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Lecture 5
Faten alamri
examples about
distribution function
• .
• (binomial- Poisson- exponential- normal )
Binomial
Example 1
If the probability that airplane will hit a target is
0.8, than we know five more airplane will hit the
target to find the following
1) The distribution for all airplane will hit the
target?
2) The mean distribution and its variance?
Answer
n=5, p=0.8, q=1-0.8=0.2
Suppose x is the number of airplanes will hit
the target, then its x is the binomial density
function, is
5 
x
5 x


f(x)=  (0.8) (0.2) ; x  0,1,2,......,5
 x
  np  5(0.8)  4
  npq  5(0.8)(0.2)  0.8
2
  0.8
Example2
Find the probability of having five heads and seven
tails in 12 flips of balanced coin?
Answer
X=5, n=12, p=1\2
b(5,12,1\2)=
12 
x
12  5


(
1
\
2
)
(
1

1
\
2
)
5 


 792(1 \ 2)12  0.19
b(7,12,1\2)=
Complete ……
12 
7
12  7


(
1
\
2
)
(
1
\
2
)
7 


Example3
• If x is a binomial distribution with parameter
p=.1 and n=20, find p(x=3). by using binomial
distribution with Poisson distribution?
• answer
• *Binomial
20 
• P(x=3)= 
3
17


 .19

(.1) (.9)
3

*Poisson
  np  2
e  2 23
p ( x  3) 
 .18
3!
poisson
• Example1
If 2percent of the book bound at certain bindery
have defective binding, use the Poisson
approximation to the binomial distribution to
determine the probability that five of 400 books
bound by this bindery, will have defective
binding, where e    .00034
Answer
  400(.02)  8, e
• n=400 x=5
5 8
8
P(5;8)= e  (32,768)(.00034)  .093
5!
120
Example 5.20 in the book
Example 5.21 in the book
8
 .00034
Exponential distribution
• Example
• A light life is written in a box as average of 8760
hours lighting if it known as an exponential
distribution then find the following:
• (i) if the lam work more then 3 years before it
stops working.
• (ii) what is the probability that the lamp will stop
working before month of the beginning.
• (iii) what is the probability the lamp will stop
working in an hour from the beginning.
Answer
26280
8760
(i)p(x>26280)=1-p(x<26280)=1-[1e
]
720
(ii) p(x<720)=1-e 8760  1  e .08219  .0789
(iii)p(x<1)=1- e

1
8760
 1 e
 0.000114155
 0.000114148
Normal distribution
• Example 1
Suppose that the amount of cosmic radiation to
which a person is exposed when flying by jet
across the united states is random variable
having a normal distribution with a mean of
4.35 mrem and stander deviation of .59 mrem.
What is the probability that a person will be
exposed to more then 5.20 mrem of cosmic
radiation?
Answer
X=5.20
  4.35  
5.20  4.35
 1.44
Z=
0.59
table
P(x>5.20)=1-p(z<1.44)
=1-0.9251
=0.0749
Example 6.24 at the book
0.59
Example2
If a university students recognized by there hight
which full on a normal distribution with mean of
168 cm and stander deviation of 6 cm. We
choose randomly a student. What is the
probability that his length will be
1)Greater then 184 cm
2)Less then 156 cm
3)Between 165 and 174
Answer
X~N(168,36)
184  168
)
1) p(x>184)=1-p(x<184)=1-(
6
=1-(2.67)  table
=1-.9962
=.0038
156  168
)
• 2)p(x<156)=z (
6
•
•
=z(-2) 
=.0228
• 3)p(165<x<174)=p(
<z<
•
=p(z<1)-P(z<-2)
156  168
6
)