Random Variables and Probability Distribution Random Variable

Transcription

1 Random Vaiables and Pobability Distibution Random Vaiable Random vaiable: If S is the sample space P(S) is the powe set of the sample space, P is the pobability of the function then (S, P(S), P) is called the pobability space, In the pobability space if X : S R is a function then x is called andom vaiable Fequency function of andom vaiable (o) Pobability density function: Function f : X( S) R is defined by f() = p[ e/ x() e = ] is called the fequency function associated with andom vaiable Whee i) 0 f( ) x( S) ii) Σ f() = x( S) Aithmetic mean of the andom vaiable: Aithmetic mean of the andom vaiable x is denoted by x o all E(x) expected value of X and is defined as x=σ f() Vaiance of the andom vaiable: If x is a andom vaiable then that Ex ( ) =Σ f( ) x xs ( ). The vaiance of andom vaiable F( x x) Vaiance of the andom vaiable Ex ( ) is defined such ( σ ) and is defined as σ = = + Ex ( x) Ex ( x xx) σ = + Ex ( ) x xex ( ) σ = Ex ( ) + x Vaiable of andom vaiable σ = Ex ( ) µ σ =Σ f() µ σ + µ =Σ f() Standad deviation: It is the positive squae oot of the vaiance of the standad deviation of the andom vaiable this is denoted by σ = vaiance Note: Let X be a andom vaiable on a sample space S. If x R then we use the following symbols to denote some events in S. i) { a S: X( a) = x} = ( X = x)

2 ii) { a S: X( a) < x} = ( X < x) iii) { a S: X( a) x} = ( X x) iv) { a S: X( a) > x} = ( X > x) v) { a S: X( a) x} = ( X x) Def : Let S be a sample space and X : S R be a andom vaiable. The function F : R R defined by F( x) = P( X x), is called pobability distibution function of the andom vaiable X. We now state some popeties of pobability distibution function fo the andom yond the scope of the book. Theoem : Let F(x) be the pobability distibution function fo the andom vaiable X. then i) 0 F ( x), x R ii) F (x) is an inceasing function i.e. x, x R, x< x F( x) F( x) iii) Lt F( x) =, Lt F( x) = 0 x x Theoem 3: If X : S R is a discete andom vaiable with ange { x, x, x 3...} then Mean and Vaiance PX ( = x ) =. Def : Let X : S R be a discete andom vaiable with ange { x, x, x 3...}. If Σ x PX ( = x ) exists, then Σ x PX ( = x) is called the mean of the andom vaiable X. It is denoted by µ o x. If ( x µ ) P( X x ) Σ = exists, then Σ( x µ ) P( X = x ) is called vaiance of the andom vaiable X. It is denoted by σ. The positive squae oot of the vaiance is called the standad deviation of the andom vaiable X. It is denoted by σ. Theoem 4: Let X : S R be a discete andom vaiable with ange { x, x, x 3,...}. If mean and vaiance of X then σ + µ =Σ x P( X = x ). = µ, σ ae the Def: Let n be a positive intege and p be a eal numbe such that 0 p. A andom vaiable X with ange {0,,,.n} is said to follows (o have) binomial distibution o Benoulli distibution with n n paametes n and p if PX ( = ) = C pq fo = 0,, n whee q = - p. Theoem : If the andom vaiable X follows a binomial distibution with paametes n and p then mean of X is np and the vaiance is npq whee q = - p. Def : Let λ > 0 be ae eal numbe. A andom vaiable X with ange {0,,.} is said to follows (have) e λ λ Poisson distibution with paamete λ if PX ( = ) = fo = 0,,,.!

3 Theoem : If a andom vaiable X follows Poisson distibution with paamete λ, then mean of X is λ and vaiance of X is λ. EXERCISE 9(a). A p.d.f of a discete andom vaiable is zeo except at the points x = 0,,. At these points it has the value p (0) = 3c 3, p() = 4c 0c, p(0) = c fo some c > 0. Find the value of c. Sol. P(x = 0) + p(x = ) + p(x = ) = 3c 3 + 4c 0c + c = 3c c = 0 Put c =, then = = 0 C = satisfy the above equation C = p(x = 0) = 3 which is not possible dividing with c, we get 3c 7c + = 0 (c ) (3c ) = 0 c = o c = /3 c = p(x = 0) = 3. 3 = 4 which is not possible c = /3 x. Find the constant C, so that F(x) = C, x =,,3... is the p.d.f of a discete andom 3 vaiable X. x Sol. Given F(x) = C, x =,,3 3 We know that p(x) = C,x =,,3 3 p(x) = x= x c = x= 3 3 c = C = c 3 = 3 a a + a + a +... =,if < c 3= c= 3 x

4 3. X=x 0 3 P(X=x) 0. k 0. k 0.3 k is the pobability distibution of a andom vaiable x. find the value of K and the vaiance of x. Sol. Sum of the pobabilities = 0. + k k k = 4k = 4k = 0.6 = k = = Mean = ( ) (0.) + ( ) k + 0 (0.) + (k) + (0.3) + 3k = 0. k k k = 4k = 4(0.) = = 0.8 µ = 0.8 Vaiance (σ ) = 0 xp(x= x) µ Vaiance = 4(0.) + (k) + 0(0.) + (k) + 4 (0.3) + 9k µ = k k + 4 (0.3) + 9k µ = k (0.8) = (0.) = σ = =.6 X=x P(X=x) is the pobability distibution of a andom vaiable x. find the vaiance of x. Sol Mean (µ) = = = ( µ ) = Vaiance (σ ) = ( ) ( ) ( ) ( 0) + () + ( ) + ( 3) µ = = σ = 9. A andom vaiable x has the following pobability distibution. X=x P(X=x) 0 k k k 3k K k 7k +k Find i) k ii) the mean and iii) p(0 < x < ). Sol. Sum of the pobabilities = 0 + k + k + k + 3k + K + k + 7k + k =

5 0k + 9k = 0k + 9k = 0 0k + 0k k = 0 0k (k + ) (k + ) = 0 (0k ) (k + ) = 0 K =, 0 Since k > 0 k = 0 i) k = 0 n ii) Mean = xp( x= x) i= i i Mean ( µ ) = 0( 0) + ( k) + ( k) + 3( k) + 4( 3k) + (k ) + 7(7k + k) = 0 + k + 4k + 6k + k + k + k + 49k + 7k = 66k + 30k = = = 3.66 iii) p(0 < x < ) p(0 < x < ) = (x = ) + p(x = ) + p(x = 3) + p(x = 4) = k+ k + k + 3k = 8k 4 = 8 = 8 = 0 0 II.. The ange of a andom vaiable x is {0,, }. Given that p(x = 0) = 3c 3, p(x = ) = 4c 0c, p(x = ) = c i) Find the value of c ii) p(x < ), p( < x 3) Sol. P(x = 0) + p(x = ) + p(x = ) = 3c 3 + 4c 0c + c = 3c 3 0c + 9c = 0 C = satisfy this equation C = p(x = 0) = 3 which is not possible dividing with c, we get 3c 7c + = 0 (c ) (3c ) = 0 d = o c = /3 c = p(x = 0) = 3. 3 = 4 which is not possible c = /3 i) P(x < ) = p(x =0) 3 3 = 3.c = = 3. = ii) p( < x ) = p(x = ) = c = = 3 3

6 iii) p(0 < x 3) = p(x = ) + p(x = ) = 4c 0c + c = 9c 0c = = 3 = = The ange of a andom vaiable x is {,, 3,..} and p(x = k) = Find the value of C and p(0 < x < 3) Sol. Sum of the pobabilities = p(x = k) = n k c = k= k 3 c c c = 3 Add on both sides 3 c c + c = 3 e c = log e e c = log e c = log e P(0 < x < 3) = p(x = ) = p(x = ) c ( ) loge = c+ = loge + EXERCISE 9(b) k c (k =,, 3,..) k. In the expeiment of tossing a coin n times, if the vaiable x denotes the numbe of heads and P(X = 4), P(X = ), P(X = 6) ae in aithmetic pogession then find n. Sol. X follows binomial distibution with p= q = ( a coin is tossed ), Hint : a, b, c ae in A.P, b = a + c (o) b a = c a Given, P(X = 4), P(X = ), P(X = 6) ae in A.P 4 n 4 n n n ac 4, C, 6 n 6 n C6 ae in A.P n n n C, 4 C, C6 ae in A.P n n n C = C4 + C6 ( n! ) n! n! = +! ( n )! 4! ( n 4 )! 6! ( n 6 )! ( n! ) = 4! n n 6 ( )( )

7 n! + n! 4! ( n 4)( n )( n 6 )! 6 4! ( n 6 )! ( n! ) 4! n n 6! ( )( ) n! = + 4! n 6! n 4 n 30 ( ) ( )( ) 30+ ( n 4)( n ) ( ) ( )( ) = n 30n 4 n 30 (n 4) = [30 + n 9n + 0] n 48 = n 9n 0 n n + 98 = 0 n 4n 7n + 98 = 0 n(n 4) 7(n 4) = 0 (n 7) (n 4) = 0 n = 7 o 4. Find the maximum numbe of times a fai coin must be tossed so that the pobability of getting at least one head is at least 0.8. Sol. Let n be numbe of times a fai coin tossed x denotes the numbe of heads getting x follows binomial distibution with paametes n and p =/ given p(x ) 0.8 p(x = 0) 0.8 p(x = 0) 0. n n C0 0. n The maximum value of n is The pobability of a bomb hitting a bidge is / and thee diect hits (not necessaily consecutive) ae needed to destoy it. Find the minimum numbe of bombs equied so that the pobability of the bidge being destoyed is geate than 0.9. Sol. Let n be the minimum numbe of bombs equied and x be the numbe of bombs that hit the bidge, then x follows binomial distibution with paametes n and p = /. Now p(x 3) > 0.9 p(x < 3 ) > 0.9 p(x < 3 ) < 0. p(x = 0) + p(x = ) + p(x = ) < 0. 0 n n n n n n C0 + C + C < 0. n n(n ). + + < n n 0 n n n. + + < n n n. 0

8 n n n n + + < 0 + n+ n n n < 0 (n + n + ) < n By tial and eo, we get n 9 The least value of n is 9 n = 9 4. If the diffeence between the mean and the vaiance of a binomial vaiate is /9 then, find the pobability fo the event of successes, when the expeiment is conducted times. Sol. Given n =, let p be the paametes of the binomial distibution Mean Vaiance =/9 np npq = /9 np( q) = /9, p + q = n.p = /9 p = /9 p = /9 p = /3 q = p = /3 = /3 3 C p(x = ) = = 0. = Pob. of the event of success = One in 9 ships is likely to be wecked, when they ae set on sail, when 6 ships ae on sail, find the pobability fo (a) Atleast one will aive safely (b) Exactly, 3 will aive safely. Sol. P = pobability of ship to be wecked = /9 q = p = = Numbe of ships = n = p(x = 0) = C0 = a) Pobability of atleast one will aive safely = p(x > 0) = p(x = 0) 6 = = b) p(x = 3) = C =. C =

9 6. If the mean and vaiance of a binomial vaiable x ae.4 and.44 espectively, find p( < x 4). Sol. Mean = np =.4 () Vaiance = npq =.44 () Dividing () by () npq.44 = np.4 q = 0.6 = 3/ p = q = 0.6 = 0.4 = / Substituting in () n(0.4) =.4.4 n = = P( < x 4) = p(x = ) + p(x = 3) + p(x = 4) = 6 C q 4.p + 6 C q 3.p C q.p C 3 3 C C4 = = 6 ( ) = ( ) = = = It is given that 0% of the electic bulbs manufactued by a company ae defective. In a sample of 0 bulbs, find the pobability that moe than ae defective. Sol. p = pobability of defective bulb = /0 q = p = = n = numbe of bulbs in the sample = 0 p(x > ) = p(x ) = [p(x = 0) + p(x = ) + p(x = )] p( x = 0) = C0 = p( x = ) = C = = p( x = ) = C = = p( x > ) = k k 0 9 = Ck 0 0 k= 0 8 8

10 0 k 0 0 k = Ck = C 0 k 0 k= 0 0 k= On an aveage, ain falls on days in evey 30 days, find the pobability that, ain will fall on just 3 days of a given week. Sol. Given p = = 30 q = p = = 3 n = 7, = 3 p(x = 3) 4 3 n n 7 3 = Cq..p = C = 3. = 7 9. Fo a binomial distibution with mean 6 and vaiance, find the fist two tems of the distibution. Sol. Let n, p be the paametes of a binomial distibution Mean (np) = 6 () and vaiance (n pq) = () npq then = q = np 6 3 p= q = = 3 3 Fom () n p = 6 8 n = 6 n = = 9 3 Fist two tems of the distibution ae 9 9 p( x = 0) = C0 = and p( x = ) = C = In a city 0 accidents take place in a span of 0 days. Assuming that the numbe of accidents follows the poisson distibution, find the pobability that thee will be 3 o moe accidents in a day. Sol. Aveage numbe of accidents pe day 0 λ= = = 0. 0 The pob. That thee win be 3 o moe accidents in a day p(x 3) k e λ λ, λ= 0. k! k= 3

11 II.. Five coins ae tossed 30 times. Find the fequencies of the distibution of numbe of heads and tabulate the esult. Sol. coins ae tossed 30 times Pob. of getting a head on a coin p =, n = Pob. of having x heads x p( x = x) = Cx ( q) x x = Cx = Cx x = 0,,,3, 4, Fequencies of the distibution of numbe of heads = N.P(X = x) = 30 C x ; x = 0,,,3,4, Fequency of Having 0 head = 30 C0 = 0 Having head = 30 C = 0 Having head = 30 C = 00 Having 3 head = 30 C3 = 00 Having 4 head = 30 C4 = 0 Having head = 30 C = 0 N(H) f Find the pobability of guessing at least 6 out of 0 of answes in (i) tue o false type examination ii) multiple choice with 4 possible answes. Sol. i) Since the answes ae in tue o false type. Pob. of success p =, q = Pob. of guessing at least 6 out of ( ) 0 p x 6 = C = Ck 6

12 ii) Since the answes ae in multiple choice with 4 possible answes Pob. of success p = /4, q = 3/4 Pob. of guessing at least 6 out of p(x 6) = C k 0 k 0 3 = Ck The numbe of pesons joining a cinema ticket counte in a minute has poisson distibution with paamete 6. Find the pobability that i) no one joins the queue in a paticula minute ii) two o moe pesons join the queue in a minute. Sol. Hee λ = 6 i) pob. That no one joins the queune in a paticula minute λ 0 e λ 6 p(x = 0 ) = = e 0! ii) pob. that two o moe pesons join the queue in a minute p(x ) = p(x ) = [p(x = 0) + p(x = )] 0 λ λ λ e λ = e + 0!! 6 6 e ( 6) = e + = 7.e 6! EXAMPLE PROBLEMS. A cubical die is thown. Find the mean and vaiance of x, giving the numbe on the face that shows up. Sol. Let S be the sample space and x be the andom vaiable associated with S, whee p(x) is given by the following table X=x i P(X=x i ) Mean of x = µ =Σ (X= x i) P(X= x i) = = ( ) 6 ( 6)( 6+ ) 7 = = = 3. 6 Vaiance of x = σ = x i p(x = x i ) µ 7 ( i.e., ) σ = ( 49 = ) 6 4 ( 6)( 6+ )( 6+ ) 49 = 6 6 4

13 9 49 = = =. The pobability distibution of a andom vaiable x is given below. Find the value of k, and the mean and vaiance of x X=x i 3 4 P(X=x i ) K k 3k 4k k Sol. we have ( ) = p X= x = k+ k+ 3k+ 4k+ k = k = Mean µ of x =.p ( x = x) = ( k) =.(k) +.(K) + 3.(3k) + 4.(4k) +.(k) = k = = 3 Vaiance (σ ) = (). k + () k + (3) 3k + (4) 4k + () (k) µ = k + 8k + 7k + 64K +k 3 = k = = = If X is a andom vaiable with the pobability distibution. P(X = k) = ( k+ ) (k = 0,,, 3, ) then find C. k+ c (k = o,,, 3, ) k Sol. given p(x = k) = ( ) 0 k= 0 ( ) p x = k = ( ) 0 k+ c = c kα = k k= 0. c + = a d Hint: in A.G.P. S = + ( ) Hee a =, d =, = / c, k

14 c + = c [ + ] = c = 4 4. Let x be a andom vaiable such that p(x = ) = p(x = ) = p(x = ) = p(x =3) =/6 and p(x = 0) = /3. Find the mean and vaiance of x. Sol. Mean = ( ) + ( ) + + () = µ = 0 Vaiance (σ ) = ( ) ( ) = = Two dice ae olled at andom. Find the pobability distibution of the sum of the numbes on them. Find the mean of the andom vaiable. Sol. When two dice ae olled, the sample space S contains 6 6 = 36 sample points. S = {(, ), (, ) (, 6), (, ), (, ) (6,6)} Let x denote the sum of the numbes on the two dice Then the ange x = {, 3, 4, } Pobability Distibution of x is given by the following table. X=x i P(X=x i ) Mean of x = µ= xp( X= x) = = ( ) = = 7 36

15 6. 8 coins ae tossed simultaneously. Find the pobability of getting at least 6 heads. Sol. p = pobability of getting head = q = p = = :n= 8 p(x 6) = p(x = 6) + p(x = 7) + p(x = 8) C 6 C = C = C6 + C7 + C 8 37 = [8+ 8+ ] = The mean and vaiance of a binomial distibution ae 4 and 3 espectively. Fix the distibution and find p(x ) Sol. Given distibution is binomial distibution with mean = np = 4 Vaiance = npq = 3 npq 3 3 = q = np So that p = q = = 4 4 np = 4 n 4 4 = n = 6 P(x ) = p(x = 0) = C0 = p(x ) 6 3 = 4 8. The pobability that a peson chosen at Random is left handed (in hand witing) is 0.. What is the pobability that in a goup of ten people thee is one, who is left handed? Sol. hee n = 0 P = 0. q = p = 0. = 0.9 p(x = ) = 0 C (0.) (0.9) 0 =0 0. (0.9) 9 = (0.9) 9 = (0.9) 9 9. In a book of 40 pages, thee ae 400 typogaphical eos. Assuming that following the passion law, the numbe of eos pe page, find the pobability that a andom sample of pages will contain no typogaphical eo. Sol. The aveage numbe of eos pe page in the book is λ= = 40 9 Hee = 0

16 λ λ e p( x = ) =! 0 8/9 8 e 9 p( x = 0) = = e 0! 8/9 The equied pobability that a andom sample of pages will contain no eo is [p(x = 0)] = ( ) e 8/9 0. Deficiency of ed cells in the blood cells is detemined by examining a specimen of blood unde a micoscope. Suppose a small fixed volume contains on an aveage 0 ed cells fo nomal pesons. Using the poisson distibution, find the pobability that a specimen of blood taken fom a nomal peson will contain less than ed cells. Sol. Hee λ = 0 4 ( < ) = ( = ) p x p x 4 λ = 0 4 e λ 0 0 = = e = 0! = 0!. A poisson vaiable satisfies p(x = ) = p(x = ). Find p(x = ) Sol. Given p(x = ) = p(x = ) ( ) λ λ e p x = =, λ> 0! λ λ λ e λ e =!! λ =, ( λ > 0) e p( x = ) =! 3 4 = = 0e e