Successfully reported this slideshow.
Upcoming SlideShare
×

Normal Distribution, Binomial Distribution, Poisson Distribution

19,271 views

Published on

• Full Name
Comment goes here.

Are you sure you want to Yes No
• Hello! Get Your Professional Job-Winning Resume Here - Check our website! https://vk.cc/818RFv

Are you sure you want to  Yes  No
• In slide 14. Why is n 6? I thought it was 12.

Are you sure you want to  Yes  No

Normal Distribution, Binomial Distribution, Poisson Distribution

1. 1. Binomial Distribution and Applications
2. 2. Binomial Probability Distribution Is the binomial distribution is a continuous distribution?Why? Notation: X ~ B(n,p) There are 4 conditions need to be satisfied for a binomial experiment: 1. There is a fixed number of n trials carried out. 2. The outcome of a given trial is either a “success” or “failure”. 3. The probability of success (p) remains constant from trial to trial. 4. The trials are independent, the outcome of a trial is not affected by the outcome of any other trial.
3. 3. Comparison between binomial and normal distributions
4. 4. Binomial Distribution If X ~ B(n, p), then where successof trials.insuccessesofnumberr 11!and10!also,1...)2()1(! yprobabilitP n nnnn .,...,1,0r)1( )!(! ! )1()( npp rnr n ppcrXP rnrrnr n r
5. 5. Exam Question  Ten percent of computer parts produced by a certain supplier are defective. What is the probability that a sample of 10 parts contains more than 3 defective ones?
6. 6. Solution :  Method 1(Using Binomial Formula):
7. 7. Method 2(Using Binomial Table):
8. 8.  From table of binomial distribution :
9. 9. Example 2 If X is binomially distributed with 6 trials and a probability of success equal to ¼ at each attempt. What is the probability of a)exactly 4 succes. b)at least one success.
10. 10. Example 3 Jeremy sells a magazine which is produced in order to raise money for homeless people. The probability of making a sale is, independently, 0.50 for each person he approaches. Given that he approaches 12 people, find the probability that he will make: (a)2 or fewer sales; (b)exactly 4 sales; (c)more than 5 sales.
11. 11. Normal Distribution
12. 12. Normal Distribution  In general, when we gather data, we expect to see a particular pattern to the data, called a normal distribution. A normal distribution is one where the data is evenly distributed around the mean, which when plotted as a histogram will result in a bell curve also known as a Gaussian distribution.
13. 13.  thus, things tend towards the mean – the closer a value is to the mean, the more you’ll see it; and the number of values on either side of the mean at any particular distance are equal or in symmetry.
14. 14.
15. 15. Z-score  with mean and standard deviation of a set of scores which are normally distributed, we can standardize each "raw" score, x, by converting it into a z score by using the following formula on each individual score:
16. 16. Example 1 a) Find the z-score corresponding to a raw score of 132 from a normal distribution with mean 100 and standard deviation 15. b) A z-score of 1.7 was found from an observation coming from a normal distribution with mean 14 and standard deviation 3. Find the raw score. Solution a)We compute 132 - z = __________ = 2.133 15 b) We have x - 1.7 = ________ 3 To solve this we just multiply both sides by the denominator 3, (1.7)(3) = x - 14 5.1 = x - 14 x = 19.1
17. 17. Example 2 Find a) P(z < 2.37) b) P(z > 1.82) Solution a)We use the table. Notice the picture on the table has shaded region corresponding to the area to the left (below) a z-score. This is exactly what we want. Hence P(z < 2.37) = .9911 b) In this case, we want the area to the right of 1.82. This is not what is given in the table. We can use the identity P(z > 1.82) = 1 - P(z < 1.82) reading the table gives P(z < 1.82) = .9656 Our answer is P(z > 1.82) = 1 - .9656 = .0344
18. 18. Example 3 Find P(-1.18 < z < 2.1) Solution Once again, the table does not exactly handle this type of area. However, the area between -1.18 and 2.1 is equal to the area to the left of 2.1 minus the area to the left of -1.18. That is P(-1.18 < z < 2.1) = P(z < 2.1) - P(z < -1.18) To find P(z < 2.1) we rewrite it as P(z < 2.10) and use the table to get P(z < 2.10) = .9821. The table also tells us that P(z < -1.18) = .1190 Now subtract to get P(-1.18 < z < 2.1) = .9821 - .1190 = .8631
19. 19. Poisson distribution
20. 20. Definitions  a discrete probability distribution for the count of events that occur randomly in a given time.  a discrete frequency distribution which gives the probability of a number of independent events occurring in a fixed time.
21. 21. Poisson distribution only apply one formula: Where:  X = the number of events  λ = mean of the event per interval Where e is the constant, Euler's number (e = 2.71828...)
22. 22. Example: Births rate in a hospital occur randomly at an average rate of 1.8 births per hour. What is the probability of observing 4 births in a given hour at the hospital? Assuming X = No. of births in a given hour i) Events occur randomly ii) Mean rate λ = 1.8 Using the poisson formula, we cam simply calculate the distribution. P(X = 4) =( e^-1.8)(1.8^4)/(4!) Ans: 0.0723
23. 23.  If the probability of an item failing is 0.001, what is the probability of 3 failing out of a population of 2000? Λ = n * p = 2000 * 0.001 = 2 Hence, use the Poisson formula X = 3, P(X = 3) = Ans: 0.1804
24. 24. Example: A small life insurance company has determined that on the average it receives 6 death claims per day. Find the probability that the company receives at least seven death claims on a randomly selected day.
25. 25. Analysis method  1st: analyse the given data.  2nd: label the value of x, λ  At least 7 days, means the probability must be ≥ 7. but the value will be to the infinity. Hence, must apply the probability rule which is  P(X ≥ 7) = 1 – P(X ≤ 6)  P(X ≤ 6) means that the value of x must be from 0, 1, 2, 3, 4, 5, 6.  Total them up using Poisson, then 1 subtract the answer.  Ans = 0.3938
26. 26. Example: The number of traffic accidents that occurs on a particular stretch of road during a month follows a Poisson distribution with a mean of 9.4. Find the probability that less than two accidents will occur on this stretch of road during a randomly selected month. P(x < 2) = P(x = 0) + P(x = 1) Ans: 0.000860