 525 – 510 = 1.40
10.7
P(z > 1.40) = 1 – 0.9192 = 0.0808
Normal distribution calculator