Lecture 5 Faten alamri examples about distribution function

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 f(x)=

Example 2 Find the probability of having five heads and seven tails in 12 flips of balanced coin? Answer X=5, n=12, p=12 b(5, 12, 12)= b(7, 12, 12)= Complete ……

Example 3 • 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)=

*Poisson

poisson • Example 1 If 2 percent 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

Answer • n=400 x=5 P(5; 8)= Example 5. 20 in the book Example 5. 21 in the book

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

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 Z= P(x>5. 20)=1 -p(z<1. 44) =1 -0. 9251 =0. 0749 Example 6. 24 at the book

Example 2 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

Answer X~N(168, 36) 1) p(x>184)=1 -p(x<184)=1=1 -(2. 67) table 2)p(x<156)=z 3)p(165<x<174)=p( <z< )