Examples of Distribution Functions in Statistics
Explore examples related to distribution functions such as binomial, Poisson, exponential, and normal distributions. Understand concepts like probability, mean distribution, variance, and more through practical scenarios and calculations.
Download Presentation
Please find below an Image/Link to download the presentation.
The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author. If you encounter any issues during the download, it is possible that the publisher has removed the file from their server.
You are allowed to download the files provided on this website for personal or commercial use, subject to the condition that they are used lawfully. All files are the property of their respective owners.
The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author.
E N D
Presentation Transcript
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)= 2 , 1 , 0 ; ) 2 . 0 ( ) 8 . 0 ( = ) 8 . 0 ( 5 = = 5 x = 5 x x ,......, 5 x = = 4 np = 2 8 . 0 ( 5 ) 2 . 0 )( 8 . 0 npq = 8 . 0
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 12 5 x 1 ( \ 2 ) 1 ( 1 \ 2 ) 5 = = 12 792 ( 1 \ 2 ) 0 . 19 12 7 12 7 ( 1 \ 2 ) ( 1 \ 2 ) b(7,12,1\2)= Complete 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 (. 1 ) (. 9 ) . 19 3
*Poisson = = 2 np 2 3 2 e = = = ( 3 ) 18 . p x 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 00034 . = e
Answer = = 8= 400 00034 )(. 768 , 32 ( (. 02 ) ) , 8 00034 . e n=400 x=5 P(5;8)= 5 8 8 e = = 093 . ! 5 120 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 26280 ] e 8760 (i)p(x>26280)=1-p(x<26280)=1-[1- (ii) p(x<720)=1- 8760 e 720 08219 . = = 1 . 0789 e 1 (iii)p(x<1)=1- 000114155 . 0 = = 1 000114148 . 0 e e 8760
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 35 = . 1 = . 4 44 35 = X=5.20 Z= P(x>5.20)=1-p(z<1.44) =1-0.9251 =0.0749 Example 6.24 at the book . 0 59 . 5 20 . 4 . 0 59 table
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) 1) p(x>184)=1-p(x<184)=1- 184 ( 168 ) 6 =1-(2.67) table =1-.9962 =.0038
156 ( 168 2)p(x<156)=z =.0228 ) 6 =z(-2) 156 168 3)p(165<x<174)=p( <z< ) =p(z<1)-P(z<-2) 6
