Random Variables and Probability Distributions

Transcription

1 Radom Varables ad Probablty Dstrbutos * If X : S R s a dscrete radom varable wth rage {x, x, x 3,. } the r = P (X = xr ) = * Let X : S R be a dscrete radom varable wth rage {x, x, x 3,.}.If x r P(X = x r ) exsts, the x r P(X = x r ) s called the mea of the radom varable X. It s deoted by or x. * If ( xr µ ) P ( X = xr ) exsts, the ( x ) r µ P ( X xr ) = s called varace of the radom varable X. It s deoted by σ. The postve square root of the varace s called the stadard devato of the radom varable X. It s deoted by σ *. If the rage of dscrete radom varable X s {x, x,., x,.} ad P(X = x ) = P for every Iteger s gve the σ + µ = x P Bomal Dstrbuto: A radom varable X whch takes values 0,,,, s sad to follow bomal dstrbuto f ts probablty dstrbuto fucto s gve by r r r P( X = r) = C p q, r = 0,,,..., where p, q > 0 such that p + q =. * If the probablty of happeg of a evet oe tral be p, the the probablty of successve happeg of that evet r trals s r p. Mea ad varace of the bomal dstrbuto The mea of ths dstrbuto s X X X p = X. C q p = p, X = X = The varace of the Bomal dstrbuto s σ = pq ad the stadard devato s = (pq ) σ. The Posso Dstrbuto : Let X be a dscrete radom varable whch ca take o the values 0,,,... such that the probablty fucto of X s gve by x λ λ e f( x) = P( X = x) =, x = 0,,,... x! where λ s a gve postve costat. Ths dstrbuto s called the Posso dstrbuto ad a radom varable havg ths dstrbuto s sad to be Posso dstrbuted.

2 Very Short Aswer Questos. A probablty dstrbuto fucto of a dscrete radom varable s zero except at the pots x = 0,,. At these pots t has the value p (0) = 3c 3, p() = 4c 0c, p() = c for some c > 0. Fd the value of c. Sol. P(x = 0) + p(x = ) + p(x = ) = 3c 3 + 4c 0c + c = 3c c = () Put c =, the 3 0+ = = 0 C = satsfy the above equato C = p(x = ) = 4-0= - whch s ot possble. Dvdg () wth c, We get 3c 7c + = 0 (c ) (3c ) = 0 c = or c = /3 c = p(x = 0) = 3. 3 = 4 whch s ot possble c = /3

3 . Fd the costat C, so that varable X. x F(x) = C, x =,,3... s the p.d.f of a dscrete radom 3 Sol. Gve F(x) = C, x =,,3 3 x We kow that x p(x) = C, x =,,3 3 x= p(x) = c = x= 3 x 3 c = C = c 3 = 3 c 3 = c = 3 3. X=x 0 3 P(X=x) 0. k 0. k 0.3 k Is the probablty dstrbuto of a radom varable x. fd the value of K ad the varace of x. Sol. We kow that p ( x ) = = 0. + k k k = 3k + 0. =

4 3k = 0. = 0.4 k = = X=x 0 3 P(X=x) 0. k 0. k 0.3 k X.p(x ) -0. -k 0 k 0. 3k X.p(x ) 0.4 k 0 k. k Mea = x p ( x ) = = 0. k k k = 4k = 4(0.) = = 0.8 µ = 0.8 Varace (σ ) = x p(x = x ) µ = Varace = 4(0.) + (k) + 0(0.) + (k) + 4 (0.3) + k µ = k k + 4 (0.3) + k µ = k (0.8) = (0.) = σ = =.

5 4. X=x P(X=x) Is the probablty dstrbuto of a radom varable x. fd the varace of x. Sol. X=x P(X=x) X.p(x ) X.p(x ) Mea (µ) = x p ( x ) = 3 3 = = ( µ ) = 0 Varace(σ )= x p(x = x ) µ = = = σ =

6 . A radom varable x has the followg probablty dstrbuto. Sol. X=x P(X=x) 0 k k k 3k K k 7k +k Fd ) k ) the mea ad ) p(0< x < ). We kow that p ( x ) = = 0 + k + k + k +3k + K + k + 7k + k = 0k + k = 0k + k = 0 0k + 0k k = 0 0k (k + ) (k + ) = 0 (0k ) (k + ) = 0 K =, 0 Sce k > 0 k = 0 ) k = 0 ) X=x P(X=x) 0 k k k 3k K k 7k +k X.p(x ) 0 k 4k k k k k 4k +7k Mea = x p( x = x ) = = 0 + k + 4k + k + k + k + k + 4k + 7k

7 = k + 30k = = = 3. ) p(0 < x < ) p(0 < x < ) = p(x = ) + p(x = ) + p(x = 3) + p(x = 4) = k+ k + k + 3k = 8k = 4 8 = 8 = 0 0. I the expermet of tossg a co tmes, f the varable x deotes the umber of heads ad P(X = 4), P(X = ), P(X = ) are arthmetc progresso the fd. Sol. X follows bomal dstrbuto wth Prob. Of gettg head s p = q = Gve, P(X = 4), P(X = ), P(X = ) are A.P a C 4 4 4, C, C are A.P C 4, C, C are A.P 4 C = C + C (! )!! ( ) ( ) ( ) = +!! 4! 4!!!

8 (! )! +! ( )( ) ( )( )( ) ( ) = 4! (! ) ( )( ) 4!!! = + 4! ( )! ( 4)( ) 30 ( ) ( )( ) ( )( ) = ( 4) = [ ] 48 = = = 0 ( 4) 7( 4) = 0 ( 7) ( 4) = 0 = 7 or 4 4! 4! 4!! 7. Fd the maxmum umber of tmes a far co must be tossed so that the probablty of gettg at least oe head s at least 0.8. Sol. Let be umber of tmes a far co tossed x deotes the umber of heads gettg x follows bomal dstrbuto wth parameters ad p =/ gve p(x ) 0.8 p(x = 0) 0.8 p(x = 0) 0. C0 0. The maxmum value of s 3.

9 8. The probablty of a bomb httg a brdge s / ad three drect hts (ot ecessarly cosecutve) are eeded to destroy t. Fd the mmum umber of bombs requred so that the probablty of the brdge beg destroyed s greater tha 0.. Sol. Let be the mmum umber of bombs requred ad x be the umber of bombs that ht the brdge, the x follows bomal dstrbuto wth parameters ad p = /. Now p(x 3) > 0. p(x < 3 ) > 0. p(x < 3 ) < 0. p(x = 0) + p(x = ) + p(x = ) < C C + + C < 0. ( ). + + < < < < 0 ( + + ) < By tral ad error, we get The least value of s =

10 . If the dfferece betwee the mea ad the varace of a bomal varate s / the, fd the probablty for the evet of successes, whe the expermet s coducted tmes. Sol. Gve =, let p be the parameters of the bomal dstrbuto Mea Varace =/ p pq = / p( q) = /, p + q =.p = / p = / p = / p = /3 q = p = /3 = /3 p(x = ) = 3 C = 0. = 7 43 Prob. of the evet of success = Oe shps s lkely to be wrecked, whe they are set o sal, whe shps are o sal, fd the probablty for (a) At least oe wll arrve safely (b) Exactly, 3 wll arrve safely. Sol. P = probablty of shp to be wrecked = / q = p = = 8 Number of shps = = 8 p(x = 0) = C0 = a) Probablty of at least oe wll arrve safely = p(x > 0) = p(x = 0) = =

11 b) p(x = 3) = C C = =. If the mea ad varace of a bomal varable x are.4 ad.44 respectvely, fd p( < x 4). Sol. Mea = p =.4 Varace = pq =.44 () () Dvdg () by () pq.44 = p.4 q = 0. = 3/ p = q = 0. = 0.4 = / Substtutg () (0.4) =.4 =.4 = 0.4 P(<x 4)= p(x = ) + p(x = 3) + p(x = 4) = C q.p + C3q.p + C4q.p C C3 C4 = + + = ( ) 3 = ( ) = = 3 3 = 3 3

12 . It s gve that 0% of the electrc bulbs maufactured by a compay are defectve. I a sample of 0 bulbs, fd the probablty that more tha are defectve. Sol. p = probablty of defectve bulb = /0 q = p = = 0 0 = umber of bulbs 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 0 k k = C k 0 0 k 0 0 = Ck = C 0 k 0 k= 0 0 k= 3 0

13 3. O a average, ra falls o days every 30 days, fd the probablty that, ra wll fall o just 3 days of a gve week. Sol. Gve p= = q = p = = 3 30 = 7, r = 3 p(x = 3) 4 3 r r 7 3 = Cr. q.p = C = 3. = 7 4. For a bomal dstrbuto wth mea ad varace, fd the frst two terms of the dstrbuto. Sol. Let, p be the parameters of a bomal dstrbuto Mea (p) = Ad varace ( pq) = () () pq the = p From () p = q = p = q = = = 3 8 = = Frst two terms of the dstrbuto are p( x = 0) = C0 = ad 3 3 p( x = ) = C =

14 . I a cty 0 accdets take place a spa of 0 days. Assumg that the umber of accdets follows the posso dstrbuto, fd the probablty that there wll be 3 or more accdets a day. Sol. Average umber of accdets per day 0 λ = = = 0. 0 The prob. That there w be 3 or more accdets a day p(x 3) k= 3 k e λ λ, λ = 0. k! Short Aswer Questos & Log Aswer Questos. The rage of a radom varable x s {0,, }. Gve that p(x = 0) = 3c 3, p(x = ) = 4c 0c, p(x = ) = c ) Fd the value of c ) p(x < ), p( < x 3) Sol. P(x = 0) + p(x = ) + p(x = ) = 3c 3 + 4c 0c + c = 3c 3 0c + c = 0 C = satsfy ths equato C = p(x = 0) = 3 whch s ot possble dvdg wth c, we get 3c 7c + = 0 (c ) (3c ) = 0 d = or c = /3 c = p(x = 0) = 3. 3 = 4 whch s ot possble

15 c = /3 ) p(x < ) = p(x =0) = 3.c = = 3. = 7 ) p( < x ) = p(x = ) = c = = 3 3 ) p(0 < x 3) = p(x = ) + p(x = ) = 4c 0c + c = c 0c = = 3 = =. The rag of a radom varable x s {,, 3 } ad p(x = k) = (k =,, 3,..) Fd the value of C ad p(0 < x < 3) k c k! Sol. Sum of the probabltes = p(x = k) = k= k c k = 3 c c c = 3 Addg o both sdes 3 c c + c = 3 e c = log e e c = log e

16 c = log e P(0 < x < 3) = p(x = ) = p(x = ) ( log ) c e = c + = loge + 3. Fve cos are tossed 30 tmes. Fd the frequeces of the dstrbuto of umber of heads ad tabulate the result. Sol. cos are tossed 30 tmes Prob. of gettg a head o a co p =, = Prob. of havg x heads x = = x ( ) ( ) p x x C q x x Cx = = Cx x = 0,,,3,4, Frequeces of the dstrbuto of umber of heads = N.P(X = x) = 30 C x ; x = 0,,,3, 4, Frequecy of Havg 0 head = 30 C0 = 0 Havg head = 30 C = 0

17 Havg head = 30 C = 00 Havg 3 head = 30 C3 = 00 Havg 4 head = 30 C4 = 0 Havg head = 30 C = 0 N(H) f Fd the probablty of guessg at least out of 0 of aswers () True or false type examato ) Multple choce wth 4 possble aswers. Sol. ) Sce the aswers are true or false type. Prob. of success p =, q = Prob. of guessg at least out of p( x ) = C 0 0 k = C 0 ) Sce the aswers are multple choce wth 4 possble aswers Prob. of success p = /4, q = 3/4 Prob. of guessg at least out of k 0 k k p(x ) = C = C

18 . The umber of persos jog a cema tcket couter a mute has posso dstrbuto wth parameter. Fd the probablty that ) o oe jos the queue a partcular mute ) two or more persos jo the queue a mute. Sol. Here λ = ) prob. That o oe jos the queue a partcular mute p(x = 0 ) = λ 0 e λ = e 0! ) prob. that two or more persos jo the queue a mute p(x ) = p(x ) = [p(x = 0) + p(x = )] = e λ 0 λ λ e λ + 0!! ( ) e = e +! = 7.e. A cubcal de s throw. Fd the mea ad varace of x, gvg the umber o the face that shows up. Sol. Let S be the sample space ad x be the radom varable assocated wth S, where p(x) s gve by the followg table X=x 3 4 P(X=x ) X.p(x ) / / 3/ 4/ / / X.p(x ) / 4/ / / / 3/ Mea of x = µ = Σ (X = x ) P(X = x )

19 = = ( ) ( )( + ) 7 = = = 3. Varace of x = σ = x p(x = x ) µ ( 3 4 ) 4 = ( )( + )( + ) 4 = 4 4 = = = 7. The probablty dstrbuto of a radom varable x s gve below. Fd the value of k, ad the mea ad varace of x X=x 3 4 P(X=x ) K k 3k 4k k Sol. we have ( ) r= p X = x = k + k + 3k + 4k + k = k = X=x 3 4 P(X=x ) K k 3k 4k k X.p(x ) k 4k k k k X.p(x ) k 8k 7k 4k k

20 Mea µ of x = r.p ( x = x) = r ( rk) r r =.(k) +.(K) + 3.(3k) + 4.(4k) +.(k) = k = = 3 Varace (σ ) = = x p(x = x ) µ = k + 8k + 7k + 4K +k 3 = k = 3 4 = = 8. If X s a radom varable wth the probablty dstrbuto. P(X = k) = (k = 0,,, 3, ) the fd C. ( k + ) c, k Sol. gve p(x = k) = ( k + ) 0 k= 0 0 k= 0 ( ) p x = k = ( ) c (k = o,,, 3, ) k k + c = c kα = k. c + =

21 Ht: A.G.P. S a dr + r r ( ) Here a =, d =, r = / + = c c [ + ] = c = 4. Let x be a radom varable such that p(x = ) = p(x = ) = p(x = ) = p(x =3) =/ ad p(x = 0) = /3. Fd the mea ad varace of x. Sol. Mea = ( ) + ( ) + + ( ) = µ = 0 Varace (σ ) = ( ) + ( ) = = 0 3

22 0. Two dce are rolled at radom. Fd the probablty dstrbuto of the sum of the umbers o them. Fd the mea of the radom varable. Sol. Whe two dce are rolled, the sample space S cotas = 3 sample pots. S = {(, ), (, ) (, ), (, ), (, ) (,)} Let x deote the sum of the umbers o the two dce The the rage x = {, 3, 4, } Probablty Dstrbuto of x s gve by the followg table. X=x P(X=x ) X.p(x ) /3 /3 /3 0/3 30/3 4/3 40/3 3/3 30/3 /3 /3 Mea of x = µ = x p( X = x ) = = ( ) 3 = = cos are tossed smultaeously. Fd the probablty of gettg at least heads. Sol. p = probablty of gettg head = q = p = = : = 8 p(x ) = p(x = ) + p(x = 7) + p(x = 8)

23 7 8 8 C C7 = C7 + 8 = C C7 C8 37 = [ ] =. The mea ad varace of a bomal dstrbuto are 4 ad 3 respectvely. Fx the dstrbuto ad fd p(x ) Sol. Gve dstrbuto s bomal dstrbuto wth mea = p = 4 Varace = pq = 3 pq 3 = p 4 3 q = 4 So that p = q p = = = 3 = 4 P(x ) = p(x = 0) = C0 4 4 = 3 = 4 p(x ) 3 = 4

24 3. The probablty that a perso chose at Radom s left haded ( had wrtg) s 0.. What s the probablty that a group of te people there s oe, who s left haded? Sol. Here = 0 P = 0. q = p = 0. = 0. p(x = ) = 0 C (0.) (0.) 0 =0 0. (0.) = (0.) = (0.) 4. I a book of 40 pages, there are 400 typographcal errors. Assumg that followg the passo law, the umber of errors per page, fd the probablty that a radom sample of pages wll cota o typographcal error? Sol. The average umber of errors per page the book s λ = = 40 Here r = 0 λ r λ e p( x = r) = r! 8/ 8 e p( x = 0) = = e 0! 0 8/ The requred probablty that a radom sample of pages wll cota o error s [p(x = 0)] = ( e 8/ )

25 . Defcecy of red cells the blood cells s determed by examg a specme of blood uder a mcroscope. Suppose a small fxed volume cotas o a average 0 red cells for ormal persos. Usg the posso dstrbuto, fd the probablty that a specme of blood take from a ormal perso wll cota less tha red cells. Sol. Here λ = 0 4 ( < ) = ( = ) p x p x r r= 0 4 = r= 0 e λ r λ r! 4 = r= 0 e 0 r 0 r!. A posso varable satsfes p(x = ) = p(x = ). Fd p(x = ) Sol. Gve p(x = ) = p(x = ) r λ e λ p( x = r ) =, λ > 0 r! r λ λ λ e λ e =!! λ =, ( λ > 0) e p( x = ) =! 3 = = 0e 4 e