Download The Normal Distribution

LECTURE 5 OF 6
SUBTOPIC
12.2 CONTINUOUS PROBABILITY
DISTRIBUTION
1
At the end of the lesson,
students are able to:
(a)Understand the normal distribution

N  ,
2

(b) Find the mean and variance for the
normal distribution.
(c) Standardize the normal random
variable and use the tables.
2
The Continuous Random Variables
40
30
20
10
160 165 170 175 180 185 190
The histogram will approach a well defined smooth
curve called probability density function (p.d.f ) if
the sample size of a continuous variable is
increased and the class intervals are made as small
3
as possible.
The Normal Distribution
X
N ( , )
2
X  continuous random va riable
  the mean value of X in the population
  the sta ndard deviation of X in the population
 2  the va riance of X in the population
Examples of continuous random variable :
-The height of children of a certain age.
- The diameters of metal cylinders.
- The time taken by students to run 100 m. 4
The Normal Distribution
X
N ( , )
2
X is a continuous random
variable with p.d.f :
a b
1
f ( x) 
e
 2
 ( x   )2
2 2
,   x 
The PROBABILITY is represented by the AREA
of the shaded section of the graph.
b
P(a  x  b)   f ( x) dx
a
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Properties of The Normal Distribution
1.The p.d.f curve is characterized by a
symmetrical bell shaped curve and it is
symmetric about the mean.
So the mean, mode and median coincide.
2. The curve extends to infinity in either direction
  X   ,       and   0 .
3. The total area under the curve is 1 .



f ( x) dx  1

6
4. The maximum value for f(x) is approximately
1
when mean = mode = median= .
 2
5. The probability that X lies in the interval from
a to b is equal to the area under the curve
bounded by the ordinates x = a and x = b .
b
P(a  x  b)   f ( x) dx
a
0a b
7
The Standard Normal Distribution
Z
N (0,1)
where Z = the standard normal random variable ,
2

mean( ) = 0 and variance(  ) = 1
To find the probability of X ,continuous random
variable ,we need to change the normal dist.
into the standard normal dist. by using the
transformation formula.
8
Transformation Formula
Z
X 

Z  st andard normal random va riable
X  continuous random variable
  the mean value of X in the population
  the st andard deviation of X in the population
9
Transformation Formula
Z
X
N ( , )
2
P( X  x)

x
X 

Z
N (0,1)
P( Z  z )
0 z
10
How to use the Standard Normal Tables
Use the symmetrical property of the graph to find
the probability from the tables.
P( Z  0)  0.5 , P( Z  0)  0.5
Total
area
under
the curve
=1
z=0
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Symmetrical Properties
1. P( Z  a)  P( Z  a)
Equal Area
=
-a 0
0 a
Read the probability from the table with
z=a.
The standard normal table shows probability for
12
P
(
Z

z
)
the ‘upper end’ area that is
.
2. P ( Z   a )  1  P ( Z   a )
 1  P( Z  a)
Equal Area
=
-a 0
P( Z  a)
0 a
P( Z  a)
13
3. P(b  Z   a)  P( a  Z  b )
 P ( Z  a )  P ( Z  b)
Equal Area
=
-b -a 0
P(b  Z   a)
0a b
P ( a  Z  b)
14
4. P(a  Z  0)  P(0  Z  a )
 P( Z  0)  P( Z  a)
 0.5  P( Z  a)
Equal Area
=
-a
0
0
a
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Standard Normal Probability Tables
The table shows
0.01positive
0.02
0.03
z 0.00 only
values of Z.Use the
0.0 0.5000 0.4960 0.4920 0.4880
0.1 0.4602symmetrical
0.4562 0.4522 0.4483
0.2 0.4207
0.4168 0.4129
0.4090
property
of the
0.3 0.3821 0.3783 0.3745 0.3707
graph
. 0.3336
0.4 0.3446 0.3409
0.3372
0.04
0.05
0.06
0.07
0.08
0.09
0.4840
0.4443
0.4052
0.3669
0.3300
0.4801
0.4404
0.4013
0.3632
0.3264
0.4761
0.4364
0.3974
0.3594
0.3228
0.4721
0.4325
0.3936
0.3557
0.3192
0.4681
0.4286
0.3897
0.3520
0.3156
0.4641
0.4247
0.3859
0.3483
0.3121
i.e: P(Z > 0.05)= 0.4801
P( Z   0.48)  P( Z  0.48) = 0.3156
The complete table shows the probability for values
of z from 0.00 to 4.09.
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For z > 4.09, P(Z > 4.09) is less than 0.00002
Example 1
If Z N(0,1) , find the probability of the following
(a ) P( Z  2.5)
( b) P( Z  2.5)
(c) P( Z   0.6 )
(d ) P(  2.23  Z   1.23)
(e) P( Z  1.2)
(f ) P( Z  1.78 )
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Solution
(a) P( Z  2.5) 
0.00621
0 2.5
(b) P( Z  2.5)
1  P  Z  2.5
 1  0.00621
 0.99379
0 2.5
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(c) P( Z   0.6)
1  P( Z   0.6)
1  P( Z  0.6)
 1  0.2743
-0.6 0
 0.7257
(d ) P(2.23  Z   1.23)
 P(1.23  Z  2.23)
 P( Z 1.23)  P( Z  2.23)
 0.1093  0.0129
 0.0964
-2.23 -1.23
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(e) P( Z 1.2)
 P( Z 1.2)  P( Z   1.2)
 2 P( Z 1.2)
 2(0.1151)
-1.2 0 1.2
 0.2302
( f ) P( Z 1.78)
 P(1.78  Z  1.78)
 1  2 P( Z 1.78)
 1  2(0.0375)
 0.925
-1.78 0 1.78
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Example 2
The random variable Z has a standard normal
distribution. Determine the value of a if given:
(i) P( Z  a)  0.3745
(ii) P( Z  a)  0.6217
(iii) P( Z  a)  0.8472
(iv) P( Z  a)  0.2714
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Solution
(i ) P( Z  a)  0.3745
From the table :P ( Z  0.32)  0.3745
a  0.32
(ii) P( Z  a)  0.6217
Since the probability > 0.5, so a < 0. Thus,
0.6217
1 - 0.6217=0.3783
a 0
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P( Z  a)  0.3783
From the table :
P ( Z  0.31)  0.3783
 a   0.31
(iii) P( Z  a)  0.8472
P(  a  Z  a)  0.8472
1  2 P( Z  a)  0.8472
0.31
From the table :
P( Z 1.43)  0.0764
 a  1.43
1  0.8472
P( Z  a) 
 0.0764
2
23
(iv) P( Z  a)  0.2714
P(Z  a)  P  Z   a   0.2714
2 P( Z  a)  0.2714
P( Z  a)  0.1357
From the table :
P( Z  1.1)  0.1357
 a  1.1
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Example 3
The random variable X has a normal distribution
with mean 10 and standard deviation 2. Find :
(a) P( X  13)
(b) P  9  X 13 
(c) the value of a if P (a  X  12)  0.15
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Solution
X
2
N (10, 2 )
(a) P( X  13)
X  

 PZ 

 

Z 
X 

Given :
  10 ,   2
13  10 

 PZ 

2 

 P( Z 1.5)
 0.0668
26
(b) P 9  X 13 
13  10 
 9  10
 P
Z

2 
 2
 P( 0.5  Z  1.5)
-0.5 0 1.5
 1  P( Z  0.5)  P  Z  1.5 
 1  0.3085  0.0668
 0.625
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(c) the value of a if P(a  X  12)  0.15
P( a  X 12 )  0.15
12  10 
 a  10
P
Z 
  0.15
2 
 2
 a  10

P
 Z 1  0.15
 2

a  10 

PZ 
  P( Z 1)  0.15
2 

a  10 

PZ 
  0.15  0.1587  0.3087
2 

28
a  10 

PZ 
  0.3087
2 

From the table :
P( Z  0.5)  0.3085
a  10
 0.5
2
a  2(0.5)  10
 a  11
29
Example 4
The heights of matriculation students in KMPP is
normally distributed with a mean of 170 cm and
a standard deviation of 5 cm.
(a)What is the probability that a student, selected
at random, has a height between 174 cm and
178 cm?
(b) Out of a group of 150 matriculation students,
how many would be expected to have a
height less than 164 cm?
30
Solution
Z 
X=height of students in cm
X 

  170 ,   5
178  170 
 174  170
Z 
 a  P(174  X  178)  P 

5
5


 P(0.8  Z 1.6)
 P( Z  0.8)  P  Z 1.6
 0.2119  0.0548
 0.1571
31
164  170 

 b  P( X 164)  P  Z 

5


 P( Z  1.2)
 P( Z 1.2)
 0.1151
Number of students expected to have
a height less than 164 cm is
150 (0.1151)  17.265
 17 students
32
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Example 5
A machine is supposed to cut up logs into pieces
that are each 2 m. However , the machine
is an old one, and while the pieces that it produces
do have an average length of 2 m, 10% of the
pieces that it produces are less than 1.95 m.
Assuming that the lengths produced are
normally distributed, determine the proportion
of the pieces that are longer than 2.10 m.
34
Solution
X = the length of a log
Given : X
N (2.0,  ) , P( X  1.95)  0.10
2
Fi nd P( X  2.10).
 1.95  2.0 
P( X  1.95)  P  Z 
  0.10



From the table : P( Z  1.28)  0.1003
1.95  2.0

 1.28
35
1.95  2.0

 1.28
  0.0390
2.10  2.0 

P( X  2.10)  P  Z 

0.0390 

 P( Z  2.56)
 0.00523
 0.5% of the pieces of log have length
greater than 2.10 metres.
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