1. Lecture 5
Faten alamri

2. .examples about distribution function
(binomial- Poisson- exponential- normal )

3. 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
The distribution for all airplane will hit the target?
The mean distribution and its variance?

4. 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 f(x)=

5. 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)= b(7,12,1\2)= Complete ……

6. 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
P(x=3)=

7. *Poisson

8. 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

9. Answer n=400 x=5 P(5;8)= Example 5.20 in the book

10. 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.

11. Answer
(i)p(x>26280)=1-p(x<26280)=1-[1- (ii) p(x<720)=1- (iii)p(x<1)=1-

12. 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?

13. Answer
X=5.20 Z= P(x>5.20)=1-p(z<1.44) = = Example 6.24 at the book
table

14. 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

15. Answer X~N(168,36) p(x>184)=1-p(x<184)=1- =1-(2.67) table=
=.0038

16. 2)p(x<156)=z
=z(-2)
=.0228
3)p(165<x<174)=p( <z< )
=p(z<1)-P(z<-2)