Sample Paper 2 _HCI 2010_solution

8/10/2019 Sample Paper 2 _HCI 2010_solution

1/8

1

Qn Solution

1(a)

2

1d

(3 2 )x

x+=

1(3 2 )

( 1)(2)

xc

+

+

=1

2(3 2 )c

x

+

+

1(b) 11 33

0dxe x =

13

1 3

0

1e

3

x

= ( )1

e 13

2(i) 52 2 2(5 ) 5V x x x x= =

3

2d 5

10d 2

Vx x

x=

Whend

0d

V

x= ,

1

25

4 02x x

=

0x = (NA) or 16

x 16- 16 16

+

d

d

V

x

+ 0

Hence max. capacity at 16x = i.e. 16h = .

Max. capacity of container 2(16) (5 16) 256= = cm3

2(ii) When 4x = ,d d d

.d d d

V V x

t x t=

=3

25

10(4) (4) (0.02)2

= 0.4 cm3s

-1

3(i)2

21

1y

x= +

+

2 2

d 4

d (1 )

y x

x x

=

+

Whenx= 0,y= 1.Wheny= 0,

2

2

2

21

1

1 2

1

1

x

x

x

x

=+

+ =

=

=


2/8

2

The coordinates of points A,B and C are (0,1)A , (1, 0)B and ( 1, 0)C

AtB,d

1d

y

x=

Thus equation of tangent atBis 1y x= + .

When 0x = , 1y x= + = 1.

Thus the tangent atBpasses through (0,1)A .

At C,d

1d

y

x=

Thus equation of tangent at Cis 1y x= + .

When 0x = , 1y x= + = 1.

Thus the tangent at Cpasses through (0,1)A .

3(ii) AtB, gradient of tangent m1= 1 At C, gradient of tangent m2= 1

Since 1 2 1m m = , the two tangents are perpendicular to each other and hence angleBAC

= o90 .

4 At point of intersection,

11kx

x=

2 1 0kx x + = Since there is only one common point, equation has only one real distinct root.

Hence 2( 1) 4( )(1) 0k =

14

k =

Thus 21

1 04x x + =

2( 2) 0x =

2x =

4(i)

y = 1

10


3/8

3

4(ii)Required area =

2

1

1 1 1(2) 1 d

2 2x

x

= [ ]2

1

1ln

2x x

=1

ln 22

4 14

m>

5(a) 2lg lg 2 lg( 1)x x+ = + 2lg 2 lg( 1)x x = +

22 1x x = +

1 1 4(2)( 1)

2(2)x

=

11 or - (NA)

2x =

5(bi)

5(bii) 2 ln 4x x x < 2 ln 4 0x x x <

0.212 2.03x < <

Replacexby ex , 2(e ) e ln(4e )x x x < 2e e ln 4 ln ex x x < + 2e e ln 4x x x < + 2e e ln 4x x x <

Thus2

e e ln 4 0.21159 e 2.0313x x x

x < < < 1.55 0.709x < <

(1, ln4)

2.030 0.212


4/8

4

Qn Solutions

6

(i) Quota sampling

(ii) One disadvantage is that it is biased as the interviewer only choose interviewees whom heknows.

(iii) To obtain a sample of 50 students using systematic sampling, the committee could choose atrandom one of the first 10 students who left the function hall and subsequently choose every10th student after the first student has been chosen. E.g. if the 3rd student is chosen, thensubsequently the 13

th, 23

th, 33

th, 43

th, 53

th, 63

th, 493

thstduent will be interviewed.

7

(i)P(H) =

600 20.4

1500 5= =

(ii)

730 210 170 1110P( ' ) 0.74

1500 1500H L

+ + = = =

(iii)

210 21P( ) 0.553

210 170 38

H L = = =

+

Or:

P( )P( )

P( )

210 1500 210.553

1500 380 38

H LH L

L

=

= = =

Since21

P( ) P( )38

H L H= or ( ) ( ) ( )P H L P H P L ,HandLare not independent within

the sample.

8 0.5 > $1000

A0.5 $1000

0.7 > $1000B

0.3 $1000

0.8 > $1000C

0.2 $1000

Given thatP(the customer carries out transactions exceeding $1000) = 0.725

(0.5) 2 (0.7) (1 3 )(0.8) 0.725

0.5 0.075

0.15

p p p

p

p

+ + =

=

=

p

2p

13p


5/8

5

8(i) P(customer carries out transactions exceeding $1000 at branchA) = 0.15(0.5)= 0.075

(ii) P( visits branch A customer carries out transaction exceeding $1000)

P(customer visits branch A and carries out transaction exceeding $1000)=

P(customer carries out transaction exceeding $1000)

0.075=

0.725

3= = 0.103

29

9

(i)

LetXbe no. of gradeAapples out of 12.

(12,0.7)

P( 8) 0.231

X B

X = =

(ii)

Let Ybe the no. of gradeAapples out of 50.

(50,0.7)

Since is large such that = 35 > 5 and (1 )= 15 > 5,

(35,10.5) approximately.

P( 30) P( 29.5) 0.955CC

Y B

n np n p

Y N

Y Y

> =

10 (i) Using G.C. 47.5x = 833 = 47.6 ( 3 s.f. )

unbiased estimate of population variance = s2

= ( )2

22.68142 514.4468132 514= = ( 3 s.f. )

(ii)

Ho : = 60H1 : < 60 at 5% level of significance

Using G.C. pvalue = 0.027203Since pvalue = 0.027203 < 0.05, we reject Ho and conclude that there is sufficientevidence at 5% level that the average time for students to be ready for morningassemble is less than 1 minute.

(iii) Population is normal with known variance.

11(i) 1.6 1.2 1.4 2.3 2.6 2.9 2.7 3.0 2.6 2.27

10

k+ + + + + + + + +=

k= 2.4

(ii)


6/8

6

(iii)

Using G.C. the product moment correlation coeffr = 0.9360025 = 0.936 ( 3 s.f. )

Since r is close to 1, we may say that xandyhave a strong positive linear correlation.However, a linear model is only appropriate if a linear relationship betweenxandycan alsobe observed from the scatter diagram of the data points.

(iv)

0.13710144 0.83043478

0.137 0.830

y x

y x

= +

= +

For each increase of 1 m2of shelf space, sales will increase by $ 13.70 or $ 13.71

(v) When x= 13, y= $ 261.27535 = $ 261.28

(vi) Since x= 20 falls outside the range of data on which we obtained the regression line,

extrapolation of the observed data points is not advisable, and thus, the estimate of thevalue of yis not reliable when x= 20.

12

(i)

2 2(320,10 ) and (315, )X N Y N

Given that P( 325) 0.25Y =

325 315P( ) 0.25

10P( ) 0.25

10P( ) 0.75

100.6744897

14.826 14.8 (to 3 s.f.)

Z

Z

Z

=

=

< =

=

=

(ii)1 2

1 2

2 (10,1079.243866)

P( 2 0) 0.6196 0.620 (to 3 s.f.)

X X Y N

X X Y

+

+ > =


7/8

7

(iii) We assume thatXand Yare independent random variables.

(iv) Let nbe the sample size.210

320,

P(315 325) 0.95

315 320 325 320P 0.95

10 10

P 0.952 2

1 2P 0.952

Z 0.0252

From GC, P( 1.95996) 0.025

1.959962

15.4

X Nn

X

Zn n

n nZ

nZ

nP

Z

n

n

< < >

< < >

< < >

< >

<

The lease value of n is 16.Anwer is not dependent on the Central Limit Theorem asXis normally distributed

Xis also normally distributed.

13

(a)

( )21 2 60..... ~ N 60 1 , 60 0.8S X X X = + + + by G.C.

( )55 65 0.5802596 0.580P S< < = = ( 3 s.f. )

(b)(i) unbiased estimate of is

( 60) 75060 60 65

150

x xx

n n

= = + = + =

unbiased estimate of 2 is ( )

( )( )2

22601

601

xs x

n n

=

( )

22 1 750

5762149 150

13.50336 13.5 3 s.f .

s

=

= =

(ii)13.50336

~ ( 60 , )150

X N approx. by C.L.T since n= 150 (large)

(iii) ( 59.8 ) 0.25249 0.252P X < = = ( 3 d.p. )

or 0.25252 0.253= = ( 3 d.p. )


8/8

8

(iv)13.50336

~ ( 60 , )X Nn

approx.

( )60.5 0.04

60.5 600.04

13.50336

60.5 601 0.04

13.50336

60.5 600.96

13.50336

P X

P Z

n

P Z

n

P Z

n

>

165.5458n > least n = 166

Sample Paper 2 _HCI 2010_solution

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Transcript of Sample Paper 2 _HCI 2010_solution