السلام عليكم و رحمة الله و بركاته

السلام عليكم و رحمة الله و بركاته
Probability and Statistics By Dr. Khaled El Naggar

Probability of an event
If the number of outcomes belonging to an event E is NE, and the total number of outcomes is N, then the probability of event E is defined as P (E) = NE/N

P (E1∪E2)= P(E1) + P(E2) – P(E1 ∩ E2)
Axioms of Probability A probability function P is defined on subsets of the sample space Ω to satisfy the following axioms 1. Non-Negative Probability: P (E) ≥ 0 2. Union Property of any two events E1, E2: P (E1∪E2)= P(E1) + P(E2) – P(E1 ∩ E2)

3. Mutually-Exclusive Events (E1∩E2 is empty):
P (E1∪E2) = P (E1) + P (E2) 4. The Sample space : P(Ω ) = 1

Report A weighted die is tossed such that the Probability of even sides facing up twice Probability of odd sides facing up, let the event A is sides facing up less than 5 and the event B is sides facing up greater than 4, find the Probabilities of events A and B.

Solved examples 1- Two dice are thrown simultaneously, what is the probability that the sum of the two numbers that turn up is less than 11?

Solution: P(sum of the two numbers that turn up is less than 11) = 1-P(sum of the two numbers that turn up is greater than or equal 11) = 1 – [3/36] = 11/12

2- 4 dice are thrown, what is the probability that the same number appears on each of them?

Solution: When 4 dice are thrown simultaneously, so the total number of possible outcomes is 64 = The chances that all the dice show same number (1,1,1,1), (2,2,2,2), (3,3,3,3), (4,4,4,4), (5,5,5,5), (6,6,6,6)} is 6, therefore Probability = Event/ Sample space = 6/ 64 = 1/ 63 = 1/216.

3- There are 9 beads in a bag. 3 beads are red, 3 beads are blue, and 3 beads are black. If two beads are chosen at random, what is the probability that they are both blue?

Solution: The probability both are blue is (3/9)(2/8) = (1/3)(1/4) = 1/12

4- A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?

Solution: P(not blue) = 2[P(RY)+P(RG)+P(YG)]+P(RR)+P(GG)+P(YY) = 6[(2/8)(2/7)]+3[(2/8)(1/7)]=15/28

5- A fair coin is tossed, and a fair six-sided die is rolled. What is the probability that the coin come up heads and the die will come up 1 or 2?

Solution: The space S is given by {(1,H), (2,H),…,(6,H), (1,T), (2,T),…,(6,T)}, therefore P{(1,H), (2,H)} = 2/12 = 1/6

6- If two fair, six-sided dice are rolled, what is the probability that the sum of the numbers will be 5?

Solution: This event is given by {(1,4), (2,3), (3,2),(4,1)}, therefore the probability that the sum of the numbers will be 5 equal 4/36 = 1/9

Problems A bag contains 3 red marbles, 3 blue marbles, and 3 green marbles. If a marble is randomly drawn from the bag and a fair, six-sided die is tossed, what is the probability of obtaining a red marble and a 6? A fair, six-sided die is rolled. What is the probability of obtaining a 3 or an odd number?

A large basket of fruit contains 3 oranges, 2 apples and 5 bananas
A large basket of fruit contains 3 oranges, 2 apples and 5 bananas. If a piece of fruit is chosen at random, what is the probability of getting an orange or a banana? If a coin is tossed n times, find probability of at least all coins show two heads

In a class of 30 students, there are 17 girls and 13 boys
In a class of 30 students, there are 17 girls and 13 boys. Five are A students, and three of these students are girls. If a student is chosen at random, what is the probability of choosing a girl or an A student? An unbiased coin is tossed 10 times. Find the probability that there are fewer than 3 heads tossed.

Conditional Probability
Given a Probability space (Ω, F, P) and two events A, B ∈ F with P(B) > 0, the conditional probability of A given B is defined by P(A/B) = P(A ⋂ B)/P(B)

Report Two fair dice are rolled. Find the probability that the sum rolled is at least ten, given: A) No information at all B) At least one die comes up six C) The same number appears on both dice

P(A/B) = P(B/A) P(A)/P(B)
Bayes' rule Bayes' theorem is often completed by noting that, according to the Law of total Probability, P(B) = P(A ⋂ B) + P(Ac ⋂ B) = P(B/A) P(A) P(B/ Ac) P(Ac), where Ac is the complementary event of A (often called "not A"). This results in the analogous form: P(A/B) = P(B/A) P(A)/P(B)

Example We have two urns, I and II. Urn I contains 2 black balls and 3 white balls. Urn II contains 1 black ball and 1 white ball. An urn is drawn at random and a ball is chosen at random from it. What is the probability of choosing black ball from Urn II?

P(urn II/B) = P(B/urn II) P(urnII)/P(B) Where: P(B) =
Solution: P(urn II/B) = P(B/urn II) P(urnII)/P(B) Where: P(B) = P(B/urn I)P(urnI) + P(B/urn II) P(urnII) =(2/5)(1/2) + (1/2)(1/2) = 0.45 Therefore P(urn II/B) = (0.5)(0.5)/0.45 =0.6

Independent Events If P (E) > 0 and P (F) > 0, then E & F are independent iff P (E ∩ F ) = P (E)P (F ) Two events E and F are independent if both E and F have positive probability and if P (E|F ) = P (E) and P (F |E) = P (F )

Examples 1- A coin is tossed twice, let the event E
that head turn up in the 1st toss and the event F that tail turn up in the 2nd toss. Are E and F independent?

Solution: E = {HT, HH}, F = {HT, TT}
E ⋂ F = {HT} , Ω = {HT,TH, HH, TT }. Therefore P(E)P(F) = (2/4)(2/4) = 0.25 Thus P(E ⋂ F) = ¼ = P(E)P(F) Hence E and F are independent F

2- Three shops A, B, C produce identical articles such that the output of shop A equal twice the output of shop B, the output of shop B equal one third the output of shop C with defective rate 1%, 3%, 5% respectively. An article is chosen at random, what is the probability that the chosen article is an excellent piece, then what is the probability that the excellent piece produced by shop A.

P(A) = 2P(B) and P(B) = P(C)/3
Solution: Since P(A) = 2P(B) and P(B) = P(C)/3 But P(A) + P(B) + P(C) = 1 Let P(C) = 3P, thus P(A) = 2P,P(B) = P Therefore 2P+P+3P = 1,so P = 1/6 Hence P(A)=1/3, P(B) = 1/6, P(C) = 1/2

Let the defective articles are denoted by D, thus P(D/A) = 0.01,
P(D/B) = 0.03, P(D/C) =0.05 Let the excellent articles are denoted By E, thus P(E/A) = 0.99, P(E/B) = 0.97 P(E/C) = 0.95 Therefore P(E) = P(E/A) P(A) + P(E/B) P(B) + P(E/C) P(C)

P(A/E) = P(E/A)P(A)/P(E)
Hence P(E) =0.99(1/3) (1/6) (1/2) = 0.97 And P(A/E) = P(E/A)P(A)/P(E) Thus P(A/E) = [0.99(0.33)]/0.97=0.34

Problems Two fair dice are rolled.
a) What is the (conditional) probability that one turns up two spots, given they show different numbers? b) What is the (conditional) probability that the first turns up six, given that the sum is k, for each k from two through 12?

c) What is the (conditional) probability that at least one turns up six, given that the sum is k, for each k from two through 12? A class consists of 60% men and 40% women. Of the men, 25% are blond, while 45% of the women are blond. If a student is chosen at random and is found to be blond, what is the probability that student is a man?

In a bolt factory, machines A,B,C manufacture such that machine A produce twice that of machine B which produce half that of machine C, 2%, 4%, 5% are defective bolts respectively , a bolt is drawn at random and it is a defective quality , what is the probability that it was produced by machine A, B, C.

A standard deck of 52 cards mixed well, one card is drawn at random, if A is the event that an ace is taken out and B is the event that a red card is taken out. Are A and B independent events? At a middle school, 48% of the students ride the bus to and from school. 30% of the students ride the bus and buy their lunch at school. If a student rides the bus, find the probability that the student buys her lunch at school.

A jar contains 3 red, 5 green, 2 blue and 6 yellow marbles
A jar contains 3 red, 5 green, 2 blue and 6 yellow marbles. A marble is chosen at random from the jar.After replacing it, a second marble is chosen. What is the probability of choosing a green and a yellow marble?

A class consists of 60% men and 40% women
A class consists of 60% men and 40% women. Of the men, 25% are blond, while 45% of the women are blond. If a student is chosen at random and is found to be blond, what is the probability that student is a man?

Discrete random variable
The probability distribution: P(X) of a discrete random variable is a list of probabilities associated with each of its possible values xi, i = 1,2, …,n Such that P(x1) + P(x2) P(xn) =1

Cumulative distribution function:
Cumulative distribution function F(x) is defined to be: F(x) = P(X ≤ x) Mode and Median: Mode is the value of x corresponding to maximum probability Median is the value of x for which the C.d.f. equal ½

µ = E(X) = x1 P(x1) + x2 P(x2) + …
Expected value : If X is a discrete random variable with possible values x1, x2, x3, ..., xn , and P(xi) denotes P(X = xi), then the expected value of X is defined by: µ = E(X) = x1 P(x1) + x2 P(x2) + … + xn P(xn) Similarly E(X2) = x12 P(x1) + x22 P(x2) + … + xn2 P(xn)

the variance of the random variable X is defined to be:
Var(X) = E(X2 ) - [E(X)]2 , where E(X) is the expected value of the random variable X and standard deviation is square root of the variance. Note: E(ax+b) = a E(x) + b (prove) Also V(ax+b) = a2 Var(X)

Examples If two dice are thrown, let the r.v. X is
the sum of the numbers facing up, find the expected value, variance and C.d.f.

Solution: Throw two dice; X = the sum of the numbers facing up such that X = 2, 3, …, 12 where X = 2 is the event {(1,1)}, X = 3 is the event {(1,2), (2,1)}, X = 4 is the event {(1,3), (2,2), (3,1)}, X = 5 is the event {(1,4), (2,3), (3,2), (4,1)},

X = 6 is the event {(1,5), (2,4), (3,3), (4,2), (5,1)}, X = 7 is the event {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}, X = 8 is the event {(2,6), (3,5), (4,4), (5,3), (6,2)}, X = 9 is the event {(3,6), (4,5), (5,4), (6,3)}, X = 10 is the event {(4,6), (5,5), (6,4)}, X = 11 is the event {(5,6), (6,5)}, X = 12 is the event {(6,6)}.

Therefore: P(X=2) = 1/36, P(X=3) = 2/36, P(X=4) = 3/36, P(X=5) = 4/36, P(X=6) = 5/36, P(X=7) = 6/36, P(X=8) = 5/36, P(X=9) = 4/36, P(X=10) = 3/36, P(X=11) = 2/36, P(X=12) = 1/36

Thus E(X) = 2(1/36) + 3(2/36) + 4(3/36) + 5(4/36) + 6(5/36)+7(6/36) + 8(5/36) + 9(4/36) + 10(3/36) + 11(2/36) + 12(1/36) = 252/36 = 7. E(X2)=22(1/36) + 32 (2/36) + 42 (3/36) + 52 (4/36) + 62 (5/36) + 72 (6/36) + 82 (5/36) + 92 (4/36) (3/36) (2/36) (1/16) = 1974/36 . Var(X) = E(X2)–(E(X))2 = 1974/36 – 72 = 210/36.

C.d.f. can be calculated such that:
F(X=2) = 1/36, F(X=3) = 3/36, F(X=4) = 6/36, F(X=5) = 10/36, F(X=6) = 15/36, F(X=7) = 21/36, F(X=8) = 26/36, F(X=9) = 30/36, F(X=10) = 33/36, F(X=11) = 35/36, P(X=12) = 1

If two dice are thrown, let the r.v. X is
the maximum of the 2 scores , find the expected value, variance and C.d.f.

Solution: Throw two dice; X = the maximum of the 2 scores such that X = 2, 3, …, 12 where X = 1 is the event {(1,1)}, X = 2 is the event {(1,2), (2,1), (2,2)}, X = 3 is the event {(1,3),(2,3),(3,1),(3,2),(3,3)}, X = 4 is the event {(1,4),(2,4),(3,4), (4,1),(4,2),(4,3),(4,4)},

X = 5 is the event {(1,5),(2,5),(3,5),(4,5),(5,1),(5,2), (5,3),(5,4),(5,5)}, X = 6 is the event {(1,6),(2,6),(3,6),(4,6),(5,6),(6,1), (6,2),(6,3),(6,4),(6,5),(6,6)},

Therefore: P(X=1)= 1/36, P(X=2) = 3/36, P(X=3) = 5/36, P(X=4) = 7/36, P(X=5) = 9/36,P(X=6) = 11/36, Thus E(X) = (1)(1/36)+(2)(3/36)+ (3)(5/36)+(4)(7/36)+(5)(9/36)+ (6)(11/36) = 161/36

Var(X) = E(X2) – (E(X))2 = 791/36 – (161/36)2 = 2555/1296 ≈ 2
And E(X2) = 12 (1/36) +22 (3/36) (5/36) + 42 (7/36) + 52 (9/36) + 62 (11/36) = 791/36 Thus Var(X) = E(X2) – (E(X))2 = 791/36 – (161/36)2 = 2555/1296 ≈ 2

C.d.f. can be calculated such that:
F(X=1) = 1/36, F(X=2) = 4/36, F(X=3) = 9/36, F(X=4) = 16/36, F(X=5) = 25/36, F(X=6) = 1

Binomial distribution
the probability of the r.v. X is given by P(X=x) = ncx pxqn-x whose expected value E(X) = n p and the variance Var(X) = npq

This distribution best describes all situations where a "trial" is made resulting in either "success" or "failure," such as when tossing a coin, or when modeling the success or failure of a surgical procedure.

Examples Suppose you toss a fair coin 12 times. What is the probability of getting exactly 7 Heads, find expected value and the variance? Solution: n = 12, p = 0.5, then P(X=7)=12c7 (0.5)7(0.5)5, E(X) = np = 6, Var(X) = npq = 3

Suppose the probability that a college freshman will graduate is 0. 6
Suppose the probability that a college freshman will graduate is 0.6. Three sisters (triplets) enter college at the same time. What is the probability that at most 2 sisters will graduate? n=3, p=0.6, P(X≤2)= Σ 3cx (0.6)x(0.4)3-x, x = 0,1,2 Or P(X≤2)=1–P(X=3) = 1- 3c3 (0.6)3(0.4)0 = 0.784

Poisson Distribution If n is the number of trials is an average rate where λ = np, then the probability of the random variable X is given by P(X=x) = λx e-λ /x! whose expected value equal its variance such that E(X)=Var(X)=np

A discrete random variable X is said to follow a Poisson distribution with parameter m. If n and the probability of success (p) is too small, then we use Poisson probability, but if n is small, then we use binomial probability

Report Suppose that in one year the number of industrial accidents follows a Poisson distribution with mean 3. What is the probability that in a given year there will be at least 1 accident?

Examples Historically, schools in a Dekalb County close 3 days each year, due to snow. What is the probability that schools in Dekalb County will close for 4 days next year? Solution: λ = 3, then P(X=4) =34 e-3 /4! = 0.168

An expert typist makes, on average, 2 typing errors every 5 pages
An expert typist makes, on average, 2 typing errors every 5 pages. What is the probability that the typist will make at most 5 errors on the next fifteen pages? Solution: λ = 3×2 = 6, X is the number of errors, then P(X = 5) = 65 e-6 /5! =0.161

The average number of children per Spanish couples was 1. 34 in 2005
The average number of children per Spanish couples was 1.34 in Suppose that one Spanish couple is randomly chosen.Find the probability that they have no children, also find the probability that they have fewer children than the Spanish average. Solution: λ=1.34, P(no children)= P(X=0)= (1.34)0 e-1.34 /0! = 0.262 and P(X= 1) = (1.34)1 e-1.34 /1! =0.351

How can we differentiate one probability question from another?
Binomial Probability Only two things can happen (success- failure) There are a finite number of trials n. Poisson Probability You are given a rate, usually a time interval or there is an average given which ties in with this rate. As n tends to infinity and P tends to 0

Problems The ABC Company manufactures toy robots. About 1 toy robot per 100 does not work. You purchase 35 ABC toy robots. What is the probability that exactly 4 do not work? An unbiased coin is tossed 10 times. Find the probability that there are fewer than 3 heads tossed.

At Covington High School, 20% of sixth-form mathematics students retake their C1 module. There are 15 students in Mr Smither’s class; find the probability that precisely four of them retake this module.

Suppose Nancy has classes 3 days a week
Suppose Nancy has classes 3 days a week. She attends classes 3 days a week 80% of the time, 2 days 15% of the time, 1 day 4% of the time, and no days 1% of the time, let X is the NO of attendance days, find standard deviation and cumulative density function.

A multiple choice test contains 20 questions
A multiple choice test contains 20 questions. Each question has four or five choices for the correct answer. Only one of the choices is correct. With random guessing, does this test have a binomial probability distribution?

Hypergeometric Distribution
In Probability theory and Statistics, the hypergeometric distribution is a discrete probability distribution that describes the number of successes in a sequence of n draws from a finite population without replacement, just as the binomial distribution describes the number of successes for draws with replacement.

Hypergeometric probability
Given x, N, n, and k, we can compute the hypergeometric probability such that a population consists of N items, k of which are successes. And a random sample drawn from that population consists of n items, x of which are successes, then the hypergeometric probability is: h(x; N, n, k) = [kCx] [N-kCn-x]/ [NCn]

N: The number of items in the population
k: The number of items in the population that are classified as successes n: The number of items in the sample x: The number of items in the sample that are classified as successes kCx: The number of combinations of k things, taken x at a time

Examples suppose we randomly select 5 cards from an ordinary deck of playing cards. Let the random variable is the number of red cards, what is the probability of selecting exactly 3 red cards?

Solution: Since N =52, n = 5, k = 26, then the probability of getting exactly 3 red cards would be an example of a hypergeometric probability, which is indicated by the following notation: P(X=3)= [26C3] [26C2]/[52C5]

Suppose a shipment of 100 VCRs is known to have 10 defective VCRs
Suppose a shipment of 100 VCRs is known to have 10 defective VCRs. An inspector chooses 12 for inspection. He is interested in determining the probability that, among the 12, at most 2 are defective. Solution: N = 100, k=10, n = 12, so P(2 ≥ x)= Σ [10Cx] [90C12-x]/[100C12], x=0,1,2

You are president of an on-campus special events organization
You are president of an on-campus special events organization. You need a committee of 7 to plan a special birthday party for the president of the college. Your organization consists of 18 women and 15 men. You are interested in the number of men on your committee. What is the probability that your committee has more than 4 men?

Solution: This is a hypergeometric problem because you are choosing your committee from two groups (men and women), n = 7, N = 33, k = 15 Thus P(x >4) = Σ [15Cx] [18C7-x]/[33C7], x=5,6,7

Continuous random variable
For any continuous random variable withProbability density function f(x), we have that: all xʃf(x)dx = 1

f(X ≤ t)= F(t) = aʃt f(x)dx
Cumulative distribution function If X is a continuous r.v. with p.d.f. f(x) defined on a<x< b, then the cumulative distribution function (c.d.f.), written F(t) is given by: f(X ≤ t)= F(t) = aʃt f(x)dx

P(s < X < t) = P(X < t) - P(X < s)
The c.d.f. can be used to find out the probability of a random variable being between two values: P(s < X < t) = the probability that X is between s and t. But this is equal to the probability that X < t minus the probability that X < s. Hence: P(s < X < t) = P(X < t) - P(X < s) = F(t) - F(s)

Expectation and Variance
With discrete r.v., we had that the expectation was xP(X=x), where P(X=x) was the p.d.f. To find the expectation of a continuous random variable, we integrate rather than sum, i.e. E(X)= all xʃx f(x)dx , also E(X2) = all xʃx2 f(x)dx, hence Var(X) = E(X2) – [E(X)]2

Mode and Median The median m for a continuous random variable X is a value which satisfies P(X<m) = P(X>m) = ½, while mode is the value of x at maximum probability.

Examples X is a continuous random variable with P.d.f. f(x) = 0.5x, 0<x<2. What is the value of the number m,for which P(X<m) = 0.5? Solution: m is called the median of the P.d.f. f(x), such that: 0ʃm 0.5x dx= 0.5, thus 0.25m2 = 0.5, hence m = is the median.

X is a continuous random variable with probability density function given by f(x) = cx for 0 < x < 1, where c is a constant. Find c. Solution: Since 0ʃ1 c x dx= 1, hence c/2= 1, therefore c = 2.

A continuous random variable X which can assume between x = 2 and 8 inclusive, has a density function given by f(x)=c(x+3) where c is a constant. (a) Calculate c (b) P(3<X<5) (c) P(X≥4)

Solution: a) Sinceall xʃf(x)dx = 1, f(x) will be density functions, thus: 2ʃ8 c(x+3) dx= 1, therefore c = 1/48 b)P(3<X<5) =3ʃ5 [(x+3)/48] dx= 7/24 c) P(X≥4) = 4ʃ8 [(x+3)/48] dx = ¾

Moment generating Function
For any continuous Pd.f. f(x), a<x<b, then m.g.f. is denoted by E(etx) such that: E(etx) = aʃb etx f(x) dx, where E(x) = d/dt [E(etx)]t=0 E(x2) = d2/dt2[E(etx)]t=0

Moment about zero (μr`)
For any continuous Pd.f. f(x), a<x<b, then μr`= aʃb xr f(x) dx, where: μ0`= aʃb f(x) dx = 1, μ1`= aʃb x f(x) dx = E(x) μ2`= aʃb x2 f(x) dx = E(x2) Hence μr`= aʃb xr f(x) dx = E(xr)

μ2= aʃb (x- μ)2 f(x) dx = Var(x),
Moment about mean (μr) For any continuous Pd.f. f(x), a<x<b, then μr = aʃb (x- μ)r f(x) dx, where: μ0 = aʃb f(x) dx = 1, μ1 = aʃb (x- μ) f(x) dx = 0 μ2= aʃb (x- μ)2 f(x) dx = Var(x), where μ = E(x).

μ0`= 1, μ1`= 0ʃ1 x(2x) dx = 2/3 = E(x),
Example f(x)=2x is P.d.f. x ∈ [0, 1],find M.G.F. and first 3 moments about zero and about mean. Solution: E(etx) = 0ʃ1 etx (2x) dx =2[(et/t)- (et/t2) + 1/t2] μ0`= 1, μ1`= 0ʃ1 x(2x) dx = 2/3 = E(x), μ2`= 0ʃ1 x2(2x) dx =1/2 = E(X2)

var(x) = E(X2) – [E(X)]2=(1/2)-(2/3)2
Since for any probability distribution: μ0 = 1 , μ1 = 0, μ2 = var(x) Therefore var(x) = E(X2) – [E(X)]2=(1/2)-(2/3)2 =0.056

Exponential distribution
The probability density function (p.d.f.) of an exponential distribution is: f(x, λ) = λ e-λx , x>0 The cumulative distribution function is given by: F(x, λ) = 1- e-λx , x>0

Mean and variance The mean or expected value of an exponentially distributed r.v. X with rate parameter λ is given by E(X)=1/ λ While the variance of X is given by Var(X) =1/ λ2

Moment generating function of exponential distribution
The moment generating function of a exponential distribution is defined by: E(etx)= 0ʃ etx (λ e-λx)dx = λ/(λ-t)

E(x2)=d2/dt2[E(etx)]t=0=[2λ/(λ-t)3]t=0 =2/ λ2
We can calculate E(x) and Var(x) using M.G.F. such that: E(x)=d/dt[E(etx)]t=0= [λ/(λ-t)2] t=0=1/ λ And E(x2)=d2/dt2[E(etx)]t=0=[2λ/(λ-t)3]t=0 =2/ λ2 Therefore Var(x) = E(X2) – [E(X)]2 = 1/ λ2

Gamma distribution If the random variable X is gamma-distributed with parameters α and β, then the P.d.f. of X is: P(X) = [βα/ α] xα-1 e-βx Where α = 0ʃ xα-1 e-x dx, X and the parameters α and β must be positive.

Mean and Variance The mean of a gamma-distributed variable is α/β, while the variance is α/β2 (derive). If α > 1, then there is a mode which is (α-1) / β. The cumulative distribution function is given by: F(x) = 0ʃx [βα/ α] yα-1 e-βy dy

Moment generating function of Gamma distribution
The moment generating function can be expressed by: E(etx) = 0ʃ etx [βα/ α] xα-1 e-βx dx = βα /(β-t)α We can calculate E(x) and Var(x) using M.G.F. such that: E(x)=d/dt[E(etx)]t=0=[α βα /(β-t)α+1]t=0 = α/β

Var(x) = E(X2) – [E(X)]2 = α/β2
E(x2)=d2/dt2[E(etx)]t=0 =[α(α+1)βα/(β-t)α+2]t=0 = [α(α+1)/β2 ]t=0 Therefore Var(x) = E(X2) – [E(X)]2 = α/β2

The probability density function of the
Uniform distribution The probability density function of the continuous uniform distribution is: f(x) = 1/(B-A), A<x<B The values at the two boundaries a & b are usually unimportant. The cumulative distribution function is given by: F(X) = Aʃx 1/(B-A) dx = (x-A)/(B-A)

The mean is expressed by E(X) = AʃB x/(B-A)dx= (a + b)/2 And
Mean and Variance The mean is expressed by E(X) = AʃB x/(B-A)dx= (a + b)/2 And E(X2) = AʃB x2/(B-A)dx= (a2+ab+b2)/3 Therefore the variance is given by: Var(x) = E(X2) – [E(X)]2 = (b−a)2/12

Moment generating function of Uniform distribution
The moment generating function can be expressed by: E(etx) = AʃB etx/(B-A) dx = [etB – etA] /(B-A)t

Examples On the average, a certain computer part lasts 10 years. The length of time the computer part lasts is exponentially distributed. What is the probability that a computer part lasts more than 7 years? Solution: Since the mean= 1/ λ = 10, therefore λ = 0.1, hence: P(X > 7) = 7ʃ (0.1 e-0.1x)dx = e-0.7.

P(X > 30) = 30ʃ (x e-x/5)dx = 7e-6
Let X be a r.v. with gamma distribution with α= 2, β =1/5, find P(X> 30). Solution: P(X > 30) = 30ʃ (x e-x/5)dx = 7e-6

The amount of time, in minutes, that a
person must wait for a bus is uniformly distributed between 0 and 15 minutes, inclusive. What is the probability that a person waits fewer than 12.5 minutes? Solution: P(x< 12.5) = 0ʃ12.5 [1/(15-0)] dx = 0.83

f(X) = c(1-x2), -1<x<1.
Problems 1-Let X be a random variable with P.d.f f(X) = c(1-x2), -1<x<1. What is the value of c? What is the cumulative distribution function of X. 2- Let X be a random real number between 0 and 5, such that X has a uniform distribution, find P(0<x<4.5).

مع تحياتى و تمنياتى بالتوفيق
د. خالد لنجار

السلام عليكم و رحمة الله و بركاته

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السلام عليكم و رحمة الله و بركاته

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