Binomial Distribution

Transcription

1 Overview Example: coi tossed three times Defiitio Formula Recall that a r.v. is discrete if there are either a fiite umber of possible values e.g. y 1,..., y or at most there is oe for every iteger e.g. y 1, y 2,.... Also recall that the probability distributio of a discrete r.v. is described by the probability of each possible value of the r.v. i.e. py = PY = y for all possible y. Some particular probability distributios occur ofte because they are useful descriptios of certai chace pheomeo uder study. For example, biomial distributio for discrete r.v. is very useful i practice. Toss a coi idepedetly three times ad record the umber of times that the coi laded o heads. Let Y = # of heads. The the sample space is: 1st toss 2d toss 3rd toss Y H H H 3 H H T 2 H T H 2 H T T 1 T H H 2 T H T 1 T T H 1 T T T 0 Suppose p = probability of heads ad q = probability of tails. The q = 1 p ad PY = 3 = p p p = p 3 PY = 2 = p p q + p q p + q p p = 3p 2 q PY = 1 = 3pq 2 PY = 0 = q 3 A biomial distributio must satisfy three 3 coditios: 1. There are trials, each of which ca result i oe of the two outcomes, either success or failure. i.e. the trials are dichotomous or Beroulli. 2. The probability of a success is costat p for all trials. 3. The trials are idepedet. Defie r.v. Y = # of successes i trials. The Y is said to have a biomial distributio with parameters ad p ad is writte as: Y B, p Let Y B, p. The formula for the biomial distributio is: where! PY = r = r! r! pr q r r = 0, 1, 2,..., are the possible umbers of successes. q = 1 p is the probability of failure. p r = p p p r times. q r = q q q r times. r =! r! r! is read as choose r. A quick review of factorial operatio k! = k k 1 1 4! = = 24 5! = = 120 0! = Examples Oe more example Properties Proportios Suppose Y B3, 1 2. PY = 1 = 3! = 3 1!2! Suppose Y B7, PY = 3 = 7! 3!4! = 3, p = 0.5 = = ! = 7, p = 0.6 Toss a fair die three times. What is the probability that there is at least oce a 6? Let Y B3, 1 6. The Pat least oe 6 = 1 Po 6 = 1 PY = 0. Sice PY = 0 = 3! = !3! 6 6 we have Pat least oe 6 = = = 3, p = Y B, p is a discrete r.v. The expectatio of Y is The variace of Y is µ Y = EY = p σ 2 Y = V ary = pq For example, suppose Y B10, 1 4, EY = = 10 4 = 2.5, V ary = = Aother example, suppose Y B100, 1 4, EY = = = 25, V ary = = I some situatios, oe is iterested i the proportio of successes amog trials. Defie where Y B, p. The W = Y EW = p, V arw = pq usig properties of biomial distributio i.e. expectatio ad variace of Y

2 z PZ>=1.5 PZ<=1.5 P0<=Z<=1.5 Overview Defiitio Defiitio ad properties Biomial distributio is a very importat probability model ad is ofte very useful. For example, cure rate of a ew drug, proportio of lakes that are suitable habitats for fish, survival rate of seedligs, etc. However, biomial distributio is ot suitable for all situatios. Are the trials dichotomous? Is the success probability costat? Are the trials idepedet? Cosider the fair die game, # of rats cured withi the same litter, etc. Always thik about the three assumptios before usig a biomial distributio. Recall that a r.v. is discrete if there are either a fiite umber of possible values e.g. y 1,..., y or at most there is oe for every iteger e.g. y 1, y 2,.... I cotrast a r.v. is cotiuous if the possible values are over a etire iterval of umbers. The probability distributio of a cotiuous r.v. is described by a probability desity curve such that area uder the curve correspods to probability. The most frequetly used type of cotiuous distributio is ormal distributio, which is also kow as Gaussia distributio. Normal distributio is a very importat distributio for modelig cotiuous data, because data that are iflueced by may small ad urelated radom effects are approximately ormally distributed e.g. a studet s height. A ormal distributio is defied by a bell-shaped curve with a probability desity fuctio fy = 1 e 1 2σ 2y µ2 2πσ fy y A r.v. Y is said to have a ormal distributio with parameters µ ad σ 2 ad is writte as: Y Nµ, σ 2 Total area uder the distributio curve is 1. The distributio curve is symmetric about the ceter µ. Values aroud µ are more probable. The area i the rage of µ ±σ is about 2/3 of the total area The distributio curve has a well-defied bell shape. The parameters µ ad σ 2 provide complete iformatio about the distributio curve. I fact, EY = µ, V ary = σ Stadard Stadard Stadard Stadard Defiitio ad properties More properties Fidig probabilities Examples A r.v. Z has a stadardized ormal distributio if Z Nµ, σ 2 with expectatio µ = 0 ad variace σ 2 = 1 ad is writte as: Z N0, 1. The distributio curve is symmetric about the ceter µ = 0. The area betwee 1 ad 1 i.e. µ ± σ is about 2/3. The area betwee 2 ad 2 i.e. µ ± 2σ is Suppose Z is a stadard ormal r.v. i.e. Z N0, 1. PZ 0 = 0.5. P < Z < = 1. PZ = 1 = 0 ad i geeral PZ = c = 0 for ay c. Hece PZ < 0 = PZ 0 = 0.5. Remark: Suppose Y B3, 0.5 the PY 0 PY > 0! Table A o page 408 of the course otes bluebook gives PZ z for ay give value z 0 also kow as the upper tail probability. The z values are o the outside of Table A ad the probabilities are i the iside of Table A. Read the z value off x.x from the row idex ad 0.0x from the colum idex. PZ 1.5 = PZ 1.5 = 1 PZ 1.5 = = P0 Z 1.5 = PZ 0 PZ 1.5 = = How to fid PZ z for ay give value z? Drawig pictures helps! fz

3 PZ>= 0.6 P 1.7<=Z<=1.7 mu = 20, sigma = 3 P20<=Y<= P 1.7<=Z<=1.7 P X < = 0.9 PZ<=?=0.90 z = 1.28 mu = 20, sigma = 3 PY<= Stadard Stadard Stadard Stadard Fidig quatiles PZ 0.6 = 1 PZ 0.6 = 1 PZ 0.6 = = P 1.7 Z 1.7 = 2 P0 Z 1.7 = = P 1.7 Z 1.7 = 1 PZ 1.7 or Z 1.7 = 1 2 PZ 1.7 = = = What value z would give PZ z = 0.90? This z value would also give PZ z = = Look up Table A for the z value that correspods to Whe z = 1.28, PZ z = Whe z = 1.29, PZ z = Thus z value sought after is a umber betwee 1.28 ad More directly the table o page 92 of the bluebook gives z = Iterpretatio: The z value sought after is the cut-off value where 90% of the populatio lies below i.e. the 0.9 th populatio quatile or the 90 th populatio percetile Geeral Geeral Geeral Geeral Defiitio ad stadardizatio Example A r.v. Y is said to have a ormal distributio with parameters µ ad σ 2 ad is writte as: Y Nµ, σ 2 For example, Y N20, 3 2 stads for ormal distributio with µ = 20, σ 2 = 9. FACT: Suppose Y Nµ, σ 2, the a very useful trasformatio is Z = Y µ N0, 1 σ For Y N20, 3 2 i.e. µ = 20, σ 2 = 9, we have P20 Y 23 = P20 µ Y µ 23 µ 20 µ = P Y µ 23 µ σ σ σ = P Z 3 3 = P0 Z 1 = PZ 0 PZ 1 = = Give a value y, the term z = y µ σ is called the z-score correspodig to y. It represets the umber of stadard deviatios y is from µ. For Y N20, 3 2, 23 is 1σ away from µ = is 0σ away from µ = 20 i.e. at µ = is 4 3σ away from µ = is 1σ away from µ = 20. Suppose Y N20, 3 2, what is PY 15.5? Y µ PY 15.5 = P 15.5 µ σ σ = P Z 3 = PZ 1.5 =

4 P X < = 0.9 mu = 60, sigma = 10 PY<=?=0.90 z = Geeral Biomial ad s Fidig quatiles Key R commads Overview Defiitios Suppose Y N60, 10 2, what value y would give PY y = 0.90? From previous discussio, we kow that Thus PZ = 0.90 Y µ P σ = 0.90 PY µ 1.282σ = 0.90 PY µ σ = 0.90 Thus y = µ σ = = y is the stadard deviatio above the mea. > # Y~B3,0.5, compute PY=1 > dbiomx=1,size=3,prob=0.5 [1] > # Y~B7,0.6, compute PY=3 > dbiomx=3,size=7,prob=0.6 [1] > # Y~B3,1/6, compute PY>0 > pbiomq=0,size=3,prob=1/6,lower.tail=f [1] > # of by complemet > 1-dbiomx=0,size=3,prob=1/6 [1] > # Z~N0,1, compute PZ>=1.5, PZ<=1.5, P0<=Z<=1.5, > porm1.5, lower.tail=f [1] > porm1.5, lower.tail=t [1] > porm0, lower.tail=f-porm1.5, lower.tail=f [1] > # Z~N0,1, compute PZ>=-0.6 > porm-0.6, lower.tail=f [1] > # Z~N0,1, compute P-1.7<=Z<=1.7 > 1-2*porm1.7, lower.tail=f [1] > # Z~N0,1, compute PZ<=?=0.90 > qorm0.9, lower.tail=t [1] > # Y~N20,9, compute P20<=Y<=23, PY<=15.5 > pormq=20,mea=20,sd=3,lower.tail=f-porm23,mea=20,sd=3,lower.tail=f [1] > pormq=15.5,mea=20,sd=3,lower.tail=t [1] > # Y~N60,100, compute PY<=?=0.90 > qormp=0.9,mea=60,sd=10,lower.tail=t [1] Recall that a radom variable r.v. is a variable that depeds o the outcome of a chace situatio. The probability distributio of a r.v. is described by the probability of all possible outcomes of the r.v. Now we put data ad probability models together by viewig a observatio obs as a outcome of a r.v. We cosider obs as outcomes of r.v. s. The r.v. s Y 1,Y 2,...,Y formig the sample data are called a radom sample of size from a give distributio if 1 the Y i s are idepedet r.v. s; 2 the Y i s are idetically distributed i.e. every Y i has the same probability distributio. Coditios 1 ad 2 are sometimes abbreviated to the Y i s are iid. Further assume: 3 For each i, EY i = µ, V ary i = σ 2. A statistic is ay quatity whose value ca be calculated from sample data. It is a r.v. For example, sample mea is a statistic: Ȳ = Y i = Y 1 + Y Y The we have uder 1 3: EȲ = µ, σ2 V arȳ = As icreases, V arȳ decreases Sample mea FACT 1 FACT 2 FACT 3 By 2 ad 3, 1 EȲ = E Y 1 + Y Y = 1 EY 1 + Y Y = 1 EY 1 + EY EY = 1 µ + µ + + µ = µ By 1, 2, ad 3, 1 V arȳ = V ar Y 1 + Y Y = 1 2V ary 1 + Y Y = 1 V ary1 + V ary2 + + V ary 2 = 1 2σ2 + σ σ 2 = 1 2 σ2 = σ2 Suppose Y 1, Y 2,..., Y is a radom sample from Nµ,σ 2 distributio. The Ȳ Nµ, σ2 For example, if Y 1, Y 2,..., Y 40 are iid N3, 4, the Ȳ N3, 4 = N3, We may write Ȳ Nµ,σ2, where the variace of Ȳ is Ȳ σ 2 Ȳ = σ2 ad the correspodig stadard deviatio of Ȳ is σ Ȳ = σ also kow as the stadard error of the mea. Suppose Y 1,Y 2,...,Y is a radom sample from Nµ,σ 2 distributio. The Y i Nµ,σ 2 For example, if Y 1,Y 2,...,Y 40 are iid N3, 4, the 40 Thik of Y i = Ȳ : E Y i = µ, Y i N120, 160 V ar Y i = σ 2. Suppose Y 1, Y 2,...,Y is a radom sample from a distributio with mea µ ad variace σ 2. If is large, the the distributio of Ȳ is closely approximated by Ȳ Nµ, σ2 The larger is, the better the approximatio. This fact is also kow as the cetral limit theorem CLT for the sample mea

5 = 60, p = Example: fair die game FACT 4 A quick summary o CLT Recall the fair die game example: Y = wiig $ with probability distributio p0 = 1/2, p9 = 1/6,p15 = 1/3 ad mea µ = 6.5, σ 2 = If we repeat the game 1000 times idepedetly, the for Y 1, Y 2,..., Y 1000, we have EȲ = 6.5, V arȳ = 1000 By CLT, Ȳ is approximately N 6.5, For example, PȲ 7 P Z = PZ 2.32 = Suppose Y 1, Y 2,..., Y is a radom sample from a distributio with mea µ ad variace σ 2. If is large, the the distributio of Y i is closely approximated by Y i Nµ,σ 2 The larger is, the better the approximatio. This fact is also kow as the cetral limit theorem CLT for the sample sum. Example: fair die game Note that = 1000,E Y i = = 6500, V ar Y i = = By FACT 4, Y i is approximately N6500, The, P Y i 7000 P Z = PZ 2.32 = Radom sample, statistic, sample mea. If Y i iid Dµ, σ 2 σ2, EȲ = µ, V arȳ =. If Y i iid Nµ, σ 2 σ2, Ȳ Nµ,. If Y i iid Dµ, σ 2 σ2, approximately Ȳ Nµ, CLT. Similarly for Y i with mea µ ad variace σ 2. The CLT is a remarkable result. It says that regardless of the shape of the origial distributio, the takig of averages or sums results i ormal distributio. Hece we oly eed to kow the populatio mea ad stadard deviatio to have a approximate distributio of Ȳ or Y i. Why is ormal distributio so useful? May atural pheomea give rise to ormal variables; may variables ca be trasformed to ormal distributio; ad CLT. Simulatios. How big must be? Usually whe 30, the CLT works well. But if Y i has a highly skewed distributio, the we eed larger, i which case we may cosider trasformatio to a symmetric distributio first. A applicatio of the CLT is to approximate a biomial distributio by a ormal distributio Normal approximatio of biomial Normal approximatio of biomial Normal approximatio of biomial Recall that if Y B, p, the µ = p, σ 2 = pq where q = 1 p. Let Y B60, 0.4. Thus = 60, p = 0.4, ad µ Y = 24, σ 2 Y = Compute PY 20 = PY = 0 + PY = PY = 20? Now, thik of Y as the sum of = 60 iid r.v. s, each of which is a 0/1 variable. Let Y i = 1 for a success with probability p ad Y i = 0 for a failure with probability q. The Y = Y i is the umber of successes i = 60 trials. By the CLT o sum FACT 4, we ca approximate the distributio of Y by ormal distributio with mea ad variace µ Y = p = 24 σ 2 Y = pq = 14.4 = Hece if Y B60, 0.4, the Y NA N24, Now, we approximate PY 20 by PY 20 PY NA 20 YNA = P = PZ = The exact probability is PY 20 = > pbiomq=20, size=60, prob=0.4, lower.tail=t > [1] The discrepacy is largely due to the fact that we are approximatig a discrete r.v. by a cotiuous r.v. We ca make the ormal approximatio better by usig cotiuity correctio cc. PY 20 PY NA 20.5 YNA = P = PZ = Normal approximatio is OK if p 5 ad q 5. Always check this before usig ormal distributio to approximate biomial distributio. I the example, Y B60, 0.4, p = 24,q = 36, thus it is OK to use ormal approximatio. Cosider proportio of success deoted as W before For Y B60, 0.4, ˆp = Y Eˆp = p = 0.4, V arˆp = pq = = Hece by CLT o average FACT 3, ˆp NA N0.4, Thus for the example Y B60, 0.4, Pˆp 0.48 Pˆp NA 0.48 ˆpNA = P = PZ = =

6 Sample variace Sample variace Example Suppose Y i are iid Nµ, σ 2. We kow that the sample mea Ȳ Nµ, σ2. What about the sample variace also a statistic FACT S 2 = 1 1 Y i Ȳ 2? If Y 1, Y 2,..., Y form a radom sample from Nµ, σ 2, the V 2 1S2 = χ 2 σ 2 1 where χ 2 1 is a chi-squared distributio with 1 degrees of freedom d.f.. Shape of the χ 2 1 distributio curve depeds o the d.f. V 2 0 The distributio curve is skewed i.e. asymmetric. The mea ad variace of V 2 χ 2 1 are EV 2 = 1, V arv 2 = 2 1. Thik of V 2 as a scaled versio of S 2. Table B is i terms of V 2. Hece cautio: always scale to V 2 before usig Table B. Sice we have ES 2 = σ 2, σ2 S 2 = 1 V 2 V ars 2 = 2σ4 1. Oe reaso to use 1 i S 2 is to esure ES 2 = σ 2. As icreases, S 2 estimates σ 2 with more precisio. Let Y 1, Y 2,...,Y 7 be a radom sample from N6, 3 2 i.e. = 7. What is PS 2 > 21? 1S PS > 21 = P > σ 2 σ 2 = PV 2 > = PV 2 > 14 where V 2 χ 2 6. From Table B, we have So ad I R, PV 2 > = 0.05, PV 2 > = > pchisqq=14, df=6, lower.tail=f > [1] < PV 2 > 14 < < PS 2 > 21 < 0.05 I the example, if Y i N200, 3 2, the PS 2 > 21 is the same, because it does ot deped o µ. Agai cautio: rescale S 2 to V 2 before usig Table B. More o the d.f. for sample variace S 2 = 1 Y i 1 Ȳ 2 Cosider = 3 ad all the deviatios. Oce we kow that it follows that Y 1 Ȳ = 3, Y2 Ȳ = 1, Y 3 Ȳ = 2. Hece the complete iformatio about deviatios is cotaied i 1 of them. I geeral, d.f. represets the amout or pieces of iformatio available about the spread. We use Ȳ,S2 istead of media ad IQR, because they are easier to deal with