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# Probability and Queueing Theory(question with answer)

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Probability and Queueing Theory(question with answer)And,and,with

Subject Name: Probability and Queueing Theory Subject Code: MA2262 Question Bank 2012 - 2013

UNIT I : RANDOM VARIABLES

PART - A

1. 2If the r.v has the mgf , determine the variance of X. Mt；？x2t12322tttt????????Mt11，，，，，？？？？；？x????????t2t2222??????????21??2??

23????1t1t3t?? ，？？？？1??????21!22!43!??????

r??t11???，，；，，，，coefficientofinM(t),,??rX12r!22??

1112??；？Var(X).，，，，，，，21244

(r1)!2. Find the mgf of the r.v whose moments are ?r，，.r

2

rr012????????????tt(r1)!1!t2!t3!t?M(t)，，，，？？？????????????Xrr012r!r!0!1!2!2222r0r0，，?????????? 232tttt4????????，？？？？，，，12341.????????22222????????2t；？

2x3. xeth，?f(x),x0A continuous r.v X has the pdf . Find the r moment of X about the 2

origin.

???2x??xe1rrrxr2，？???EXxf(x)dxxdxexdx，，，，，?????r??22 ??00，?

；？；？11r3r2!，；？，？

14. 22Let such that t ? 1, be the mgf of r.v X. Find the mgf of Mt；？x1t

Y = 2X +1.

t1e??tt MtMteM2te.，，，，；？；？；？Y2X1X??，，1t12t???t2t

Find the mgf of X whose pdf is given by f(x) = 1 - |x| , - 1 < X < 1. 5.

?01ttee2？，txtxtx Mtef(x)dxe(1x)dxe(1x)dx.，，？？，，；？x???2t，?，10

The mean of a Binomial distribution is 20 and standard deviation is 4. Find the parameters of 6.

the distribution.

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Subject Name: Probability and Queueing Theory Subject Code: MA2262 Question Bank 2012 - 2013

4npq4npq16(20)q16q，?，?，?， np = 20 and 5

411p1q1.np20n100，，，，，?，?， 100 and are the parameters. 555

One percent of jobs arriving at a computer system need to wait until weekends for scheduling, 7.

owing to core-size limitations. Find the probability that among a sample of 200 jobs there are

no jobs that have to wait until weekends.

p = 0.01, n = 200, ? = np = 2, X is the no. of jobs that have to wait

，?，，x2x20ee(2)e(2)?2P(Xx)P(X0)e0.1353.，，，?，，，， x!x!0!

The number of monthly breakdown of a computer is a r.v having a Poisson distribution with 8

mean equal to 1.8. Find the probability that this computer will function for a month with

only one breakdown.

Mean = ? = np = 1.8

，?，，x1.8x1.81ee(1.8)e(1.8)? P(Xx)P(X1)0.2975.，，，?，，，x!x!1!

Let one copy of a magazine out of 10 copies bears a special prize following Geometric 9.

distribution, Determine its mean and variance.

Sol:

191q pq?Mean10,Variance9021010pp

If the probability is 0.05 that a certain kind measuring device will show excessive drift, what 10.

is the probability that the sixth of these measuring devices tested will be the first to show

excessive drift?

x-15p = 0.05, q = 0.95, x =6, P(X=x) = p q = (0.05) (0.95) = 0.0387.

11. 4If X is a Uniformly distributed r.v with mean 1 and variance , find P(X<0). 3

2baab；？4，?？，1ab2Mean = and variance = ，?，，ba42123

By solving the above eqns. We get a = -1 and b = 3

11 f(x)inaxb,3x3ba6001110 . ！，，，，P(X0)f(x)dxdxx：~??1444，，11

If a r.v „X‟ is uniformly distributed over (-3,3), then compute P ( | X 2 | < 2). 12.

11 f(x)inaxb,3x3ba6

P ( | X 2 | < 2) = P ( -2 < X 2 < 2) = P ( 0 < X < 4)

331113f(x)dxdxx.，，， = ：~??066200

The time required to repair a machine is exponentially distributed with parameter 13.

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Subject Name: Probability and Queueing Theory Subject Code: MA2262 Question Bank 2012 - 2013

1? =. What is the conditional probability that a repair takes at 11h given that its direction 2

exceeds 8h?

xx????22e1e1.5??f(x)dxdxe0.2231 P( X ? 11 / X > 8) = P(X > 3) = ????221/233????Suppose the length of life of an appliance has an exponential distribution with mean 10 years. 14.

What is the probability that the average life time of a random sample of the appliances is

atleast 10.5?

x11，?x10?，，?(?，(,f(x)e,x0f(x)e,x01010 x??11.0510(，，，，P(X10.5)f(x)dxedxe0.3499??1010.510.5

Write down the moment generating function of Gamma distribution. 15.

1 ，！Mt,t1；？x?1t；？

If a r.v X follows Gamma distribution with variance 3, find P( | X| < 1). 16.

? = variance = Var(X) = 3

，?，x1x2x2??exex?exx0?x0?x0????，，f(x)(3)；?() 2???

???0otherwise0otherwise0otherwise???

1

P ( |X| < 1) = P ( -1< X < 1) = f(x)dx?10111x2，？5e2ex，？，？，，f(x)dxf(x)dx0dx0.0803. ???22100

Write the Physical conditions of Binomial distribution. 17.

(a) There are two possible outcomes (b) Probability of success is constant for each and every

trial

(c ) No. of Trials are independent and finite.

Find the expected value of the number of times one must throw a die until the outcome one 18

has occurred 4 times.

1r4,p are the parameters of Negative binomial distribution 6

r4 E(x)Mean241p；？6

If X is a random variable with cdf as F(x), show that the random variable Y = F(x) is 19.

uniformly distributed in (0,1).

Y = F(x), F is a monotonic function. R = {0,1} y

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Subject Name: Probability and Queueing Theory Subject Code: MA2262 Question Bank 2012 - 2013

dyd, as F(x) is cdf of X with the pdf of f(x). (F(x)f(x)dxdx

1,0y1?dx11g(y)f(x)f(x)f(x) ?dy0,otherwisedyf(x)?

dx

State Weibull Distribution. 20.

The continuous random variable X has a Weibull distribution, if its density function is

given by

??1?x???xe,x(0? f(x)where?(0and?(0??0,elsewhere?

What are the applications of Weibull distribution? 21.

(i) It can be applied to reliability problems like time to failure of a component.

(ii) It provides a close approximation to the distribution of the lifetime of a

component.

(iii) The failure rate is not a constant but is proportional to the rate of change of a

time exponent.

PART B

The incidence of an occupational disease in an industry is such that the worker have 25% 1

change of suffering from it. What is the probability that out of 5 workers, at the most two

contract that disease.

X : number of workers contracting the diseases among 5 workers

Then, X is a binomial variate with parameter n=5 and

p = P[a worker contracts the disease] = 25/100 = 0.25

x5-xThe probability mass function (p.m.f) is P(x) = 5C(0.25) (0.75), x=0,1,2,….5 x

The probability that at the most two workers contract the disease is

P[X?2]p(0)p(1)p(2)

0514235C(0.25)(0.75)5C(0.25)(0.75)5C(0.25)(0.75)012

0.23730.39550.2637

0.8965

In a large consignment of electric lamps, 5% are defective. A random sample of 8 lamps is 2

taken for inspection. What is the probability that it has one or more defectives? X: number of defective lamps

Then, X is B(n=8, p=5/100 = 0.05)

x8xThe p.m.f. is p(x) = 8C(0.05)(0.95),x0,1,2,...8x

P[sample has one or more defectives] = 1-P[no defectives] = 1-p(0)

08 = 1 - 8C(0.05)(0.95)= 1 0.6634 = 0.3366 0

In a Binomial distribution the mean is 6 and the variance is 1.5. Then, find (i) P[X=2] and 3

(ii) P[X?2].

Let n and p be the parameters. Then,

Mean = np = 6

Variance = npq = 1.5

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Subject Name: Probability and Queueing Theory Subject Code: MA2262 Question Bank 2012 - 2013

Variancenpq1.51 Meannp64

Therefore, q = ? and p = ?

Therefore, Mean = n*3/4 = 6

That is, n = 24/3 = 8

x8-xThe p.m.f is ------- p(x) = 8C (3/4) (1/4), x=0,1,2,…….8 x

26(i) P[X=2] = 8C (3/4) (1/4) =252/65536=0.003845 2

(ii)08 17 26 P[X?2] = p(0)+p(1)+p(2) = 8C (3/4) (1/4)+ 8C (3/4) (1/4)+ 8C (3/4) (1/4)012

= 277/65536 = 0.004227

The number of accidents occurring in a city in a day is a Poisson variate with mean 0.8. Find 4

the probability that on a randomly selected day. (i)There are no accidents (ii)There are

accidents

Let X: number of accidents per day.

Then, X ~ P(λ=0.8).

The p.m.f. is

0.8xe(0.8)p(x),x0,1,2,3,.... x!

(i) Probability that a particular day three are no accidents is P[no accidents] =

P[X=0]=p(0)

0.80e(0.8)0.8e0.449 = 0!

P[accidents occur] = 1-P[no accidents] = 1-p(0) = 1-0.449 = 0.551 The number of persons joining a cinema queue in a minute has Poisson distribution with 5

parameter 5.8. Find the probability that (i) no one joins the queue in a particular minute (ii)2 or more persons join the queue in the minute.

Solution:

Let X : number of persons joining the queue in a minute Then, X ~ P(λ=5.8).

5.8xe(5.8)p(x),x0,1,2,3,....The p.m.f is x!

(i) P[no one joints the queue] = P[X=0] = p(0)

5.80e(5.8)= 0!

5.8e= =0.003

[?2]1[2]1{(0)(1)} P[two or more join] = PXPXpp

5.805.81??(5.8)(5.8)ee5.8??1115.8e ??0!1!??

10.003?6.810.02040.9796

2 percent of the fuses manufactured by a firm are expected to be defective, Find the 6.

probability that a box containing 200 fuses contains (i) defective fuses (ii) 3 or more

defective fuse

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Subject Name: Probability and Queueing Theory Subject Code: MA2262 Question Bank 2012 - 2013

Solutions:

2 percent of the fuses are defective. Therefore, probability that a fuses is defective is

p = 2/100 = 0.02

Let X denote the number of defective fuses in the box of 200 fuses. Then, X is

B(n = 200, p = 0.02)

Here, p is very small and n is very large. Therefore, X can be treated as Poisson variate with

parameter λ=np = 200 * 0.02 = 4.

4xe4p(x),x0,1,2,3,...The p.m.f. is ---- x!

P[box has defective fuses] = 1-P[no defective fuses]

-40 = 1 p(0) = 1 e[(4)/0! = 1 0.0183 = 0.9817

P[3 or more defective fuses] = 1-P[less than 3 defective fuses]

-4 = 1 [p(0)+p(1) +p(2)] = 1 e[1+4+8] = 1 0.0183*13

= 1-0.2379 = 0.7621

The probability that a razor blade manufactured by a firm is defective is 1/500. Blades are 7.

supplied in packets of 5 each. In a lot of 10,000 packets, how many packets would

-0.01(e=0.99)

Let X be the number of defective blades in a packet of 5 blades. Then, X is B (n = 5, p =

1/500)

Since p is very small and n is sufficiently large, X is treated as Poisson variate with

0.01xe(0.01)p(x),x0,1,2,3,...parameter λ=np = 5*(1/500) = 0.01, x!

-0.010(i) P[ no defective blades] = p(0) = e(0.01)/0! = 0.99

The number of packets which will be free of defective blades is

N * P[no defective blades] = 10000*0.99 = 9900

-0.011(ii) P[one defective blade] = p(1) = e(0.01)/1! = 0.0099

The number pf packets which will have one defectives blade is

N * P[one defective blade] = 10000*0.0099 = 99

A random variable X has a uniform distribution over (-3, 3). Compute 8.

1(1)P(X2)(2)PX2(3)FindkforwhichP(X(k)；？3

X is uniformly distributed over (-3,3)

1?,3x3?f(x)6? ?0otherwise?

222115??(1)(2)()PXfxdxdxx????666 ??333

2122(22)2) ；？3) Given PXPxdx?632

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Subject Name: Probability and Queueing Theory Subject Code: MA2262 Question Bank 2012 - 2013 311 dx?3k2?k1?63 k

The daily consumption of milk in excess of 20000 litres in a town is approximately 9.

exponentialy distribution with parameter 1/3000. The town has a daily stock of 35,000 litres.

What is the probability that of 2 days selected at random the stock is insufficient for both days?

If Y denotes daily consumption of milk then X = Y 2000 follows an exponential

1x/3000distribution with parameter 1/3000. Then we have . f(x)e;x(0

3000

P(stock insufficient for one day) = P(Y>35000) = P(X+20000 > 350000)

?1x/30005 = (15000)PX(edxe?300015000

2510Therefore, P(stock insufficient for two days ) = ；？ee

The length of the shower in a tropical island in a rainy season has an exponential distribution 10.

with parameter 2, time being measured in minutes. What is the probability that it will last for atleast one more minute?

2xf(x)2e;x?0Let X be the length of the shower, then

P(Shower will last for at least one more minute provided shower has lasted for 2 minutes) =

?2x2 [ by memoryless property].? P(X?3/X?2)P(X?1)2edxe0.1353?1

In a factory the daily consumption of electric power in millions of kilowatt hours can be 11.

treated as a random variable having Erlang distribution with parameters ? = 1 / 3 and k = 4.

If the power supplied to the factory is 12 million kilowatt hours, what is the probability that this power supply will be inadequate on any given day?

Let X be the power consumption in millions of kilowatt hours. Then pdf of X is given by f(x)

41????x3??33f(x)xe;x?0

3!

4x?11??33P(X12)xedxP(the power supply is inadequate) = ( = 0.4335 ???3!3??12

In a certain locality the daily consumption of water in millions of litres, can be treated as a 12.

random variable having a gamma distribution with an average of 3 million litres. If the pumping station of the locality has a daily supply capacity of a millions litres. What is the probability that this water supply will be inadequate on any given day?

It is given that X be the daily water consumption of the locality in million litres and

X follows gamma distribution with average of 3 million litres.

Therefore Mean = ? = 3 million litres.

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Subject Name: Probability and Queueing Theory Subject Code: MA2262 Question Bank 2012 - 2013

x?1x2exexThe density function of X is ();?0?() fxxfx

(?1)!(31)!P(Water supply of 4 million litres inadequate on a given day)

?2xex4 = P(X(4)P(4X?)13e0.2376?2!4

????If X is uniformly distributed in , find the probability density function of ,??13. 22??

Y = tanX

????Given X is uniformly distributed in ,??22??

A random variable X is said to have a continuous uniform distribution over an interval (a,

b) if its probability density function is

??1?,x??11???,axb,x?????22f(x) ba22????22???0,otherwise0,otherwise???0,otherwise?

??-1xGiven y = tanx, tany = x and , if

22

???????? x?ytan，?,x?ytan?????2222????

The pdf of Y is

??dx1111?????f(y)f(x)?i.ef(y),?x??????22dy?????1y1y??

???11??,?x????2i.e.f(y)??1y???0,otherwise?

The life time of a component measured in hours in weibull distributed with parameters14.

?0.2,?0.5. Find (a) the mean life time of the component (b) the probability that such

a component will be more than 200 hours. 2(a)E(x)(0.2)(3)50hours

? 0.50.5(0.2)x(b)P(X(200)(0.1)xedx0.059?200

In a company 5% defective components are produced. What is the probability that at least 5 15.

components are to be examined in order to get 3 defectives? Here p = 0.05, q = 0.95

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Subject Name: Probability and Queueing Theory Subject Code: MA2262 Question Bank 2012 - 2013

?3x3= Required probability = P(X = 5) + P(X = 6) + ……. = (x1)C(0.05)(0.95)?2

x5

43x3= = 0.9995 1(x1)C(0.05)(0.95)?2

x3

UNIT II TWO DIMENSTIONAL RANDOM VARIABLES

PART - A

1. The joint probability mass function of (X,Y) is P(x,y) = k ( 2 x+3y), X = 0,1,2 and Y = 1,2,3. Find k.

33

We know that P(xy)1??ij,jj，，11

Y/X 1 2 3

0 3k 6k 9k 18k

1 5k 8k 11k 24k

2 7k 10k 13k 30k

15k 24k 33k 72k

1 72k = 1 ；，k72

2. If X and Y are independent continuous random variables, Show that E(Y/X) = E (Y).

E(Y/X) yf(y/x)dy(conditionalp.d.f)y?xy

yf(y)dy(marginalprobability) y?y

E(y)

3. fxyxy,2,01，！！！XYLetandhave joint density function. Find the marginal density ；？

YXxfunction. Find the conditional density function given .

XMarginal density function of is given by

?

fxfxfxydy，，,；？；？；？X?，?

111 ，，，fxydydyy,22 ；？；？??xxx

，，！！21,01.xx ；？

YMarginal density function ofis given by

?

fyfyfxydx，，,；？；？；？Y?，?

y

，，！！22,01dxyy. ?0

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Subject Name: Probability and Queueing Theory Subject Code: MA2262 Question Bank 2012 - 2013

fxy,；？21yConditional distribution function ofgiven is. YXxf，，，；？xfxxx211，，；？；？4. If Y = -2x + 3, Find the Covariance of(X,Y)

Cov(X,Y) = E(XY) E(X) E(Y)

= E(X(-2X+3)) E(X){E(-2X+3)}

2 = E (-2X + 3X) - E(X) {-2XE(X)-3}

22 = -2E(X) +3 E( X) + 2 (E(X)) - 3E(X)

22 = 2(E(X)) 2 E(X) = -2 var(X)

5. Write the statement of central limit theorem.

XIf is the mean of the random sample of size n taken from a population having the mean and

X2?the finite variance , then is a random variable whose distribution function Z

?x

approaches that of the standard normal distribution as n --> ?

6. Find the value of k if f(x,y) =k (1-x) (1-y) for 0

By the property of joint density function, we have

11

fxydxdykxydxdy(,)1(1)(1)1，?，，，????R00

111122??xxy?，，？，?，，？，kxyxydxdykxxydy(1)11?????22??0000 1111yy?????，，？，?，？，kydykydy111??????2222????00

122??yyy111?，？，?，？，?，kkk1()14??224224??0

7. If the joint P.d.f of (X,Y) is given by f(x,y) = 24y(1-x), 0 ? X,Y ? 1 Find E(XY).

1111

2EXYxyxdxdyxyxydxdy()4(1)4()，，，，????

0000

1123xyxyyy1，，，，4()4()dydy??0 232300

122??yy1，，?，4()Exy??463??0 18. Find the variance of X Y from the following data.. ??(，，，5,4,xyxy8

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