lecture 7-3-2014 - normal distribution ii(1)
TRANSCRIPT
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Special Continuous ProbabilityDistributions
Normal Distribution II
Instructor: Dr. Gaurav Bhatnagar
22001: PROBABILITY AND STATISTICS
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Normal Distribution - Definition
Probability estimation for Normal distribution.
a b
f(x) P(a
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Normal Distribution - Definition
Probability estimation for Normal distribution.
a
f(x)
P(X
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Normal Distribution - Definition
Probability estimation for Normal distribution.
a
f(x)
P(X> a)
P( ) = ( ) 1 ( ) 1 P( )a
X a f x dx F a X a
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Standard Normal Distribution - Definition
A random variable is said to have a standard normal
distribution if the mean and standard deviation are 0 and 1
respectively.
The probability density function is given by
Notation:
Z
2
21( ) , (2)
2
z
f z e z
~ (0,1)X N
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Standard Normal Distribution - Definition
Moments:
Mean: r= 1
Variance:
21
21
' ( ) (3)2
zr r
r z f z dz z e dz
21
2
1
1' 0 (4)
2
z
ze dz
2 22
2 1( ) ( ) ' ' 1 (5)V X E X E X
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Standard Normal Distribution - Definition
Moment Generating Function:
2
2
1
2
1
2
( )
1
2
(6)
zt
Z
zzt
t
M t E e
e e dz
e
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Limitation of Normal Distribution
Each normal distribution with its own values ofand
would need its own calculation of the area under various
points on the curve.
100 200
(= 100, = 50)
(= 0, = 1)
0 2
Possible Solution: Standardizing the normal distribution
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Standardizing or Z-score
Suppose:X ~ N(,)
Form a new random variable (Z) by subtracting the mean
() fromXand dividing by the standard deviation (), i.e.,
It is clear that the mean and standard deviation of new
random variable (Z) is 0 and 1, i.e.,Z ~ N(0,1)
This process is called standardizing the random variable
X.
Z-scores make it easier to compare data values
measured on different scales.
XZ
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Meaning of Z-score
Suppose marks of Probability and Statistics among
students are normally distributed with a mean of 25 and a
standard deviation of 10. If a student scores a 45, whatwould be the z-score?
- 45 252
10
Xz
Soln:TheZ-score is
The z-score would be 2 which means her score is two
standard deviations above the mean.
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Meaning of Z-score
Suppose mid-sem 1 marks of students in Probability and
Statistics has a mean of 9 and a standard deviation of 4
whereas their Signal, System and Networks marks has amean of 12 and a standard deviation of 8. For which course
18 marks have higher standing?
18 92.25
4
Pz
18 120.75
8Sz
Soln:TheZ-score is
The P&S mark would have the highest standing since it is 2
standard deviation above the mean while the SSN mark is
only 0.75 standard deviation above the mean.
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P( ) = ( ) ( ) ( )b
a
a Z b f z dz F b F a
a b
f(z)
P(a
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Cumulative Distribution Function:
Analytical solution is very hard to obtain. Therefore, the
above equation needs to solve numerically. The numericalvalues lie from 0 to 3 are reported in a Table usually called
Z-tables.
Typically negative values are not reportedo Symmetrical, therefore area below negative value = Area above
its positive value
Always helps to draw a sketch!
Probabilit ies and z scores: z tables
21
21( ) ( ) 2
zu
F z P Z z e du
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Draw a picture of standard normal distribution
depicting the area of interest.
Re-express the area in terms of shapes like the one
on top of the Standard Normal Table
Look up the areas using the table.
Do the necessary addition and subtraction.
Strategies for finding probabilities for N(0,1)
P(Z>b)=P(Z
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Strategies for finding probabilities for N(0,1)
P(0
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P(Z>0.78)?
Strategies for finding probabilities for N(0,1)
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P(-1.2
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Meaning of Z-score Revisited
Suppose mid-sem 1 marks of students in Probability and
Statistics has a mean of 9 and a standard deviation of 4
whereas their Signal, System and Networks marks has amean of 12 and a standard deviation of 8. For which course
18 marks have higher standing?
Soln:TheZ-score is
The P&S mark would have the highest standing.
18 92.25
4
( ) 0.9878
P
P
z
P Z z
18 120.75
8
( ) 0.7734
S
S
z
P Z z
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Example 1
IfX~N(109, 13), Then find out (i) P (X141)
(i)
(ii)
120( 120) ( 120 )
109 120 1090.85 0.8023
13 13
XP X P X P
XP P Z
141( 141) ( 141 )
109 141 1092.46 1 ( 2.46)
13 13
1 0.9931 0.0069
XP X P X P
XP P Z P Z
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Example 2
The P&S exam is given to 147 students. The scores
have a normal distribution with a mean of 78 and a
standard deviation of 5.
(i) How many students have scores between 82 and 90?
(ii) What percent of the students have scores above 60?(iii) How many students have scores above 70?
Ans.:(i) 0.2037
(ii) 0.6406
(iii) 0.9452
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Normal and Binomial Distribution
Theorem: IfXis a random variable with distribution B(n,
p), then for sufficiently large n, it can be approximated by
Normal Distribution, i.e.,
2~ (0,1), with and (1 )X
Z N np np p
~ ( , ) ~ , (1 )X B n p X N np np p
Observations:
(i)
(ii) The normal distribution is a good approximation for the
binomial distribution whennp 5andn(1 p) 5.
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Normal and Binomial Distribution
Example 1: Is it possible to obtain normal distribution
approximation from the binomial distribution where n= 20
andp= 0.25?
220 0.25 5
(1 ) 20 0.25 0.75 3.75
3.75 1.94
npnp p
Solution: According to theorem, define the following
parameters:
Sincenp= 5 5andn(1 p)= 15 5, based on the previous
theorem, we can conclude thatB(20,0.25) ~ N(5,1.94).
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Normal and Binomial Distribution
Example 2: A sample of 100 items is taken at random
from a batch known to contain 40% defectives. What is the
probability that the sample contains: (i) at least 44
defectives (ii) exactly 44 defectives?
2
100 0.40 40(1 ) 100 0.4 0.6 24
24 4.89
np
np p
Solution: According to theorem, define the following
parameters:
Sincenp= 40 5andn(1 p)= 60 5, based on the previous
theorem, we can conclude thatB(100,0.4) ~ N(40,4.89).
Ans.:(i) 0.2376 (ii) 0.584