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Standard Normal Distribution
• Normal Distribution
f ( x) 
1

2
e
( x )2
2
,  X  
When x = 0 and σ = 1 then
f ( x) 
1
2
e
( x)2
2
,  X  
2
A normal random variable with   0 and   1
is called a standard normal variable and it’s denoted as Z.
Standard Normal Distribution
• Areas under the normal curve:• The
curve
of
any
continuous
probability function or density function
is constructed by integrate the function
between x1, x2 as:
Standard Normal Distribution
• Areas under the normal curve:-
Standard Normal Distribution
P( x1  X  x 2 ) 
x2

x1
1
2
e

1
( x )2
2
dx
The above function can be transforming
all the observations of it into a new set of
observations to a normal random variable
Z with mean = 0 and variance = 1, and this
transformation written
Standard Normal Distribution
as:
P( z1  Z  z 2 ) 
z2

z1
Z 
1
2
X  

e
1
 ( x )2
2
dz
Standard Normal Distribution
• Find
• P( z < 1.74) = 09591
Standard Normal Distribution
• Example 2:
• Given a standard normal distribution,
find the area under the curve lies:
• To the right of z=1.84.
• Between z = -1.97 and z = 0.86.
Standard Normal Distribution
• Example 2:
Standard Normal Distribution
1. Area left z = 1.84 = 1 – area right of
z = 1.84 = 1 – 0.9671 = 0.0329
2. Area between z = -1.97 and z = 0.86 is
equal to the area of z = 0.86 – area of
z = - 1.96 = 0.8051 – 0.0244 = 0.7807
Standard Normal Distribution
• Example 3:
• Given a standard normal distribution,
find the value of k such that:
• P (z > k) = 0.3015
• P (k < z < - 0.18) = 0.4197
Standard Normal Distribution
• Example 3:
‫‪Standard Normal Distribution‬‬
‫= ‪• P (z > k) = 1 - P (z < k) = 1 – 0.3015‬‬
‫‪0.6985‬‬
‫نبحث في الجدول عن هذه المساحة ونجد أنها تقابل القيمة‬
‫‪Z = 0.52k  0.52‬‬
‫‪• P (k < z < -0.18) we should find the area‬‬
‫‪of z = 0.18 = 0.4286‬‬
‫نبحث عن هذه المساحة في الجدول ونجد أن‬
‫‪• z = - 2.37 k  2.37‬‬
Standard Normal Distribution
• Example 4:
• Given a random variable X having a
normal distribution with
  50 and   10
, find the probability that X assumes a
value between 45 and 62.
Standard Normal Distribution
• Example 4:
Standard Normal Distribution
• Solution
• x1 = 45 , x2 = 62
X 
45  50
z1 

 0.5

10
X   62  50
z2 

 1 .2

10
P (45 < X < 62) = P ( -0.5 < z < 1.2)
Then
P ( -0.5 < z < 1.2) = P (z<1.2) – P(z < -0.5)
= 0.8849 – 0.3085 = 0.5764
Standard Normal Distribution
• Example 5:
• Given that X has a normal distribution
with find the probability that X assumes
a value greater than 362.
Standard Normal Distribution
• Solution:
• P (X > 362) = 1 – P (X < 362)
z
X 

362  300

 1.24
50
Then
P(X>362)= 1 – P (z < 1.24)
= 1 – 0.8925= 0.1075
Standard Normal Distribution
•
•
Example 6:
Given that a normal with find the
value of (x) that has:
1. 45% of the area to left.
2. 14% of the area to the right.
Standard Normal Distribution
•
Example 6:
Standard Normal Distribution
• Solution:
P (z < k) = 0.45
Then k = -0.13
z
X 

X  z    6  (0.13)  40  39.22
Standard Normal Distribution
• Solution:
14% of the area to the right = 1 – the area
of 14% of the left = 1 – 0.14 = 0.85
P (z < k) = 0.86
Then k = 1.08
X 
z

X  z    6  (1.08)  40  46.48