Questiomaterijalns & Answers Verovatnoca

NPTEL ASSIGNMENT PROBLEMS

M2L3

Q1. During the manufacture of concrete blocks in a factory, it is found that the probability of producing a defective unit is 0.045. It is further observed that the probability of producing a defective unit in terms of dimensions of the block is 0.05 and in terms of the material quality is 0.025. What is the probability of producing a unit that is defective in terms of dimensions of the block as well as in terms of the concrete mix ?

Soln. Let us denote,

A : Event of production of a defective unit in terms of dimensions of the block.

B : Event of production of a defective unit in terms of the material quality.

Thus,

045.0)(,025.0)(,05.0)( === BAPBPAP

The probability of producing a unit that is defective in terms of dimensions of the block as well as in terms of the material quality is given by

M2L4

Q1. A city gets 70% of its required energy from thermal power and remaining from hydropower. If probability of shortage of thermal power is 0.2 and that of hydropower is 0.35, what is the probability of shortage of power to the city?

Soln. Let us denote the following events:

S: shortage of power in the city

T: energy from thermal power, H: energy from hydropower

Thus,

35.0)/(3.0)(2.0)/(7.0)(

==

==

HSPHPTSPTP

03.0045.0025.005.0)()()()( =+=+= BAPBPAPBAP

From total probability theorem, the probability of shortage of power to the city is given by

( )( ) ( )( )245.0

3.035.07.02.0)()/()()/()(

=

+=

+= HPHSPTPTSPSP

Q2. A team of four students are sent to a particular station to measure rainfall. 20% of the measurements are done by student A, who makes a mistake in measurement once in 20 times on an average. 30% of the measurements are done by student B, who makes a mistake once in 10 times on an average. 15% of the measurements are done by student C, who makes a mistake once in 20 times on an average. 5% of the measurements are done by student D, who makes a mistake once in 20 times on an average.

(a) What is the probability that a particular measurement checked at random will be found to be a wrong one ?

(b) If a particular measurement is found to be wrong, what is the probability that it is recorded by student A ?

Soln.

(a) Let 321 ,, AAA and 4A denote the events that the measurements are done by students A, B, C and D respectively. Let B denote the event that the recorded measurement is wrong.

( ) ( )( ) ( )( ) ( )( ) ( )

201/

201

1005

201/

203

10015

101/

53

10060

201/

51

10020

44

33

22

11

===

===

===

===

ABPAP

ABPAP

ABPAP

ABPAP

The probability that a particular measurement checked at random will be found to be a wrong one is given by

08.0201

201

203

201

53

101

51

201

)()/()()/()()/()()/()( 44332211

=

+

+

+

=

+++= APABPAPABPAPABPAPABPBP

(b) The wrong measurement may have been recorded by either of the students A, B, C or D. The probability that the wrong measurement is recorded by student A is given by

M3L2

Q1. The pmf of a random variable X is given by

X 0 25 50 75 100

( )xpX 0.5 2/C 5/C 8/C 10/C (a) Under what condition is this function a valid pmf ?

(b) What is the probability of X being greater than 25?

Soln.

(a) For a valid pmf, the following two properties must be satisfied.

( ) 1=xpossibleallX xp and ( ) 0xpX

From the first condition,

( )54.0,

1925.05.0,

110852

5.0

=

=+

=++++

CorCor

CCCC

As the value of C is positive, the second condition is also satisfied.

(b) The probability of X being greater than 25 is given by

( ) ( ) ( )( ) 2295.054.0425.0425.01085

50 ===++= CCCCxPX

( ) ( ) ( )( ) ( )

( )( )( )( ) ( )( ) ( )( ) ( )( ) 8

120/120/120/320/15/310/15/120/1

5/120/1

/

// 4

1

111

=

+++=

=

=i

ii APABP

APABPBAP

Q2. A random variable X has a pdf of the form

( ) ( ) 30939

2 += xxxCxf X

(a) Check the validity of the pdf .

(b) What is the probability of X being greater than 1.5?

Soln.

(a) For a valid pdf, the following two properties must be satisfied.

( ) 1=xpossibleall

X xf and ( ) 0xf X


( )

( )

( )

( )

( )4.0,

15.229

,

1392

3333

9,

192

339

,

1939

,

1

23

3

0

23

3

0

2

3

0

=

=

=

+

=

+

=+

=

Cor

Cor

Cor

xxxC

or

dxxxCor

xf X

As the value of C is positive, the second condition is also satisfied.

Thus, the given function is a valid pdf.

(b) The probability of X being greater than 1.5 is given by

( ) ( ) ( )5.05.01

9394.015.115.1

5.1

0

2

==

+==> dxxxxPxP XX

M3L3

Q1. A certain probability density function is expressed as

( )

elsewherexforexforxf

x

X

0002.0

25.0

=

>=

==

(a) Find the value of .

(b) Formulate the CDF.

Soln.

(a) For a valid pdf, the following two properties must be satisfied.

( ) 1=xpossibleall

X xf and ( ) 0xf X


( )

2.0

8.025.010

8.025.0

8.0

12.0

1

0

25.0

0

25.0

0

25.0

0

=

=

=

=

=+

=

or

or

eor

dxeor

dxeor

dxxf

x

x

x

X

As the value of is positive, the second condition is also satisfied.

(b) For formulation of CDF,

( )( )

( )x

x

xx

x

x

X

X

e

e

e

dxe

xXPxFxForxXPxFxFor

=

=

+=

+=

=>====

8.0118.02.0

25.02.02.0

2.02.0

][,02.0][,0

5.20

25.0

0

25.0

Thus the CDF can be expressed as

( )

elsewherexforexforxF

x

X

008.0102.0

25.0

=

>=

==

Q2. Consider the continuous probability density function given by, ( ) .025.0 axxf = (a) What is the value of a ?

(b) Determine ( )2/aXP > . Soln.

( ) .025.0 axxf = (a) Since f(x) is a valid pdf,

( )

4,125.0,

125.0.,

1

0

=

=

=

=

aor

aor

dxor

dxxfa

(b) Now, the probability

>

2aXP is given by

( )( )( )

( )5.0

225.0121

21224

2

=

=

=

=>=

>=

>

FXP

XP

XPaXP

Q3. X is a random variable with pdf

( ) ( ) 1016 = xxxxf (a) Check that this is a valid pdf.

(b) Obtain the CDF.

(c) Determine a number b such that ( ) ( )bXPbXP = 2 Soln. The pdf of RV X is given by

( ) ( ) 1016 = xxxxf For a valid pdf, the following two properties must be satisfied.

( ) 1=xpossibleall

X xf and ( ) 0xf X

Checking the first condition,

( ) ( )

1616

31

216

326

16

1

0

32

1

0

=

=

=

=

=

xx

dxxxdxxf X

Also in the range 10 x , the first condition is satisfied.

Hence f(x) is a valid pdf.

(c) The CDF is given by

( ) ( ) ( )

320

32

0

23

326

16

xx

xx

dxxxdxxfxFx

xx

=

=

==

(c) To find b such that,

( ) ( )( ) ( )[ ]( ) ( )[ ]( )

615.0,

0296,3223,

23,12,12,

2

23

32

=

=+

=

=

=

==

berrorandtrialBy

bbor

bbor

bForbFbFor

bXPbForbXPbXP

Q4. Each of the following functions represents the CDF of a continuous random variable. In each case,

( )( ) bxxF

axxF==

10

Where [a,b] is the indicated interval.

In each case, determine the pdf f(x) and verify that it is a valid pdf.

(a) ( ) 505

= xxxF

(b) ( ) 10sin2 1 = xxxF

Soln.

(a) The CDF is given by ( ) 505

= xxxF

So, the pdf is

( ) ( )51

==

dxxdF

xf

For a valid pdf, the following two properties must be satisfied.

( ) 1=xpossibleall

X xf and ( ) 0xf X


( ) 155

1 5

0

5

0

=

==

xdxxf X

Checking the second condition,

( ) xxf >= 051

Thus f(x) is a valid pdf.

(b) The CDF is given by ( ) 10sin2 1 = xxxF

So, the pdf is

( ) ( ) ( )xxxxdxxdF

xf

=

==

11

21

112 2/1

For a valid pdf, the following two properties must be satisfied.

( ) 1=xpossibleall

X xf and ( ) 0xf X


( ) ( )

[ ]10

22

sin21

21

11

10

1

1

02

1

0

=

==

=

=

z

dxz

dxxx

xf X

Checking the second condition,

( ) ( ) 10011 >

= x

xxxf

Thus f(x) is a valid pdf. Q5. A random variable X has a density function

( ) 03 >= xcexf x (a) Find the constant c.

(b) Find the CDF.

(c) Find ( )21 XP (d) Find ( )3XP (e) Find ( )1XP Soln. ( ) 03 >= xcexf x (a) Since f(x) is a valid pdf,

( )

[ ]3,

103

,

13

,

1.,

1

0

0

3

0

3

=

=

=

=

=

cor

ec

or

ecor

dxceor

dxxf

x

x

(b) The CDF is given by

( )

[ ] xxxx

x

x

eeee

dxexF

303

0

3

0

3

13

3

3

==

=

=

(c) The required probability is given by

( ) ( ) ( )( ){ } ( ){ } 047.011

12211323

==

= ee

FFXP

(d) The required probability is given by

( ) ( ) ( ) ( ){ } 00012.01131313 33 ==== eFXPXP (e) The required probability is given by

( ) ( ) ( ){ } 9502.0111 13 === eFXP

Q6. The failure free working time (i.e, time between breakdowns) in hours of a small equipment is given by

( ) 06.0 6.0 >= xexf x Find the probability that the time between two successive breakdowns exceeds 10 hours.

Soln.

The CDF is given by

xx

x

X

x

XX

e

dxexFor

dxxfxF

6.00

6.0

0

1

6.0)(,

)()(

=

=

=

The probability that the time between two successive breakdowns exceeds 10 hours is

( )[ ] ( ) 0025.011)10(1]10[

6106.0106.0====

= eee

FXP X

M4L5

Q1. The hydraulic head loss in a 2.5 km long and 750 mm diameter pipe due to friction is given by the Darcy Weisbach equation

2

2V

gDLfhL =

Where L and D are the length and diameter of the pipe, f is the friction factor assumed to be 0.016, v is the velocity of flow in the pipe. If v has an exponential distribution with mean velocity 2m/s, derive the density function for the headloss hL.

Soln.

As, v has an exponential distribution with mean velocity 2m/s,

( ) 021 2 =

vevfv

V

As,

( ) ( )( ) ( )22 718.2

75.081.922500016.0

,016.0,75.0,2500

vvh

fmDmL

L ==

===

( )L

LV

LVLH

LL

L

hhfhfhf

theoremlfundamentafromThushdh

dv

hvThus

L 718.221

718.2718.2

,,

718.221

718.2,

+

=

=

=

However, ( ) 00 =

=

=

L

h

LLH

h

LLH

LV

LLH

heh

hfor

eh

hfor

hfh

hf

L

L

L

L

L

M5L12

Q1. The joint pdf between streamflows in two rivers is given by

( ) 20;103

,2 += yxxyxyxf

Determine the following

(a)

21XP (b) ( )XYP (c)

21|

21 XYP

Soln.

The joint pdf between the two streamflows

( ) 20;103

,2 += yxxyxyxf

Marginal density function of X

( ) ( ) 103

2263

,2

2

0

22

2

0

2 +=

+=

+==

xx

xxyyxdyxyxdyyxfxg

Now, the marginal CDF of X

( )33

233

23

2223

0

23

0

2 xxxxdxxxXGxx

+=

+=

+=

(a) The probability

65

21

31

21

321

211

211

21

23

=

=

=

=

G

XPXP

(b) The probability

( )

247

67

3

3

1

0

3

0

1

0

2

1

0 0

2

==

+=

+=

=

=

dxx

dxxyx

dydxxyxXYP

xy

y

x

(c) The marginal density function of X

( ) 103

22 2 += xxxxg

Now,

( )

( )

41

62

32

63

322

3,

21|

2/1

0

23

2/1

0

23

2/1

0

2

2/1

0

2

2/1

2/1

y

xx

yxx

dxxx

dxxyx

dxxg

dxyxfxyh

x

x

x

x

+=

+

+

=

+

+

==

=

=

=

=

Therefore,

82

42/

21|

21|

20

2

0

yy

yy

dyxyhxyH

y

y

+=

+=

=

Thus, the probability

( ) ( )325

84/12/12

21|

21

=

+=

XYP

Q2. Given the joint pdf

( ) ( ) 10;20431

,

2

+= yxyxyxf

Find

=

The conditional density of X|Y

( ) ( )( )yhyxfyxg ,| =

Now,

( ) ( )( )

( )2

2

0

222

2

0

2

2

0

2

3121

23

241

43

431

y

yxx

dxxyx

dxyxyh

+=

+=

+=

+=

Therefore,

( ) ( )( )( )

( ) 2312

431,| 2

2 x

yyx

yhyxfyxg =

+

+==

Now,

( ) ( )442

||2

00

2

0

xxdxxdxyxgyxGxxx

=

===

The probability

643

41

21

41

31|

21

41 22

=

=

=

( )( )

( )

( ){ }( )

[ ]( ) 5027.21

1,

11,

11,

1,

11

,

1,

1,

21

112

1

0

1

1

0

1

1

0

1

0

1

0

1

0

1

0

1

0

=

=

=+

=

=

=

=

+

+

=

=

+

+

ekor

eeekor

eekor

dyeekor

dykeor

dxdykeor

dxdyyxf

yy

yy

x

x

yx

yx

Q4. The joint pdf of (X,Y) is given by

( ) xyxeyxf y >>= ;0, (a) Find the marginal pdf of X

(b) Find the marginal pdf of Y

(c) Evaluate ( )4|2 YXP Soln.

( ) xyxeyxf y >>= ;0, (a) Marginal pdf of X

( ) ( )

[ ]( )

0

,

>=

=

=

=

=

xe

ee

e

dye

dyyxfxg

x

x

x

y

x

y

(b) Marginal pdf of Y

( ) ( )

[ ]0

,

0

0

>=

=

=

=

yye

xe

dxe

dxyxfyh

y

yy

yy

(c) The conditional density of 4| YX

( )( )

( ) 44

44

4

4

0

4

4

0

4

5114

,

4|

=

+

+===

e

ee

ee

ee

dyye

dye

dyyh

dyyxfyxg

xx

y

x

y

x

Now, conditional CDF of 4| YX

( ) ( )

4

40

4

4

04

4

511

51

51

4|4|

+=

=

=

=

e

xee

e

xee

dxe

ee

dxyxgyxG

x

xx

x x

x

Thus,

( ) ( )( )

0885.051

121

4|214|214|2

4

42

=

+=

==

e

ee

YGYXPYXP

Q5. Observations on average annual discharge (represented as random variable X) on a stream are as given below:

Obs. No. Discharge (m3/s)




1 110 7 41 13 125 19 140

2 95 8 31 14 270 20 96

3 85 9 20 15 410 21 63

4 71 10 18 16 460 22 55

5 63 11 20 17 405 23 56

6 52 12 42 18 250 24 100

The volume of runoff in an adjacent stream is functionally represented as 5ln3 += XY

If the discharge X follows lognormal distribution, find ( )10YP . Soln.

The discharge X in the stream follows lognormal distribution.

021.1,97.130,25.128 ====X

SCVSX XX

Xln follows lognormal distribution.

( )

( ) 4968.41021.125.128ln

1ln

2

2ln

=

+=

+=

X

XCV

X

( )[ ]( )[ ] 8452.01021.1ln

1ln2

2ln

=+=

+= XX CV

So,

( )( ) ( )[ ]

( )5356.2,4904.18~5ln3,8452.03,54968.43~5ln3,

8452.0,4968.4~ln

NXYorNXYThus

NX

+=

++=

Thus, the probability

( ) ( )

( )

9996.00004.01

35.315356.2

4904.18101

10110

=

=

=

=

=

ZP

ZP

YPYP

Q6. In a certain catchment, the following rainfall-ruoff data has been observed for 16 years.

Year Rainfall (cm) Runoff (cm) Year Rainfall (cm) Runoff (cm)

1 42.39 13.26 9 47.08 22.91

2 33.48 3.31 10 47.08 18.89

3 47.67 15.17 11 40.89 12.82

4 50.24 15.5 12 37.31 11.58

5 43.28 14.22 13 37.15 15.17

6 52.60 21.20 14 40.38 10.40

7 31.06 7.70 15 45.39 18.02

8 50.02 17.64 16 41.03 16.25

Rainfall is represented by a random variable X and runoff by Y. If X and Y are independent, find the joint pdf ( )yxf YX ,, and joint CDF ( )yxF YX ,, . Soln.

Let,

( )( )

6275.1411

,0

9406.4211

,0

===

===

Ywhereyeyf

Xwherexexf

yY

x

X

As X and Y are independent, the joint pdf is given by

( ) ( ) ( )( )( )

( )

+

+

=

=

=

=

6275.149406.42

,

11.6281

,

yx

yx

yx

YXYX

e

e

ee

yfxfyxf

Again, the joint CDF is given by

( ) ( )

( )( )

( )( )

=

=

=

=

=

=

+

6275.149406.42

0

0

0 0

0 0,

11

11

1

1

,

yx

yx

xx

y

x

yx

x yyx

x yyx

YX

ee

ee

ee

dxee

dxee

dydxeyxF

Q7. The joint distribution of random variables X and Y is given by:

( ) ( ) 0,0, = + yxeyxf yx U and V are functions of random variables X and Y and are given by:

YXVYXU +== ,

Find ( )vuf , .

Soln.

UUVYUX

UVYor

YYUYXVNowYUXSo

YXU

+==

+=

+=+=

=

=

1

1,

,

,

Thus, ( )( ) VUUVYX =

+

+=+

11

Now,

( ) ( ) ( )vuJyxfvuf ,,, = where,

( ) ( ) ( )

( ) ( )( ) ( ) ( )233

2

2

1111

11

11,

u

v

u

uv

u

v

uu

v

u

u

u

v

v

yu

yv

x

u

x

vuJ+

=

++

+=

++

++=

=

Thus,

( ) ( ) ( )

( ) ( ) ( )positivealwaysareVandUSinceuve

u

ve

vuJyxfvufv

v

22 11

,,,

+=

+=

=

M6L1

Q1. The total number of rainstorms in a year in a particular river basin is a normally distributed random variable. The mean number of rainstorms per year is 35 and standard deviation is 5.5. (a) What is the probability that the number of rainstorms in a certain year is between 25 and 40 ? (b) What is the probability that the number of rainstorms in a certain year is less than 10 ? (c)What is the 95% dependable number of rainstorms in a year?

Soln.

(a) The probability that the number of rainstorms in a certain year is between 25 and 35 is

( ) ( ) ( )

( ) ( )( ) ( )[ ]

( )9656.018186.082.1191.0

82.191.05.53525

5.53540

25404025

=

==

=

=

Soln.

(a) Coefficient of variation (CV) =0.25

Standard deviation s = (CV) ( )X ( )( ) KNm5.125025.0 ==

When a concentrated load of 25 KN is applied at the free end of the cantilever beam, the maximum moment occurs at the fixed end and it is given by ( )( ) KNm25105.2 = If the moment capacity is less than the developed moment (i.e, 25KNm), then the beam will fail. Thus, the probability of failure is given by

( )

( )( )

0227.09773.0121

25.125025

2525

==

M6L2

Q1. Based on experience, a contractor estimates the expected time of completion of a job A to be 30 days. However, because of uncertainties in the material supply and weather conditions, the job may not be finished in exactly 30 days. The contractor is 90% confident that the job will be completed within 45 days. If the number of days required to complete the job be a Gaussian variable X,

(a) Find the mean and standard deviation.

(b) What is the probability that X will be less than 50 ?

(c) What is the probability of X taking on a negative value ? Based on the result, is the assumption of normal distribution of X reasonable ?

(d) Assuming X to be a lognormal variable with the same mean and variance as those in the normal distribution of part (a), what is the probability that X will be less than 50 ?

Soln.

(a) As the expected time of completion of job A is 30 days, mean days30=

As the contractor is 90% confident that the job will be completed within 45 days, ( )

daysor

or

ZPor

XP

63.11,

29.13045,

9.03045,

9.045

=

=

=

=

Thus, standard deviation = 11.63 days.

(b) The probability that X will be less than 50 is given by

( )( )9573.0

72.163.11305050

=

=

=

ZP

ZPXP

(c) The probability of X taking on a negative value is

( )( )

( )

005.09950.01

58.2158.263.113000

=

=

==

Q2. The time between breakdowns of a road construction equipment can be modeled as a lognormal variate with a mean of 5 months and standard deviation of 1.2 months. If the desired probability of the equipment being operational at any given time be 90%

(a) How often should the equipment be scheduled for maintenance ?

(b) If a certain equipment is in good operating condition at the time it is scheduled for maintenance, what is the probability that it can operate for at least another month without its regular maintenance ?

Soln.

(a) Coefficient of variation 24.052.1

==CV

For the log transformed variables,

Standard deviation

( )[ ]( )[ ] 2366.0124.0ln

1ln2

2ln

=+=

+= XX CV

Mean

( )

( ) 5814.1124.05ln

1ln

2

2ln

=

+=

+=

X

XX

CV

The desired probability of the equipment being operational at any given time be 90%. Let the equipment be scheduled for maintenance after X90 months.

( )( )

monthsXorXor

Xor

XZPor

XXPorXXP

36.3,2132.1ln,

29.12366.0

5184.1ln,

1.02366.0

5184.1ln,

9.0ln1,9.0ln

90

90

90

90

90

90

=

=

=

=

==>

(b) The probability that the equipment fails after ( ) months36.4136.3 =+ is ( )( ) ( )( )

( )

( )( )[ ]

( )5841.0

194.0194.011

194.012366.0

5184.136.4ln1

36.4ln136.4ln

=

==

=

=

=>

ZPZP

ZP

ZP

XPXP

Now, let us denote the following events,

A: The equipment fails after 4.36 months

B: The equipment does not fail before 3.36 months

Therefore the conditional probability

( ) ( ) ( )( )( )( )

( )649.0

9.05841.01

//

=

=

=

BPAPABPBAP

Thus, if the equipment is in good operating condition at the time it is scheduled for maintenance i.e, 3.36 months. The probability that it can operate for at least another month i.e, 4.36 months without its regular maintenance is 0.649

Q3. In a certain location, it is observed that the depth H to which a pile can be driven in the soil without encountering rock stratum is a lognormal variate with mean 9.5 m and standard deviation 1.9 m.

(a) What is the probability that the depth H will be between 5 m and 15 m ?

(b) What is the probability that the depth H will be at least 8 m ?

Soln.

(a) The coefficient of variation 2.05.99.1

==CV


Mean is

( )

( ) 23.212.05.9ln

1ln

2

2ln

=

+=

+=

X

XX

CV

The standard deviation is

( )[ ]( )[ ] 198.012.0ln

1ln2

2ln

=+=

+= XX CV

The probability that the depth H will be between 5 m and 15 m is

( ) ( ) ( )

( ) ( )( ) ( )[ ]

( )9911.0

9991.019920.013.3141.2

13.341.2198.0

23.25ln198.0

23.215ln515155

=

=

==

=

=

ZP

ZP

XPXP

(b) For flow in the stream B,

Coefficient of variation (CV) =0.25

Standard deviation s = (CV) ( )X ( )( ) sm /75.83525.0 3==


Standard deviation

( )[ ]( )[ ] 2462.0125.0ln

1ln2

2ln

=+=

+= XX CV

Mean

( )

( ) 525.3125.035ln

1ln

2

2ln

=

+=

+=

X

XX

CV

The stream B will overflow if the annual maximum flow exceeds the maximum capacity. Thus, the probability that stream B will overflow in a certain year is given by

( ) ( )

( )

0251.09749.01

959.112462.0

525.355ln1

55155

=

=

=

=

=>

ZP

ZP

XPXP

(c) The probability that any one of the streams will overflow in a certain year is

0931.00251.00668.0 =+

(d) The probability that none of the two streams will overflow in the next 5 years is

( )( )[ ]( )( )[ ]( ) 623.09097.0

9749.09332.00251.010668.01

5

5

5

==

=

(e) If the probability of overflow in stream A is to be 2%, let the new capacitiy of stream A be Qnew

( )( )

smQor

Qor

QZPor

QZPor

QXPorQXP

new

new

new

new

new

new

/6.70,

06.210

50,

98.010

50,

02.010

501,

02.01,02.0

3=

=

=

=

==>

(f) There is a 20% chance that the capacity of stream B may be 60 m3/s instead of 55 m3/s

If the maximum capacity of stream B is 60 m3/s, then the probability of overflow in stream B is given by

( ) ( )

( )

0104.09896.01

313.212462.0

525.360ln1

60160

=

=

=

=

=>

ZP

ZP

XPXP

As there is a 20% chance that the capacity of stream B may be 60 m3/s instead of 55 m3/s, the probability that stream B will overflow in a certain year is given by

( ) ( ) 02216.00251.08.00104.02.0 =+

Q5. The interarrival time between earthquakes in an earthquake prone region is an exponentially distributed variable. If the mean interarrival time is 10 years and one earthquake has taken place this year

(a) What is the probability that there will be no earthquakes in the next 15 years?

(b) What is the probability that another earthquake will take place within the next 2 years ?

Soln.

Mean inter arrival time = 10 years

So 1.010/1 ==

Therefore, the interarrival time between accidents can be expressed as

( ) xX exf 1.01.0 = The cumulative distribution is given by

( ) xX exF 1.01 = (a) The probability that there will be no accidents in the next 15 years is

[ ] [ ]

223.0)1(1

15115

151.0

151.0

==

=

=>

e

e

XPXP

(b) The probability that the interarrival between accidents will not exceed 2 years is

[ ]181.0

12 21.0

=

= eXP

Q6. The daily rainfall (in mm) over a catchment can be expressed as an exponential probability distribution as follows:

( ) ( ) 02.0 2.0 >= Xforexf xX (a) A rainy day is defined as a day when the total rainfall is greater than 2.5 mm. What is the probability that a particular day will be a rainy day?

(b) What is the 95th quantile value of daily rainfall ?

Soln.

(a) The cumulative distribution is given by

( ) 01 2.0 >= XforexF xX The probability that the total rainfall will be greater than 2.5 mm on a particular day is

[ ] [ ]( )

6065.011

5.215.25.22.0

=

=

=>e

XPXP

Thus, the probability that a particular day will be a rainy day is 0.6065

(b) Let x95 be the 95th quantile value of daily rainfall.

[ ]

mmxor

eor

eor

xXP

x

x

98.14,05.0,

95.01,95.0

95

2.0

2.095

95

95

=

=

=

=

The rainfall value which has non exceedence probability of 0.95 is 14.98 mm.

M6L3

Q1. In a particular earthquake prone zone, the interarrival time between successive earthquakes follows an exponential distribution. If, on an average, an earthquake occurs in this zone once in 8 years, find the expected time till the third earthquake.

Soln.

The interarrival time between successive earthquakes is given by

( ) 081 8

=

xexfx

X

The time till the third earthquake is described by the sum of three exponential distributions. It is given by the gamma distribution

( ) ( ) 8,3and01 1

==

=

xexxfx

X

So, the expected time till the third earthquake is the mean ( ) years2483 == .

Q2. In a certain river, streamflow is a random variable having gamma distribution with 25= and 55= . If the streamflow is greater than the 1500 m3/s, then the neighboring areas get inundated. What is the probability that the neighboring area will get inundated ?

Soln.

The streamflow is given by gamma distribution with 25= and 55=

( ) ( ) ( ) 025551

25551 5524

2555125

25 =

=

xexexxfxx

X

If the streamflow is greater than the 1500 m3/s, then the neighboring areas get inundated. The probability that the neighboring area will get inundated is given by

[ ] [ ]

( )3058.06942.01

255511

1500115001500

0

552425

==

=

=>

x

ex

XPXP

Q3. The interarrival time between two successive hurricanes follows an exponential distribution with a mean of 12 years. Assuming that the time between any two events of hurricanes are independent of each other, find the probability that the maximum time between two hurricanes exceeds 25 years over a sequence of 5 hurricanes.

Soln.

The interarrival time, denoted by X, follows exponential distribution with 12/1/1 == X and the maximum time Y between two earthquakes follows extreme value distribution.

In a sequence of 5 hurricanes, there are 4 interarrival periods.

The probability that the maximum interarrival time exceeds 25 years over a sequence of 5 successive hurricanes is

( ) ( ){ } 5875.0125254

1225

4=

==

eFF XY

Q4. In a certain region, the maximum annual daily precipitation has an average of 21.6 cm and a standard deviation of 2.65 cm.

(a) What is the probability that the maximum annual daily precipitation will exceed 25 cm ?

(b) What is the maximum annual daily precipitation which has return period of 50 years ?

Soln.

(a) The parameters of Gumbel distribution are calculated as

( ) 406.2065.245.06.2145.0066.2

283.165.2

283.1

===

===

SX

S

Now,

224.2066.2

406.2025=

=y

The probability that the maximum annual daily precipitation will exceed 25 cm is given by

[ ] [ ]( )[ ]( )[ ]

1025.08975.01

224.2expexp1expexp1

1

=

=

=

=

=>y

yYPyYP

(b) For T=50 years, 02.05011

===

TP

[ ][ ] [ ]

( )[ ]( )

9021.3,0202.0exp,

02.01expexp,1,

02.0

=

=

=

>==>

yoryor

yoryYPyYPNow

yYP

So, ( ) cmx 468.28406.20066.29021.3 =+= The maximum annual daily precipitation which has return period of 50 years is 28.468 cm.

Q5. The service life (in hours) of a soil boring equipment is a random variable having Weibull distribution with 05.0= and = 0.4.

(a) How long can the equipment be expected to last ?

(b) What is the probability that the equipment will be in operating condition after 6000 hours ?

Soln.

(a) The equipment be expected to last for

( ) ( )( ) hours06.59453234.385.17884.0

1105.0 4.01

==

+=

(b) The probability that the equipment will be in operating condition after 6000 hours is given by

[ ] [ ]( )( )

1959.011

6000160004.0600005.0

=

=

=>e

XPXP

M6L4

Q1. The waste generated from a manufacturing plant is treated daily so that there is 90% probability that the treated effluent will meet the pollution control standards on a given day.

(a) What is the probability that the treated effluent will not meet the pollution control standards in exactly 2 of the next 7 days ?

(b) What is the probability that that the treated effluent will not meet the pollution control standards in atmost 2 of the next 7 days?

Soln.

(a) The probability that the treated effluent will not meet the pollution control standards on a given day is ( ) 1.09.01 =

The probability that the treated effluent will not meet the pollution control standards in exactly 2 of the next 7 days is given by

( ) ( ) ( ) 124.09.0)1.0(2 27227 === CXP (b) The probability that that the treated effluent will not meet the pollution control standards in

atmost 2 of the next 7 days is given by

( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )

974.0124.0372.0478.09.0)1.0(9.0)1.0(9.0)1.0(

210252

2761

1770

07

=++=

++=

=+=+==

CCCXPXPXPXP

Q2. A tower was built to a certain height against a design wind speed of 100 kmph which has a return period of 20 years.

(a) What is the probability that design wind speed will be exceeded within the return period ?

(b) If the design wind speed is exceeded, then the probability of damage to the tower is 70%; what is the probability that the tower will be damaged within three years ?

Soln.

(a) The probability that the design wind speed will be exceeded in a given year is

05.02011

===

Tp

The probability that the design wind speed will be exceeded within the return period is

( ) 6415.03585.0105.011 20 === (b) For the tower to be damaged in 3 years, the design wind speed may be exceeded 0,1,2 or 3

times in the 3 year period.

So the probability of no damage in 3 years is

8986.000006.00406.08574.0

)05.0()3.0()]95.0()05.0(3[)3.0(])95.0)(05.0(3)[3.0()95.0(1 332223

=

+++=

+++=

So the probability of damage in 3 years is

( ) 1014.08986.01 == Q3. The drainage system of a city has been designed against a 50 year rainfall. If the design rainfall is exceeded, then the drains get flooded.

(a) What is the probability that the drains will get flooded for the first time on the second year after the drainage system is constructed ?

(b) What is the probability that the drains will get flooded for the first time within 2 years after the drainage system is constructed?

Soln.

(a) The probability of occurrence of a 50 year rainfall in a given year is

02.05011

===

Tp

The probability that the drains will get flooded for the first time on the third year after the drainage system is constructed is given by

[ ] ( )( )0196.0

02.0102.02=

==TP

(b) The probability that the drains will get flooded within 2 years after the drainage system is constructed is given by

[ ] ( )( )( )( ) ( )( )

0396.00196.002.0

98.002.098.002.0

02.0102.02

1211

2

1

1

=

+=

+=

=

=

t

tTP

Q4. In a certain stretch of a highway, the average number of accidents is 4 per year. Assuming that the occurrence of accidents is a Poisson process,

(a) what is the probability that there would be no accidents on the highway next year?

(b) what is the probability that there would be exactly 3 accidents next year?

(c) what is the probability that there would be 4 or more accidents next year?

Soln.

(a) The average number of accidents is 4 per year i.e, the mean rate of occurrence 4=

Now, ( ) 414 ==t The required number of accidents 0=x .

The probability that there would be no accidents on the highway next year is given by

( ) ( ) 0183.0!0

4!

0 40

==== ee

x

tXP tx

t

(b) The required number of accidents 3=x .

The probability that there would be exactly 3 accidents next year is given by

( ) ( ) 1952.0!3

4!

3 43

==== ee

x

tXP tx

t

(c) The required number of accidents 4=x or more.

The probability that there would be 4 or more accidents next year is given by

( )

5667.01952.01465.00733.00183.01

!41

!44

3

0

4

4

4

=

=

=

=

=

=

x

x

x

x

t

ex

ex

XP

Q5. In an extensive construction project, mean number of boulders encountered during soil boring is 2 per 50 m. Assuming that the occurrence of boulders during soil boring is a Poisson process,

(a) What is the probability that exactly 1 boulder will be encountered within the next 30 m ?

(b) What is the probability that atmost 2 boulders will be encountered within the next 50 m ?

Soln.

(a) The required number of boulders encountered 1=x in 30 m.

The average number of boulders encountered is 2 per 50 m, i.e, the mean rate of occurrence

502

=

( ) 2.130502

, =

=tNow

The probability that exactly 1 boulder will be encountered within the next 30 m is given by

( ) ( ) ( ) 3614.0!12.1

!1 2.1

1

==== ee

x

tXP tx

t

(b) The required number of boulders encountered is atmost 2 in 50 m, i.e, 2=x or less.

The average number of boulders encountered is 2 per 50 m, i.e, the mean rate of occurrence

502

=

( ) 250502

, =

=tNow

The probability that atmost 2 boulders will be encountered within the next 50 m is given by

( ) ( )

6767.02707.02707.01353.0

!22

!12

!02

!2

22

21

20

2

0

=

++=

+

+

=

=

=

eee

ex

tXPx

tx

t

Q6. If a steel component has a mean working strength of 110 MPa with a standard deviation of 20 MPa, what is the probability that a sample mean strength of a sample of 10 such steel components will be between 90 MPa and 120 MPa ?

Soln.

10,20,110,120,90 21 ===== nMPaMPaMPaXMPaX

Now,

( )( ) ( )

( )[ ][ ]

9427.00002.09429.0

9992.019429.016.319429.0

16.3)58.158.116.3,

58.110/2011012016.3

10/2011090

21

=

=

=

==

=

==

=

ZPZPZP

ZPNow

ZandZ

The probability that a sample mean strength of a sample of 10 such steel components will be between 90 MPa and 120 MPa is 9427.0 .

M7L1

Q1. The interarrival time of rainstorms in a certain catchment is expressed by an exponential distribution

( ) t

T etf

=

1

The time between successive rainstorms was observed as 22 days, 4 days, 35 days, 18 days , 14 days, 8 days.

Determine the mean inter arrival time by the

a) method of moments

b) the maximum likelihood method.

Soln.

(a) The first moment about the origin of fx(x) is

[ ]

daystXThereforeor

eteor

dtte

ii

tt

t

83.1661

,

,

,

1

6

1

0//

0

/

====

=

+=

=

=

(b) Assuming random sampling, the likelihood function of the observed values is

( )

( )

=

=

=

=

6

1

6

6

1621

1exp

exp1;,...,

ii

i

i

t

ttttL

The estimator can now be obtained by differentiating the likelihood function L with respect to .

Hence,

daysor

tor

tor

ttor

ttt

L

ii

ii

ii

ii

ii

ii

ii

83.16,61

,

61,

01exp16,

01exp1exp6

6

1

6

1

6

1

6

1

8

2

6

16

1

66

1

7

=

=

=

=

+

=

+

=

=

=

==

=

=

=

Q2. Fifty steel rod specimens are selected from a large batch and they are tested under identical conditions. The sample mean of the tensile strength of these steel rods is found to be 104 MPa. If the standard deviation of the population is known to be 16.5 MPa, determine the 99% and the 95% confidence interval of the mean tensile strength of the steel rod specimens.

Soln.

(a) For the 99% confidence interval,

1-=1-0.99=0.01

From the standard normal table,

( )( )( )

575.2,995.0,

005.01,2

1

005.0

005.0

005.0

2/

=

==

=

zor

zZPorzZPor

zZP

( ) 00.6575.250

5.16, 2/ ==

z

nNow

The 99% confidence interval of the mean tensile strength of the steel rod specimens is

( )( ) MPaei

MPa110;98,.

0.6104;0.6104 +

(c) To determine the 95% confidence interval,

( )( )

96.1,975.0,

205.01

005.0

025.0

025.0

=

=

=

zor

zZPor

zZP

( ) 57.496.150

5.16, 2/ ==

z

nNow


( )( ) MPaei

MPa57.108;43.99,.

57.4104;57.4104 +

It is more likely that the larger interval will contain the mean value than the smaller one. Hence the 99% confidence interval is larger than the 95% confidence interval.

M7L2

Q1. A random sample of 50 steel rod specimens were selected from a batch of steel rods prepared under identical conditions. The sample mean of the tensile strength of these steel rods is found to be 104 MPa and the sample standard deviation is found to be 16.5 MPa. Determine the 99% confidence interval of the mean tensile strength of the steel rod specimens.

Soln.

Here 50=n .

So, nS

X/

has a t-distribution with ( ) 491.. == nfod degrees of freedom.

For the 99% confidence interval, /2=0.005

From the t-Distribution table, we get the value of 49,005.0t for 49995.0 == fandp , 684.249,005.0 =t

( ) 26.6684.250

5.16, 1,2/ ==nt

nNow


( )( ) MPaei

MPa

26.110;74.97,.

26.6104;26.6104

99.0

99.0

=

+=

[Note: This interval is larger compared to that where the standard deviation of the population was known. This is expected because uncertainty is greater when the standard deviation is unknown.]

Q2. In a particular construction site, the in situ density of soil is measured at 15 different locations. The sample variance is found to be ( )232 /442 mKNs = . What is the 95% upper confidence limit of the population variance 2 ?

Soln.

Sample size n = 15, sample variance ( )232 /442 mKNs = So ( )2

21

sn will have chi-square distribution with ( ) 141 =n degrees of freedom.

Now, from chi-square tables, for 1505.0 == nand ,

57.614,05.01, == cc n

Hence, the 95% upper confidence limit of the population variance 2 is

( ) ( )231,

2

/86.94157.6

)442(141mKN

c

sn

n

==

Q3. While assembling laboratory equipments in a factory, it was found that 24 out of the 100 equipments assembled were defective.

(a) What is the proportion p of assembled laboratory equipments that will not be defective ?

(b) What is the 95% confidence interval of p ?

Soln.

(a) The point estimate of proportion p of assembled laboratory equipments that will not be defective is given by

76.0100

24100 =

=p

(b) The 95% confidence interval of p is

( )( )843.0;677.0,

083.076.0;083.076.0,100

)76.01(76.096.176.0;100

)76.01(76.096.176.0,.

)1(;

)1( 2/2/

or

or

ei

n

ppzp

n

ppzp

+

+

+

M7L3

Q1. The specifications for a certain timber construction project require timber with mean modulus of rupture 30 N/mm2. If 8 timber specimens selected at random have a mean breaking strength of 28.5 N/mm2 with a standard deviation of 2.5 N/mm2, test the null hypothesis =30 N/mm2 against the alternative hypothesis

697.18/5.2305.28

/0

=

=

=

ns

Xt

As 998.2697.1 > so the null hypothesis cannot be rejected at the significance level of 0.01.

Q2. A sample n1 =35 current meters has mean lifetime of 5200 hours with a standard deviation of 350 hours. Another sample n2 =35 current meters of a different make has mean lifetime of 4950 hours with a standard deviation of 435 hours..Can it be claimed that the current meters of the first make are superior in terms of the lifetime at 0.05 level of significance ?

Soln.

The null hypothesis is H0: 1 2=0.

The alternative hypothesis is H1: 1 2 >0 .

The level of significance =0.05

The criteria for rejection of the null hypothesis is

( ) 645.102

22

1

21

21 >

+

=

n

s

n

s

XXZ

Now,

( ) 65.235

43535

350049505200

22=

+

=Z

As 2.65 is greater than 1.645, so the null hypothesis must be rejected at 0.05 level of significance.

For 2.65Z = , the P value is 0.004

Q3. The field densities of soil at two regions A and B are tested. The random sample readings are as follows:

Field Density of soil (KN/m3)

Reg. A 1653 1611 1708 1730 1681 1604 1658 - -

Reg. B 1648 1665 1678 1740 1743 1690 1613 1685 1735

Use 0.01 level of significance to test whether the difference between the means of the soil densities at the two locations is significant.

Soln.

The null hypothesis is H0: 1 2=0.

The alternative hypothesis is H1: 1 2 0 .

The level of significance =0.01

The criteria for rejection of the null hypothesis is 977.2977.2 >< tort

Where

+

=

21

21

11nn

S

XXt

p

and 2.977 is the value of 005.0t for ( ) 14297 =+ degrees of freedom. Now, 3.1975,2191,6.1688,6.1663 222121 ==== ssxx

And

( ) ( ) ( )( ) ( )( )

47.45,

7.206714

3.19758219162

1121

222

2112

=

=

+=

+

+=

p

p

Sornn

snsnS

Hence

09.15039.05498.0

91

7147.45

6.16886.166311

21

21=

=

+

=

+

=

nnS

XXt

p

As 09.1 is less than 977.2 but greater than 977.2 , the null hypothesis cannot be rejected at the 0.01 level of significance.

Q4. The maximum permitted population standard deviation in the tensile strength of steel rods is 22 MPa. Use the 0.05 level of significance to test the null hypothesis = 22 MPa against the alternative hypothesis > 22 MPa, if the tensile strength of 20 randomly selected steel rods s = 25 MPa .

Soln.

Here null hypothesis is MPa22=

The alternative hypothesis is MPa22>

Level of significance =0.05

The criteria for rejection of null hypothesis is 1.302 >

where ( )20

22 1

sn =

and 30.1 is the value of ( ) 19120205.0 =for degrees of freedom. Now,

( )

( )( )( )53.24

2225120

1

2

2

20

22

=

=

=

sn

As 24.53 is less than 30.1, the null hypothesis cannot be rejected at the 0.05 level of significance.

Q5. It is to be determined whether there is less variability in the treatment efficiency of waste water through process A than that in process B. If the residual pollutant concentration of 8 random samples of the treated waste water through process A and 9 random samples of the treated waste water through process B are tested, it is found that s1=0.15 mg/L and s2=0.35 mg/L. Test the null hypothesis 12=22 against the alternative hypothesis 12

The alternative hypothesis is 123.5

where 2122 / ssF =

and 3.5 is the value of 8705.0 andforF degrees of freedom respectively.

Now,

( )( ) 44.515.0

35.02

2

21

22

===

s

sF

As 5.44 is greater than 3.5, so the null hypothesis must be rejected at the 0.05 level of significance.

Q6. The daily consumption of water in a town for a month is given below. Check whether the daily consumption follows normal distribution or not by plotting on normal probability paper. Determine the mean daily consumption and the standard deviation.

Day Cons. (ML)

Day Cons. (ML)

Day Cons. (ML)

Day Cons. (ML)

Day Cons. (ML)

Day Cons. (ML)

1 15.60 6 18.82 11 12.36 16 13.80 21 16.95 26 14.67

2 10.52 7 16.27 12 15.36 17 16.13 22 8.90 27 10.75

3 12.74 8 12.63 13 14.77 18 11.37 23 18.06 28 13.45

4 11.80 9 15.07 14 11.66 19 13.00 24 16.77 29 14.44

5 17.00 10 13.94 15 12.79 20 13.29 25 15.31 30 15.07

Soln.

The daily consumptions of water are first arranged in ascending order. Then their plotting positions are determined by m/(N+1), where N=30 and m=rank of the observed data point when arranged in ascending order of their values.

The normal probability plot is prepared and the data is found to plot almost as a straight line. Thus the daily consumption of water follows normal distribution.

Fig. Normal Probability Plot

From the plot, the value of daily consumption of water corresponding to cumulative probability of 0.5 is 14.5 ML . Thus the mean daily consumption of water is 14.5 ML.

The standard deviation is given by the inverse of the slope of the straight line. It is obtained roughly as 6.28 ML.

M7L4

Q1. The crushing strengths of M25 concrete cubes were tested in the laboratory. Based on the test results, normal distribution function appears suitable in describing the strength of the concrete cubes. Use the 2 test to determine the goodness of fit of normal distribution at the 5% significance level.

Pro

babi

lity

Daily Consumption (Million Litres)

Interval of crushing strength (KN/m3)

Observed frequency ni

30 3

Soln.

The observed and theoretical frequencies and the errors are computed in tabular form. As the parameters of the normal distribution are estimated from sample mean and sample variance, hence the degrees of freedom for the distribution is ( ) 437 ==f . At the 5% significance level and for f =4, the value of 49.923,95.0 =

Interval of crushing strength (KN/m3)

Observed frequency ni Theoretical frequency ei

( )i

ii

e

en2

30 3 1.81 0.7824

Comparing the value of 49.923,95.0 = with ( ) 159.4

2

=

i

ii

e

en, it can be can be found that

4.159

Determine 0 , 1 and 2 and evaluate the conditional variance ( )21 ,| xxYVar . Also find the correlation coefficient r2.

Soln.

The required calculations for multiple linear regression are shown in tabular form.

Obs No.

ET (Y) (mm)

W spd (x1) (kmph)

Temp (X2) (mm)

(x1i-x1mean

)2

(x2i-x2mean

)2

(x1i-x1mean)(x2i-x2mean)

(x1i-x1mean

)(yi-ymean)

(x2i-x2mean

)(yi-ymean)

1 7 12 22.30 13 7.728 10.008 16 12.51 7.63 0.399

2 6 10 24.50 31 0.336 3.248 31 3.19 6.41 0.171

3 5 8 22.30 58 7.728 21.128 49 18.07 4.10 0.808

4 11 15 21.90 0 10.11 1.908 0 1.59 10.18 0.674

5 13 19 25.60 12 0.270 1.768 5 0.78 14.63 2.656

6 12 22 26.20 41 1.254 7.168 3 0.56 17.43 29.448

7 26 25 27.80 88 7.398 25.568 136 39.44 20.47 30.557

8 11 14 23.80 3 1.638 2.048 1 0.64 9.77 1.514

9 13 18 29.00 6 15.36 9.408 4 5.88 14.59 2.539

10 11 13 27.40 7 5.382 -6.032 1 -1.16 9.78 1.482

Sum 115.000 156 250.80 258 57.21 76.220 247 81.500 70.247

Mean 11.500 16 25.080 26

From the table, the following can be calculated

( )( ) ( )( ) ( )( )( )( ) ( ) ( )( )

=+

=+

=

======

====

yyxxxxxxxxand

yyxxxxxxxxNow

syysy

xx

iiiii

iiiii

xxYiY

2212

2212112211

112211122

111

2,|

22

21

,

035.101210

247.70,28.34

91

,5.1110115

08.2510

80.250,16

10156

21

Therefore,

5.8122.5722.7624722.76258

21

21

=+

=+

Solving the above equations,

5.8249.0,8825.0

22110

21

==

==

xx

Correlation coefficient is 707.028.34

035.1012 ==r

Thus, the mean monthly evapotranspiration (in mm) is given by

( ) 2121 249.08825.05.8,| xxxxYE ++= (where x1: average wind speed (in kmph), x2: average temperature (in degree Celsius))

The conditional variance ( ) 035.10,| 21 =xxYVar and the correlation coefficient 707.02 =r .

M7L6

Q1. The coefficient of correlation between rainfall and runoff in a certain catchment is 0.61. Given the following data,

Mean Standard Deviation

Rainfall (cm) 20.5 2.6

Runoff (cm) 7.5 1.25

(a) Find the runoff when rainfall is 18 cm.

(b) Find the rainfall when the runoff is 5 cm.

Soln.

Let Y be runoff and X be rainfall.

So, for estimating the runoff, we have to use the regression of Y on X and for the purpose of estimating the rainfall, we have to use the regression of X on Y.

Now,

61.0,25.1,6.2

,5.7,5.20

=

==

==

yx

YX

Therefore, regression coefficients are :

293.06.2

25.161.0

269.125.16.261.0

|

|

==

==

XY

YX

Hence, the regression equation of Y on X is

( )494.1293.0

5.20293.05.7)(|

+=

=

=

XYXYxxyy XY

Similarly, the regression equation of X on Y is

( )98.10269.1

5.7269.15.20)(|

+=

=

=

YXYXyyxx YX

(a) When rainfall i.e, X = 18 cm,

Runoff is

( )cm

XY

698.6494.118293.0

494.1293.0

=

+=

+=

(b) When runoff i.e, Y = 5 cm,

Rainfall is

( )cm

YX

33.1798.105269.1

98.10269.1

=

+=

+=

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