Answers to some even exercises

Transcription

1 Answers to some even eercises Problem - P (X = ) = P (white ball chosen) = /8 and P (X = ) = P (red ball chosen) = 7/8 E(X) = (P (X = ) + P (X = ) = /8 + 7/8 = /8 = /9 E(X ) = ( ) (P (X = ) + P (X = ) = /8 + 7/8 = so Var(X) = σ = E(X ) E(X) = ( /9) = 77/8 Problem - X is Binomially distributed with parameters n = and p = Put q = p = 6 (a) P (X 5) = P (X = ) + P (X = ) + + P (X = 5) = ( ) p o q + = (6) + ()(6) () 5 (6) 7 =??? ( ) pq + + ( ) p 5 q 7 5 too much calculator effort for me, so I m using table II: P (X 5) = 665; that s better, much easier (b) P (X 6) = P (X 5) = 665 = 8 by part (a) (c) P (X = 7) = P (X 7) P (X 6) = = 9 from Table II (d) µ = E(X) = np = () = 8 and Var(X) = npq = 8(6) = 88 Problem -6 X is Binomial distributed with parameters n = 7 and p = 5 (a) X is Binomial distributed with parameters n = 7 and p = 5 (b) (i) P (X ) = P (X ) = 766 = 8 from table II ( ) 7 (ii) Put q = p = 85 P (X = ) = pq 6 = 7(5)(, 85) 6 = 96 or, if you prefer to use table II: P (X = ) = P (X ) P (X = ) = = 96 (iii) P (X ) = 9879 Problem 5- (a) X is binomial with parameters n = 5 and p = 7 (cf, eample( 5-) ) µ = np = ( 5(7) ) = 5, σ = np( p) = 5 5 5(7)() = 5 and P ( X ) = P (X = ) + P (X = ) = (7)() + (7) () = 7 (b) X is geometric with parameter p = (cf, eample 5- and its prelude) µ = /p = /, σ = ( p)/p = 7/ =, and P ( X ) = P (X = ) + P (X = ) = p(q + q ) = (7 + 7 ) = 5 (c) X is binomial with parameters n = and p = 55, that is, X indicates the outcome of a Benouli trial with probability p of success µ = p = 55, σ = pq = (55)(5) = 5 and P ( X ) = P (X = ) = p = 55 (d) X is a discrete random variable assuming values,,, and with probabilites,,, and, respectively µ = () + () + () + () =, σ = ( ) () + ( ) () + ( ) () + ( ) () = 89 and P ( X ) = P (X = ) + P (X = ) = + = 7 (e) X has the negative binomial distribution with paramters ( p ) = 6 and r = (cf, 5-) µ = r/p = /, σ = rq/p = ()/(6) =, and P ( X ) = P (X = ) = pq = (6)() =

2 Problem 5- The probability that a person has their birthday on a given day is p = /65 X counts the number of people you ask until you find someone with the same birthday as yours If X = k then the first k people you ask do not share your birthday Consider each person you ask as a trial, that is, either they share your birthday (success) or they don t (failure) Assuming the trials are independent then for k, P (X = k) = P (k failures then success) = q k p = (6/65) k (/65) In short, X has the geometric distribution with p = /65 (a) The density function for X is P (X = k) = (6/65) k (/65) (b) µ = /p = 65, σ = q/p = (6/65)(65 ) = 6 6 =, 86, and σ = 65 (c) P (X > k) = q k (cf, pg, 9) so P (X > ) = (6/65) = and P (X < ) = P (X 99) = P (X > 99) = q 99 = (6/65) 99 = 56 Problem 5-6 X is geometrically distributed with p = So P (X ) = P (X > 99) = q 99 = (99/) 99 = 7 Problem 6- For Poisson random variables µ = σ = λ So if X is Poisson with variance σ = then λ = and P (X = ) = λ! e λ = e Problem 6- If X has a Poisson distribution with parameter λ and P ( = ) = P (X = ) then λ! e λ = λ! e λ or, since λ, λ = 6 P (X = ) = 6! e 6 = 6! e 6 = 8e 6 9 Problem 6-6 For a 5 foot roll we ect 5 defects If X counts the number of defects in a 5ft roll then X is can be modeled as Poisson with parameter λ = 5 and P (X = ) = e λ = e Problem 6-8 Let X counts the number of those inoculated who suffer side effects from the vaccine Then X is B(n, p) with n = and p = 5 Since n is large and p small then X is approimately Poisson with parameter λ = np = 5 (a) P (X ) from Poisson distribution table Just to remark, if we stick to the Binonimal distribution to compute this probability, then (using a calculutor) P (X ) = P (X = ) + P (X = ) ( ) ( ) = (5) ( 5) + (5) ( 5) 999 = (b) Using the table for Poisson distribution P (X =, 5, or 6) = P (X 6) P (X ) = 97

3 Whereas, using the Binonimal distribution and calculutor: P (X ) = P (X = ) + P (X = 5) + P (X = 6) ( ) ( ) ( ) = (5) ( 5) (5) 5 ( 5) (5) 6 ( 5) = Problem - (a) We need c so that c d = Since c d = c 6 then take c = Set F () = if and F () = if > For < put F () = (Cumulative) distribution function is F if F () = if < 6 if < f(t) dt = t dt = 6 c c c (b) Choose c so that c 6 d = Since c 6 d = 6 d = 8 c then, once again, take c = Set F () = if and F () = if > And for < put F () = f(t) dt = t dt = (Cumulative) distribution function is if F () = + 8 if < 6 if <

4 F c c (c) Need c so that d = Since d = c then take c = / Set F () = if and F () = if > And for < put F () = f(t) dt = t / dt = (Cumulative) distribution function is F if F () = if < if < Since lim =, the pdf is unbounded (at the origin) + Problem - (a) pdf is f() = / for so E(X) = f() d = o d = = 8 5 E(X ) = f() d = 6 d = o = 8 So, µ = E(X) = 8 5 and σ = E(X ) [E(X) = 8 ( ) 8 =

5 (b) pdf is f() = (/6) for so E(X) = E(X ) = f() d = d = d = 6 6 f() d = d = 6 8 d = 5 (c) pdf is f() = /( ) for so E(X) = E(X ) = So, µ = E(X) = and σ = E(X ) = 5 f() d = f() d = d = d = d = So, µ = E(X) = and σ = E(X ) [E(X) = 5 ( / d = 5 ) = 5 Problem -6 (a) M(t) = E(e tx ) = Integrate the last integral by parts, twice, to get M(t) = e t e d = for t < ( t) e (t ) d (b) Note that M() = and M (t) = ( t) and M (t) = ( t) 5 µ = M () = and σ = M () [M () = 9 = 5

6 Problem -8 (a) Need f() d = so (b) So c = E(X) = f() d = = c = c d = ln = = lim ln = Problem - f is the pdf of a uniformly distributed random variable on the interval f F / Mean µ = ( + )/ = and variance σ = ( ( )) / = / = / Problem - M(t) is the moment generating function for a uniformly distributed random variable X on the interval 5 Hence, the pdf of X is f() = for 5 and (a) E(X) = ( + 5)/ =, (b) σ = (5 ) / = /, and (c) P ( < X 7) = 7 d = 7 = 5 6

7 Problem -6 Since θ = then the pdf for X is f() = e / for > (a) P ( < X < ) = (b) P (X > ) = e / d = e / + e / = 8 e / d = e / = (c) Remember the lack of memory property for onentially distributed random variables P (X > and X > ) P (X > ) P (X > X > ) = = P (X > ) P (X > ) = (/)e / d (/)e / d = e e / = e / = P (X > ) = from part (b) (d) For onentially distributed random variables: µ = θ = and σ = θ = (e) I d try this one if I had the data set in a computer file Problem -8 If X is measured in minutes then in ( > ) minutes from am we ect λ = (/) calls In part (a) below I ll go through again the steps we took in class to devlope the pdf for the onential distribution You could go right to the pdf of course (see pages -) (a) P (X ) = P (X > ) = P (no calls in first minutes) = λ! e λ = e λ = e (/) The cumulative distribution function F for X is then F () = P (X ) = e (/) for > and otherwise To get a pdf f for X we differentiate F : for >, f() = F () = (/)e (/) with f() = otherwise (b) P (X > ) = P (X ) = F () = e / = 6 Problem 5- P (X < 5) = 5 Γ() e / d = 6 5 e / d = 6 Problem 5- Moment generating function for the Gamma distribution with parameters α and θ is M(t) = (see page 5 in your tet), for t < λ = /θ ( θt) α This means So M (t) = αθ ( θt) α+ and M α(α + )θ (t) = ( θt) α+ µ = M () = αθ and σ = M () [M () = α(α + )θ (αθ) = αθ 7

8 Problem 5-6 W is Gamma distributed with α = and λ = 7 (a) W has probability density function (pdf) is f(w) = (7) Γ() w99 e 7w for w with mean µ = α/λ = /7 = 68 (seconds) and variance σ = α/λ = /(7) = 6 (b) and s, the sample mean and sample variance (see page 7 of your tet), computed from the given data are = i = (685) = 67 n 5 and s = (i ) = n [ n n n i = 5 [ (, 677) (67) = (c) Note, P (W 66) = f(w) dw = (7) w 99 e 7w dw = a tough integral to compute (integrate by parts Γ() 99 times?not me) So let s use the data to get an idea about the number P (X 66): from the 5 values in the data set I count 9 of them to be no larger than 66 Since 9/5 = 6 then an estimate for P (X 66) is 6 Remark: But it would be nice to see how good the 6 estimate of P (W 66) really is Because the sampled data gave nice estimates ( and s ) for µ and σ we may trust that 6 is good But it would be nice to compare the actual value of P (X 66) We can do this with some help to evaluate the above tough to evaluate integral and, as it turns out, we can get help First, substituting t = 7w, gives 66 w 99 e 7w dw = This last integral is of the form Γ(α, ) = (7)(66) ( ) 99 t e t 7 7 dt = (7) 97 t 99 e t dt 66 w 99 e 7w dw t α e t dt with = 97 and α = It s known as an incomplete Gamma function These integrals have been tabulated so we can conveniently look them up In our case: P (W 66) = (7) Γ() 66 w 99 e 7w dw = Γ() 97 t 99 e t dt = Γ(, 97) 95 Γ() Problem 5-8 W is Gamma distributed with parameters λ = 6 and α = 7 (a) The (cumulative) distribution function F for W is (see equation (5-)): (b) α F (w) = P (W w) = k= (λw) k e λw = k! 6 (6w) k e 6w k! k= 6 (6(5)) k 6 P (W 5) = F (5) = e 6(5) = k! k= 8 8 = ( + +! + 8! + 8! ! ! = (96)e 8 = 69 8 k k! e 8 k= ) e 8 8

9 Problem 6- Practice using N(, ) table, ie, Table Va in your tet (a) P (5 < Z 6) = P (Z 6) P (Z 5) = = 78 (b) P ( 79 < Z 5) = P (Z 5) P (Z 79) = P (Z 5) ( P (Z 79)) = = 69 (c) P (Z > 77) = P (Z < 77) = 966 (d) P (Z > 89) = P (Z 89) = 998 = 9 (e) P ( Z < 96) = P ( 96 < Z < 96) = P (Z < 96) P (Z 96) = P (Z < 96) ( P (Z 96)) = P (Z < 96) = (975) = 95 (f) P ( Z > ) = P (Z > ) = ( P (Z < )) = ( 8) = 7 (g) P ( Z < ) = P ( < Z < ) = P (Z < ) P (Z < ) = P (Z < ) ( P (Z < )) = P (Z < ) = (977) = 95 (h) P ( Z < ) = P ( < Z < ) = P (Z < ) P (Z ) = P (Z < ) ( P (Z < )) = P (Z < ) = (9987) = 997 Problem 6- If Z N(O, ) and α, z α is defined to be the number so that P (Z z α ) = α (see page 6 in tet) Either table Va or Vb may be used here but sometimes Vb is quicker (a) = P (Z z ) implies z 8 because P (Z 8) = (b) 5 = P (Z z 5 ) implies z 5 6 So z 5 6 (c) 85 = P (Z z 85 ) implies z So z (d) Since z α = z α then z 9656 = z 9656 = z = 8 Problem 6-6 X N(66, ) so Z = X 66 N(, ) (a) µ = 66 (b) σ = = ( 7 66 (c) P (7 < X < ) = P < X 66 ) 66 < = P ( < Z < 7) = P (Z < 7) P (Z ) = = 6 ( 8 66 (d) P (8 < X < 7) = P < X 66 ) 7 66 < = P ( 9 < Z < ) = P (Z < ) P (Z 9) = P (Z < ) ( P (Z 9)) = 679 ( 859) = = 8 Problem 6-8 The pdf of X is f() = [ σ π ( µ) σ and has a graph with a classical bell shape symmetric about the vertical line = µ Inspection of this graph shows that there should be two inflection points To find them is straightforward: solve the equation f () = To make the computations easier (but not by much) and because it s in same spirit as normalizing nomal random variables to N(, ), I m going to make the change of variable: z = z() = µ The density function reduces to σ g(z) = ) ( z π 9

10 Note that we have f() = g(z()) so that f () = g (z) dz d = σ g (z) and f () = σ g (z) Thus, g (z) = if and only if f () =, where = σz + µ, and the only thing we need check now is that the condition g (z) = implies z = ± Let s compute g : So g (z) = when π g (z) = z ) ( z π ) ( z = z π and g (z) = π ) ( z + z π ) ( z ) ( z and, since the onential terms cancel (they re never zero), z = ± Problem 6- By now, you re intuition should tell you this is true But let s do it anyway: put Y = ax + b where we first assume a > (what uninteresting thing happens if a =?) then ( P (Y y) = P (ax + b y) = P X y b ) a = (y b)/a σ ( µ) [ π σ d (now substitute t = a + b) ( ) t b = y µ σ π a dt σ a = aσ π y [ (t [b + aµ) a σ dt So P (Y y) = y σ (t µ) [ π σ dt where µ = aµ + b and σ = aσ That is Y N( µ, σ ) What happens in the above computation is a <? Well, the inequality in the last term of the first line reverses: ( P (Y y) = P (ax + b y) = P X y b ) a = σ ( µ) [ π (y b)/a σ d (again substitute t = a + b) ( ) t b = µ σ π a dt σ a y = a σ π y [ (t [b + aµ) a σ dt So, again P (Y y) = Y N(aµ + b, a σ ) σ π y (t µ) [ σ dt where µ = aµ + b and σ = a σ That is, it s still the case that