Introduction to Probability and Statistics Slides 4 Chapter 4

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1 Inroducion o Probabiliy and Saisics Slides 4 Chaper 4 Ammar M. Sarhan, asarhan@mahsa.dal.ca Deparmen of Mahemaics and Saisics, Dalhousie Universiy Fall Semeser 8

2 Dr. Ammar Sarhan Chaper 4 Coninuous Random Variables and Probabiliy Disribuions

3 4. Coninuous Random Variables and Probabiliy Disribuions Coninuous Random Variables A random variable is coninuous if () is se of possible values is an enire inerval of numbers; () P( c) for any number c. Example: () The deph of a chosen locaion; () The lifeime of a produc; () The waiing ime spen by a cusomer o receive his/her serves.

4 Probabiliy Disribuion Dr. Ammar Sarhan 4 Le be a coninuous rv. Then a probabiliy disribuion or probabiliy densiy funcion (pdf) of is a funcion f (x) such ha for any wo numbers a and b, P ( a b) f ( x) dx This gives he probabiliy ha akes on a value in he inerval [a, b]. I also gives he area under he densiy curve. b a P(a b) f (x) a b x

5 Probabiliy Densiy Funcion Dr. Ammar Sarhan 5 The pdf f (x) saisfies he following condiions:. f (x) for all values of x.. The area beween he graph of f (x) and he x-axis is equal o. P( ) f ( x) dx If is a coninuous rv hen For any even A defined on, P ( A) f ( x) dx x A P(a b) P(a < b) P(a < b) P(a < < b)

6 Dr. Ammar Sarhan 6 Example: Le be a rv wih he following pdf f ( x) 6 x < 6 oherwise 6 f (x) P( 6) dx 6 P(9 8) Because whenever a b 6, P(a b) depends only on he widh b-a of he inerval, is said o have a uniform disribuion.

7 Dr. Ammar Sarhan 7 Uniform Disribuion A coninuous rv is said o have a uniform disribuion on he inerval [A, B] if he pdf of is Example: (Prob. 5, p. 5) is he ime elapses beween he end of he hour and he end of he lecure. a) Find k. Since, k f ( x) dx 8 k x x < f ( x) oherwise k k x dx k x

8 b) Wha is he probabiliy ha he lecure ends wihin min of he end of he hour? Dr. Ammar Sarhan 8 k P( ) f ( x) dx k x dx k x /8 8.5 c) Wha is he probabiliy ha he lecure coninuous beyond he hour for beween 6 and 9 sec? min P(.5) 8.5 f ( x) dx.5 [( ) ] c) Wha is he probabiliy ha he lecure coninuous for a leas 9 sec beyond he end of he hour? min P(.5 ).5 8 f ( x) dx.5 k x k x dx [ ] dx k x k x.5.5

9 4. Cumulaive disribuion funcions and Expeced values Dr. Ammar Sarhan 9 The Cumulaive Disribuion Funcion The cumulaive disribuion funcion, F (x) for a coninuous rv is defined for every number x by F( x) x ( x) f ( y dy P ) For each x, F(x) is he area under he densiy curve o he lef of x. Example: (4.6, p. 7)

10 Using F(x) o Compue Probabiliies Dr. Ammar Sarhan Le be a coninuous rv wih pdf f(x) and cdf F(x). Then for any number a, P( > a) - F(a) and for any numbers a and b wih a < b, Example: (4.7, p. 8) P(a b) F(b) - F(a)

11 Obaining f(x) from F(x) Dr. Ammar Sarhan If is a coninuous rv wih pdf f(x) and cdf F(x), hen a every number x for which he derivaive F (x) exiss F (x) f (x). Example: (4.8, p. 9) Perceniles When we say ha an individual s es score was a he 85 h percenile of he populaion, we mean ha 85% of all populaion scores were below ha score and 5% were above. Le p be a number beween and. The (p)h percenile of he disribuion of a coninuous rv denoed by η(p), is defined by p P η ( p) ( η( p) ) F( η( p)) f ( y) dy

12 Dr. Ammar Sarhan Shaded area p F (x) p F(η(p)) η(p) x η(p) x Example: Le be a rv wih he following pdf x x f ( x) oherwise Find he (p)% percenile of? Firs, he cdf F x x x ( x) f ( y) dy y dy y x for x

13 The (p)h percenile is he soluion of he following eqn: Dr. Ammar Sarhan p F( η( p)) Then, he (p)h percenile is p [ η( p) ] η For he 5h percenile, p.5, η.5.77 p f (x) F (x).5 F(η(.5)) η(.5).77 x η(.5).77 x

14 Dr. Ammar Sarhan 4 Median The median of a coninuous disribuion, denoed by, is he 5h percenile. So ~ μ saisfies.5 F( ~ μ ). Tha is, half he area under he densiy curve is o he lef of ~ μ. Noice: For he disribuion wih symmeric pdf he median equals he poin of symmery. μ ~ f(x) f(x) f(x) μ ~ μ ~ Medians of symmeric disribuions μ ~

15 Expeced Value Dr. Ammar Sarhan 5 The expeced or mean value of a coninuous rv wih pdf f (x) is μ E( ) x f ( x) dx Example: In he previous example, find E(). E ( ) x f ( x) dx x dx y Expeced Value of h() If is a coninuous rv wih pdf f(x) and h(x) is any funcion of, hen μ [ h( )] E h( x) f ( x dx h ( ) )

16 Variance and Sandard Deviaion Dr. Ammar Sarhan 6 The variance of coninuous rv wih pdf f(x) and mean μ is V ( ) σ E[( μ) ] ( x μ) f ( x) dx. The sandard deviaion (SD) of is σ σ Shor-cu Formula for Variance V ( ) ( ) E μ

17 Example: In he previous example, find ) V(), ) E[ + ] ) V[ + ] Dr. Ammar Sarhan 7 ) V ( ) ( ) E μ We have μ E( ) and Then E 4 ( ) x f ( x) dx x dx 4 y 4 V ( ) ) E[ + ] E() + (/) + 4 ) V[ + ] V() 9 V() 9 (5/8)

18 Dr. Ammar Sarhan 8 Exponenial Disribuion A coninuous rv has an exponenial disribuion wih parameer λ> if he pdf is λ e f ( x; λ) λ x x oherwise Mean and Variance Le be a rv having he exponenial disribuion wih parameer λ. Then E( ) V ( ) λ λ Cdf of exponenial Disribuion F( x; λ) e λ x x oherwise

19 Dr. Ammar Sarhan The Memoryless propery of exponenial Disribuion Le has an exponenial disribuion wih parameer λ>. Then ) ( ) ( P P + ( ) ( ) ( ) ( ) ) ( P P P + + ( ) ( ) ( ) ( ) F F P P + + ( ) ( ) ( ) ( ) e e e e e λ λ λ λ λ + + ) ( ) ( P F Proof: 9

20 Dr. Ammar Sarhan Example: Suppose ha he response ime a a cerain on-line compuer erminal (he elapsed ime beween he end of a user s inquiry and he beginning of he sysem s response o ha inquiry) has an exponenial disribuion wih expeced response ime equal o 5 esc. Find ) Probabiliy ha he response ime is a mos sec? Since E() 5 /λ, hen λ /5. P( ) F() - e -.() - e ) Probabiliy ha he response ime is beween 5 and sec? P(5 ) F() F(5) e -.(5) -e -.() e - -e -. ) Assume ha one user is waied for sec, wha is he probabiliy ha he will ge he sysem s response wihin he nex 5sec?

21 Dr. Ammar Sarhan Applicaions of he Exponenial Disribuion The exponenial disribuion is frequenly used as a model for he disribuion of he occurrence of successive evens, such as ) Cusomers arrivals a a service faciliy, ) Calls coming o a swichboard. The reason for his is ha he exponenial disribuion is closely relaed o he Poisson process. Proposiion Suppose ha he number of evens occurring in any ime inerval of lengh has a Poisson disribuion wih parameer α (where α is he expeced number of evens occurring in uni of ime) and ha he numbers of occurrences in nonoverlapping inervals are independen of one anoher. Then he disribuion of elapsed ime beween he occurrences of wo successive evens is exponenial wih parameer λ α.

22 Example: Dr. Ammar Sarhan Suppose ha calls are received a 4-hour holine according o a Poisson process wih rae α.5 per day. Compue. The probabiliy ha more han days elapse beween wo successive calls (wo calls)?. The expeced ime beween wo successive calls.. Given ha hey have jus received a call, wha is he probabiliy ha hey will receive a call wihin he nex 6 hours? Soluion: Le denoe he number of days beween successive calls. Then ~ Exp(.5). Using his informaion, one can easily answer he above hree quesions.

23 4. The Normal Disribuion Dr. Ammar Sarhan A coninuous rv is said o have a normal disribuion wih parameers μ and σ, where - < μ < and < σ, if he pdf of is f ( x; μ, σ ) e σ π ( x μ ) /(σ ), < x < Sandard Normal Disribuions The normal disribuion wih parameer values μ and σ is called a sandard normal disribuion. The random variable is denoed by Z. The pdf is f ( x;,) e π x /, < x <

24 Dr. Ammar Sarhan 4 The cdf is z Φ ( z) P( Z z) f ( y;,) dy Sandard Normal Cumulaive Areas

25 Dr. Ammar Sarhan 5 Sandard Normal Disribuion Le Z be he sandard normal variable. Find (from able) a. P(Z <.85) Area o he lef of b. P(Z >.) - P(Z.).94 c. P(-. Z.78) P(Z.78) P(Z -.)

26 z α Noaion Dr. Ammar Sarhan 6 z α will denoe he value on he measuremen axis for which he area under he z curve lies o he righ of z α is α. z curve P(Z <-z α ) α z α Since α of he area under he z curve lies o he righ of z α, hen -α of he area lies o is lef. Thus, z α is he (-α)h percenile of he sandard normal disribuion. By symmery he area under SNC o he lef of z α is also α. The z α s are usually referred o as z criical values. Table 4. liss he mos useful perceniles and values z α.

27 Example: The z.5 is he (-.5)h percenile, so z The area under he SND curve o he lef of z.5 is also.5. Dr. Ammar Sarhan 7 z curve z z h percenile

28 Dr. Ammar Sarhan 8 Example: Le Z be he sandard normal variable. Find z if a. P(Z < z).978 Look a he able and find an enry.978 hen read back o find z.46 b. P( z < Z < z).8 P(-z < Z < z ) P(Z<z) P(Z<-z) P(Z<z) P(Z>z) P(Z<z) [-P(Z<z)] P(Z<z) +P(Z<z) P(Z < z).8+ P(Z < z) P(Z < z).8 /.966 z.

29 Nonsandard Normal Disribuions Dr. Ammar Sarhan 9 If has a normal disribuion wih mean μ and sandard deviaions σ, shorly N(μ, σ ), hen has a sandard normal disribuion. Example: Le be a normal random variable wih μ 8 and σ. Find P( 65)?

30 Example: A paricular rash shown up a an elemenary school. I has been deermined ha he lengh of ime ha he rash will las is normally disribued wih μ 6 days and σ.5 days. Find he probabiliy ha for a suden seleced a random, he rash will las for beween.75 and 9 days? Dr. Ammar Sarhan Suppose ha ~ N(6,.5), find P(.75 9)?

31 Perceniles of an Arbirary Normal Disribuion Dr. Ammar Sarhan (p)h percenile for normal(μ, σ ) (p)h percenile μ + σ sandard normal Example: The amoun of disilled waer dispensed by a cerain machine is normally disribued wih mean value 64 oz and sandard deviaion.78 oz. Wha conainer size c will ensure ha overflow occurs only.5% of he ime? Le denoe he amoun dispensed, hen ~ N(64,.78). The desired condiion is ha P( > c).5, or, equivalenly ha P( c).995. Thus, c is he 99 h percenile of N(64,.78). Since 99 h of Z is z..58, hen c 64 + (.58) (.78) oz

32 Normal Approximaion o he Binomial Disribuion Dr. Ammar Sarhan Le be a binomial rv based on n rials, each wih probabiliy of success p. If he binomial probabiliy hisogram is no oo skewed, may be approximaed by a normal disribuion wih μ np and σ npq. P( x) x +.5 np Φ npq In pracice, his approximaion is adequae provided ha boh np and nq. Example: A a paricular small college he pass rae of Inermediae Algebra is 7%. If 5 sudens enrol in a semeser deermine he probabiliy ha a mos 75 sudens pass.

33 Dr. Ammar Sarhan Le denoe he number of sudens pass he es. Then, ~ Bino(5,.7) Soluion and he desired probabiliy is P( 75 ) μ np 5 (.7) 6, and n(-p) 5 (.8) 4, σ npq 5(.7)(.8) Then, approximaely ~ N(6, ) P( 75) Φ Φ (.55).994

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