Free Essay

# Binomial Distribution

In: Other Topics

Submitted By sharis
Words 1210
Pages 5
DISTRIBUSI BINOMIAL

Distribusi binomial berasal dari percobaan binomial yaitu suatu proses Bernoulli yang diulang sebanyak n kali dan saling bebas. Distribusi Binomial merupakan distribusi peubah acak diskrit. Secara langsung, percobaan binomial memiliki ciri-ciri sebagai berikut: – – – – percobaan tersebut dilakukan berulang-ulang sebanyak n kali setiap percobaan menghasilkan keluaran yang dapat dikategorikan sebagai gagal dan sukses probabilitas sukses p tetap konstan dari satu percobaan ke percobaan lain percobaan yang berulang adalah saling bebas

En : SSS ...S GGG...G (sukses sebanyak x kali, gagal sebanyak n – x kali) x n x

Jika peluang sukses dinotasikan dengan p maka peluang gagal adalah q = 1 – p. Peubah acak X menyatakan banyaknya sukses dari n percobaan yang saling bebas. Maka peluang X pada masing – masing percobaan E adalah:

P(X)  p P(X)  qp  pq P(X)  pq
2 2

untuk E1 untuk E2 untuk E3

P(X)  q p  pq untuk E4 P(X)  qpq  pq 2 untuk E5 P(X)  pqq  pq 2 untuk E6 P(X)  p x q n  x untuk En

 2 Dapat dilihat bahwa E2 dan E3 memberikan hasil yang sama. Jumlahnya   , yaitu 1 

jumlah semua titik sampel yang mungkin menghasilkan x = 1 yang sukses dan n – x = 2 – 1 = 1 yang gagal dari 2 percobaan. Begitupun untuk E4 , E5 , dan E6 juga
 3 memberikan hasil yang sama. Jumlahnya   , yaitu jumlah semua titik sampel yang 1 

mungkin yang menghasilkan x = 1 yang sukses dan n – x = 3 – 1 = 2 yang gagal dari 3 percobaan. Secara umum, jumlah titik sampel yang mungkin untuk menghasilkan x sukses dan n – x gagal dalam n percobaan adalah banyaknya cara yang berbeda dalam
n mendistribusikan x sukses dalam barisan n percobaan, sehingga terdapat   cara.  x

Dan distribusi peluang atau Probability Mass Function (PMF) X dinyatakan pada definisi berikut:
n  n P(X  x)  f ( x)    p x q n x    p x (1  p)n  x  x  x

untuk x  1, 2,..., n dan 0  p  1 . Pembuktian distribusi Binomial merupakan suatu PMF. Bukti: Untuk membuktikan suatu peubah acak adalah PMF, maka harus ditunjukan: 1. 2. f ( x)  0

 f ( x)  1 x Akan ditunjukkan distribusi binomial memenuhi kedua syarat di atas: 1. f ( x)  0

Karena 0  p  1 dan nilai kombinasi pasti positif maka f(x) pasti positif. 2.

 f ( x)  1 x Menggunakan persamaan binomial Newton pada

 f ( x) , akan diperoleh: x  f ( x)    x  p (1  p) x x x 0

n

n  

n x

 ( p  (1  p)) n  1n  1

(1)

Dari 1 dan 2 dapat dikatakan bahwa distribusi binomial merupakan PMF.

Mean Jika X
B(n, p) ( X variabel random berdistribusi Binomial), maka nilai ekspektasi

E ( X )   x.P( x) x   x.P( X  x) x 0 n

n

n   x.   p x (1  p) n  x x 0  x n n   x.   p x (1  p) n  x x 1  x n n!   x. p x (1  p) n  x x !(n  x)! x 1   x. x 1 n

n.(n  1)! . p. p x 1 (1  p) n  x x.( x  1)!(n  x)!

 np 

(n  1)! p x 1 (1  p) n  x x 1 ( x  1)!( n  x )! n Misalkan m = n – 1 dan s = x – 1, maka persamaan di atas menjadi
E ( X )  np m! p s (1  p)m s s  0 s !(m  s )! m Berdasarkan (1),

 s !(m  s)! p (1  p) s s 0

m

m!

m s

 1 , maka

E ( X )  np  s 0

m

m! p s (1  p) m  s s !(m  s )!

 np.1  np

Sehingga didapat mean dari X, E ( X )  np

Variansi
Var ( X )  E ( X 2 )  ( E( X ))2

Dalam meencari Var(X), kita harus tahu nilai ekspektasi dari X2 :

E ( X 2 )   x 2 .P( X  x)   x 2 . x 0 x 0

n

n

n! p x (1  p) n  x x !(n  x)!

E ( X 2 )   x 2 .P( x) x   x 2 .P ( X  x ) x 0 n n   x 2 .   p x (1  p) n  x x 0  x n n   x 2 .   p x (1  p) n  x x 1  x n n!   x2. p x (1  p) n  x x !(n  x)! x 1

n

  x2. x 1 n

n

n.(n  1)! . p. p x 1 (1  p) n  x x.( x  1)!(n  x)! (n  1)! p x 1 (1  p) n  x ( x  1)!(n  x)!

 np  x. x 1

Misalkan m = n – 1 dan s = x – 1, maka persamaan di atas menjadi
E ( X 2 )  np ( s  1) s 0 m m  m m! p s (1  p) m s  np  ( s  1)   p s (1  p) m s s !(m  s)! s 0 s 

E ( X 2 )  np  ( s  1) s 0

m

m! p s (1  p) m  s s !(m  s )!

m m  np  ( s  1)   p s (1  p) m s s 0 s  m  m m s  m m s  np.   s.   p (1  p)  1.   p s (1  p) m s  s 0 s   s 0  s    np[ mp  1]

 np[( n  1) p  1]  np[ np  p  1]

Var ( X )  E ( X 2 )  ( E ( X )) 2  np[np  p  1]  (np )2  (np )2  np 2  np  (np )2  np  np 2  np (1  p )
Sehingga didapat variansi dari X, Var ( X )  np(1  p)

Contoh: 1. Probabilitas bahwa sejenis komponen tertentu yang lolos uji kelayakan adalah ¾. Tentukan probabilitas dimana 2 dari 4 komponen yang selanjutnya diuji akan dinyatakan layak!
n  n P(X  x)  f ( x)    p x q n x    p x (1  p)n  x  x  x

p = ¾, q = 1 – ¾ = ¼ untuk x = 2
42

 4 3   1  P( X  2)        2 4   4 
2

 6.

9 1 27 .  16 16 128

2. Berdasarkan data biro perjalanan PT Sentosa, yang khusus menangani perjalanan wisata turis mancanegara, 20% dari turis menyatakan sangat puas berkunjung ke Indonesia, 40% menyatakan puas, 25% menyatakan biasa saja, dan sisanya menyatakan kurang puas. Apabila kita bertemu dengan 5 orang dari peserta wisata turis mancanegara yang pernah menggunakan jasa biro perjalanan tersebut. Tentukan probabilitas: a. Tepat 2 diantaranya menyatakan biasa saja

b. Paling banyak 2 diantaranya menyatakan sangat puas

Jawab: n=5 a. p = 0.25, q = 1 – 0.25 = 0.75
5 2

5 1   3  P( X  2)        2 4   4 
2

 0.2637

b. p = 0.2, q = 1 – 0.2 = 0.8
P( X  2)  P( X  0)  P( X  1)  P( X  2) 5  5 5    (0.2) 0 (0.8)5    (0.2)1 (0.8)51    (0.2) 2 (0.8)52 0 1   2  0.32768  0.40960  0.20480  0.94208

### Similar Documents

Free Essay

#### Binomial Distribution

...Binomial Distribution This is a discrete random variable, where the process of obtaining the Binomial distribution is called “Bernoulli “ process. An experiment that often consists of repeated trials, each with two possible outcomes, which could be labeled as “success” or “failure”. This experiment is known as binomial experiment. A binomial experiment is one that possesses the following properties: 1. The experiment consists of n repeated trials. 2. Each trial has only 2 possible outcomes that can be classified as “Success” or “Failure”. 3. The probability of a success and failure , denoted by pand q, remains constant from trial to trial. 4. The repeated, trials are independent. Formula for success in the Bernoulli process. The probability of r success from n trial is : [pic] where; r - the number of success n – the number of trial p – the probability of success from one trial and q – the probability of failure from one trial. q = 1 - p [pic] with mean, [pic] and variance, [pic] It is written as X ~ B(n,p) and read as X is Binomial distribution with parameters n success and the probability of success, p. Example 7.4 One coin is tossed 5 times. If X is random variable for number of heads . Find the probability of getting (a) no heads (b) one head (c) 3 heads (d) at least 3 heads Solution X......

Words: 986 - Pages: 4

Free Essay

#### Lecture 07 - Binomial Distributions (5.2-5.3)

...Binomial Distributions How to recognize binomial random variable: 1. The sample size n is fixed. 2. The n observations (or trials) are independent. 3. There are only 2 possible outcomes for each observation. They are labeled “Success” and “Failure” 4. The probability of success is the same for each trial. Let p = success probability and 1 – p = failure probability 5. The binomial random variable is . . . X = the number of successes out of n observations. Binomial distributions are identified specifically by two parameters: n and p. X takes on values 0, 1, 2, . . . , n Notation: [pic] Mean: [pic] Variance: [pic] Standard deviation: [pic] The Binomial Probability Formula: [pic] Suppose [pic], then the probability of observing 16 successes out of 20 observations is [pic] Find the mean, variance, and standard deviation of X. Mean: [pic] Variance: [pic] Standard deviation: [pic] Table 3 in Appendix II contains probability distributions for a variety of binomial distributions. Look up the following probabilities: For X ~ B(20, 0.7), find P(X=16). Answer: P(X=16) = 0.130 For X ~ B(8, 0.45), find P(X > 5). Answer: P(X > 5) = P(6) + P(7) + P(8) = 0.070 + 0.016 + 0.002 = 0.088 More examples: 1. A research team found that 10% of...

Words: 371 - Pages: 2

#### Binomial Distributions

...Different people will have different views of the impact of a particular risk – what may be a small risk for one person may destroy the livelihood of someone else. One way of putting figures to risk is to calculate a value for it as: risk = probability of event x cost of event Doing this allows you to compare risks objectively. We use this approach formally in decision making with Decision Trees. To carry out a risk analysis, follow these steps: 1. Identify Threats: The first stage of a risk analysis is to identify threats facing you. Threats may be: * Human - from individuals or organizations, illness, death, etc. * Operational - from disruption to supplies and operations, loss of access to essential assets, failures in distribution, etc. * Reputational - from loss of business partner or employee...

Words: 909 - Pages: 4

Free Essay

#### Solutions Manual for Statistical Inference, Second Edition

...Solutions Manual for Statistical Inference, Second Edition George Casella University of Florida Roger L. Berger North Carolina State University Damaris Santana University of Florida 0-2 Solutions Manual for Statistical Inference “When I hear you give your reasons,” I remarked, “the thing always appears to me to be so ridiculously simple that I could easily do it myself, though at each successive instance of your reasoning I am baﬄed until you explain your process.” Dr. Watson to Sherlock Holmes A Scandal in Bohemia 0.1 Description This solutions manual contains solutions for all odd numbered problems plus a large number of solutions for even numbered problems. Of the 624 exercises in Statistical Inference, Second Edition, this manual gives solutions for 484 (78%) of them. There is an obtuse pattern as to which solutions were included in this manual. We assembled all of the solutions that we had from the ﬁrst edition, and ﬁlled in so that all odd-numbered problems were done. In the passage from the ﬁrst to the second edition, problems were shuﬄed with no attention paid to numbering (hence no attention paid to minimize the new eﬀort), but rather we tried to put the problems in logical order. A major change from the ﬁrst edition is the use of the computer, both symbolically through Mathematicatm and numerically using R. Some solutions are given as code in either of these languages. Mathematicatm can be purchased from Wolfram Research, and R is......

Words: 59353 - Pages: 238

Free Essay

#### Sir Nicolas

...Lin20522_appb_726-743.indd Page 726 28/10/13 11:55 AM user-f-w-198 /201/MH02018/Lin20522_disk1of1/0078020522/Lin20522_pagefiles A P P E N D I X B : TA B L E S B.1 Binomial Probability Distribution n51 Probability x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 0 1 0.950 0.050 0.900 0.100 0.800 0.200 0.700 0.300 0.600 0.400 0.500 0.500 0.400 0.600 0.300 0.700 0.200 0.800 0.100 0.900 0.050 0.950 n52 Probability x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 0 1 2 0.903 0.095 0.003 0.810 0.180 0.010 0.640 0.320 0.040 0.490 0.420 0.090 0.360 0.480 0.160 0.250 0.500 0.250 0.160 0.480 0.360 0.090 0.420 0.490 0.040 0.320 0.640 0.010 0.180 0.810 0.003 0.095 0.903 n53 Probability x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 0 1 2 3 0.857 0.135 0.007 0.000 0.729 0.243 0.027 0.001 0.512 0.384 0.096 0.008 0.343 0.441 0.189 0.027 0.216 0.432 0.288 0.064 0.125 0.375 0.375 0.125 0.064 0.288 0.432 0.216 0.027 0.189 0.441 0.343 0.008 0.096 0.384 0.512 0.001 0.027 0.243 0.729 0.000 0.007 0.135 0.857 n54 Probability x 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 0 1 2 3 4 0.815 0.171 0.014 0.000 0.000 0...

Words: 12879 - Pages: 52

#### Statistics First Five

.... . . . . . . . . . . . . . . . . . . . 11 2.3 Five Number Summaries and Box and Whisker Displays . . . . . 12 3 Probability 13 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 3.2 Random Variables . . . . . . . . . . . . . . . . . . . . . . . . . . 16 3.2.1 Deﬁnition . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 3.2.2 Expected Value . . . . . . . . . . . . . . . . . . . . . . . . 17 3.2.3 Variance and Standard Deviation . . . . . . . . . . . . . . 17 3.2.4 “Shortcuts” for Binomial Random Variables . . . . . . . . 18 1 4 Probability Distributions 19 4.1 Binomial Distributions . . . . . . . . . . . . . . . . . . . . . . . . 19 4.2 Poisson Distributions . . . . . . . . . . . . . . . . . . . . . . . . . 21 4.2.1 Deﬁnition . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 4.2.2 As an Approximation to the Binomial . . . . . . . . . . . 22 Normal Distributions . ....

Words: 11010 - Pages: 45

#### Acme Electronics Case

...Acme Electronics Case Date: March 6, 2012 To: Jetson, on behalf of Acme Electronics From: Team 4 Consulting Firm Re: Legal and statistical evaluation of problems facing Acme Per your request, we have assembled a report with a legal and statistical evaluation of the problems facing Acme. If you have any questions, feel free to contact us at any time. Group 4: Acme Electronics Case Executive Summary This report is dealing with the case of ACME Electronics vs. Otto Gunter. Gunter purchased a computer from ACME Electronics in 2002. In 2004, the hard drive crashed and he brought it in to ACME Electronics to have it replaced, as well as request for his old hard drive back to retrieve his information. ACME replaced the hard drive and gave Gunter his supposed old hard drive. Gunter spent \$800 retrieving information from the hard drive only to find out it was mixed with another customer’s. The other customer disposed of Gunter’s old hard drive, so he cannot retrieve it. Gunter is suing ACME for \$800 for retrieving information from the wrong hard drive supplied by ACME, \$5,000 for time spent reconstructing the lost information, and \$10,000 for punitive damages. The point of this report is to analyze the legal ramifications and statistics based on the situation at hand and to deduce the appropriate course of action to take based off the analysis. We analyzed a survey using......

Words: 3603 - Pages: 15

#### Truth

...The Al Gore Eﬀect: An Inconvenient Truth and Voluntary Carbon Oﬀsets∗ [Job Market Paper] Grant Jacobsen University of California-Santa Barbara 2120 North Hall University of California Santa Barbara, 93106-9210 jacobsen@econ.ucsb.edu Phone: (717) 315-5503 Fax: (805) 893-8830 I thank Matthew Kotchen, Robert Deacon, Olivier Deschenes, and Charles Kolstad for helpful comments. I also thank participants at a UCSB seminar, the Western Economics International Association’s Conference, and the University of Colorado Environmental and Resource Economics Workshop. ∗ 1 The Al Gore Eﬀect: An Inconvenient Truth and Voluntary Carbon Oﬀsets Abstract This paper examines the relationship between climate change awareness and household behavior by testing whether Al Gore’s documentary An Inconvenient Truth caused an increase in the purchase of voluntary carbon oﬀsets. I ﬁnd that in the two months following the ﬁlm’s release, zip codes within a 10-mile radius of a zip code where the ﬁlm was shown experienced a 50 percent relative increase in the purchase of voluntary carbon oﬀsets. During other times, oﬀset purchasing patterns for zip codes inside the 10-mile radius were similar to the patterns of zip codes outside the 10-mile radius. There is, however, little evidence that individuals who purchased an oﬀset due to the ﬁlm renewed them again a year later. This research has implications for how information campaigns, which are commonly used by policy-makers......

Words: 10098 - Pages: 41

#### Magic of Statistics

...your students while teaching or reviewing concepts including: – Probability and Independence – The Binomial distribution – The Normal approximation to the binomial – Hypothesis testing Semantics • Remove all ACEs from the deck • Place one red ACE and one black ACE face up • Ask each student, in order, to guess the color of a card which they cannot see • Place card onto appropriate colored ACE stack • After 10 (or so) guesses, place opposite color ACE on each stack • Repeat second step above • Reveal to students their hidden psychic ability! Probability • Define “success” as guessing right • What is the probability of “success”? – Bayesian? – Frequentist? • What is the probability of getting all n right? Independence • What is the probability of getting all n right? • Does the guess of one student affect the next student? – Series of reds often forces the next guess to be black – Does the probability of “success” change? Binomial Distribution • Recall requirements of – Dichotomous event – Independence – N trials – Constant probability of success (50%) • What is the probability of getting all n right? • What is the probability of getting n-1 out of n right (modification)? Hypothesis Testing • p-value: If the true probability of “success” is 50%, what is the chance you get all n right? – Binomial calculation – Normal approximation to the binomial • Only need n=10 until np and n(1-p) = 5 • In the end, students don’t need to understand how......

Words: 483 - Pages: 2

#### Tree Value

...Overview and examples from Finite Mathematics Using Microsoft Excel® Revathi Narasimhan Saint Peter's College An electronic supplement to Finite Mathematics and Its Applications, 6th Ed. , by Goldstein, Schneider, and Siegel, Prentice Hall, 1997 Introduction In any introductory mathematics course designed for non-mathematics majors, it is important for the student to understand and apply mathematical ideas in a variety of contexts. With the increased use of advanced software in all fields, it is also important for the student to effectively interact with the new technology. Our goal is to integrate these two objectives in a supplement for the text Finite Mathematics and Its Applications, by Goldstein, Schneider, and Siegel. The package consists of interactive tutorials and projects in an Excel workbook format. The software platform used is the Microsoft Excel 5.0 spreadsheet. It was chosen for the following reasons: • • • suited to applications encountered in a finite math course widespread use outside of academia ease of creating reports with a professional look Use of Excel 5.0 was put into effect in the author's sections of the Finite Mathematics II course in the Spring 1996 semester. It was expanded to cover the Finite Mathematics I course for the Fall semester of 1996. Using a combination of specially designed projects and tutorials, students are able to analyze data, draw conclusions, and present their analysis in a professional format. The mathematical...

Words: 3176 - Pages: 13

Free Essay

#### Us Public Health Services Case

...event of one person testing positive is independent of any other person testing positive. 2. Every person testing positive for the disease is equally likely. Solution • Let N be the total population, r be the number of infected people in the population. Hence, N-r represents the number of people in the population without the disease. • Let n denote the number of samples in a group. Therefore, [pic] be the number of groups. Let ‘p’ represent the probability that a group tests negative for the disease, that is, none of the group members carry the antigen. Probability Distribution In a population of N, there are r people who carry the disease and N-r people who are free from the disease. Samples of size n are chosen from this population. Let X denote the number of people who test positive for the disease in the sample n. Then, p follows hypergeometric distribution. [pic] For the problem at hand, k = 0. [pic] Therefore, 1-p will represent the probability that the group tests positive. Number of tests to be performed through sampling = Number of groups + Testing each person in the groups testing positive If the proposed technique is to require lesser number of chemical analyses than those required by the existing procedure [pic] [pic] [pic] f(x) was plotted on MS Excel for different values of N, n and r, as shown here: |N...

Words: 1040 - Pages: 5

Free Essay

#### On an Improved Fraction Nonconforming Control Chart

...student, University of Alabama in Huntsville Abstract. In their article published in the April 2013 edition of Control Engineering Practice, Silvia Joekes and Emanuel Pimentel Barbosa present the use of a correction factor, the Cornish-Fisher quantile correction, to improve the fraction nonconforming control chart (p-chart) when dealing with high quality processes. Introduction. The use of the standard Shewhart control chart for the fraction of nonconforming parts will produce a type I error–signaling that the process is out of control when, in fact, it is still in control–with probability 0 = 0.0027. When dealing with a process with both high quality and a large yield, the control limits for the Shewhart p-chart based on the normal distribution will become tighter, leaving a greater chance of a false alarm signal occurring. This journal entry proposes a modified version of the p-chart based on the CornishFisher expansion, which allows monitoring processes with very low values of p. This new corrected p-chart is shown using one and two terms of correction. Shewhart p-chart basics. Shewhart control charts use 3 limits for both upper and lower control limits, i.e.   3 . The use of Shewhart control charts is commonplace due to their ease of use: the process will only ever signal in control or out of control. This type of pass/fail or conforming/nonconforming relationship is used for attributes data: data that is not easily assigned some numerical value for the......

Words: 1085 - Pages: 5

#### Stats Notes

...Descriptive Statistics Descriptive statistics involves organizing, summarizing and illustrating statistical data. The objective is to show important characteristics of the data without drawing any conclusions. Inferential statistics involves using a representative subset of data (a sample) in order to draw conclusions about unknown characteristics of an entire set of data (a population). Population: The entire set of elements of interest (i.e. all humans, all working-age people in Canada, all IT companies). A population parameter is a characteristic used to describe a population. For example, Population mean ( Population standard deviation ( Population median ( The values of the population parameters would be preferable for use in decision-making but seldom will these values be known since collecting all the population elements (a census) is usually too expensive and/or time consuming. Sample: A representative subset of the entire set of elements of interest that is used to gain insight about the population. A sample statistic is a characteristic used to describe a sample. For example, Sample mean [pic] Sample standard deviation s Sample median Md It is cheaper, less time-consuming and more practical to use sample statistics as estimates for population parameters in making business decisions. How well the sample represents the population depends......

Words: 14529 - Pages: 59

Free Essay

#### Sta Course

...Solution to selected exercises from chapter 3 3-14. f X (0)  P( X  0)  1/ 6  1/ 6  1/ 3 f X (1.5)  P( X  1.5)  1/ 3 f X (2)  1/ 6 f X (3)  1/ 6 a) P(X = 2) = 1/6 b) P(0.6 < X < 2.7) = P(X = 1.5) +P(X = 2) = 1/3 + 1/6 = 1/2 c) P(X > 3) = 0 d) P(0  X  2)  P( X  0)  P( X  1.5)  1/ 3  1/ 3  2 / 3 e) P(X = 0 or X = 2) = 1/3 + 1/6 = ½ --------------------------------------------------------------------------------------------------------------------3-15. All probabilities are greater than or equal to zero and sum to one. a) P(X  1) = 1/8 + 2/8 + 2/8 + 2/8 = 7/8 b) P(X > 2) = 2/8 + 2/8 + 2/8 + 1/8 = 7/8 c) P(1  X  1) = 2/8 + 2/8 + 2/8 =6/8 = 3/4 d) P(X  1 or X = 2) = 1/8 + 2/8 +1/8 = 4/8 = ½ --------------------------------------------------------------------------------------------------------------------3-20. P(X = 0) = 0.053 = 1.25  104 P(X = 1) = 3[0.95(0.05)(0.05)] = 7.125  103 P(X = 2) = 3[0.95(0.95)(0.05)] = 0.1354 P(X = 3) = 0.953 = 0.8574 --------------------------------------------------------------------------------------------------------------------- 3-25. P(X = 15 million) = 0.6, P(X = 5 million) = 0.3, P(X = 0.5 million) = 0.1 --------------------------------------------------------------------------------------------------------------------- 3-32. x0   0, 1/ 3 0  x  1.5      F ( x)  2 / 3 1.5  x  2  where 5 / 6 2  x  3    3 x   1   f X (0)  P( X  0)  1/ 6  1/ 6  1/ 3 f X (1.5)  P( X ...

Words: 1968 - Pages: 8