solutions math s p2 stpm 2011 trial sabah

8/4/2019 Solutions Math s p2 Stpm 2011 Trial Sabah

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STPM 950/2, (EXCEL II 2011 JPN Sabah)

Marking Scheme MATHEMATICS S (Paper 2)

1. (a) The information can simplified as:

Male Female Total

Wearing spectacles 20 44 64Without glasses 28 28 56

Total 48 72 120

| = =

[1 mark]=

[1 mark]

(b)

= =

= =

[2 marks]

Therefore eventsA andB are not independent. [1 mark]

2. (a) (i) Plot scatter diagram correctly. [2 marks]

(ii) Plot scatter diagram correctly. [2 marks]

(b) Farthers height has a strong linear relationship with sons height.

Because points lie very close to each others. [2 marks]

3. 95% confidence interval for is: = [1 mark]

Thus, = 0.4 [1 mark] = 0.4 [1 mark]

= (

) [1 mark]

Hence, sample size, n = 600 [1 mark]


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*This question paper is CONFIDENTIAL until the examination is over.

4. (a)

Stem Leaf

5 1

5 9

6 2 3 3 4 4 4 4 46 5 7 8 8 8 9 9

7 0 2 2 3

7 6 7

Key: 5 9 mean 59 [2 marks]Number of observations, n = 23

Median , Q2 = (23 +1)th observation

= 12th

observation

= 67 [1 mark](b) Q1 = 6th observation = 64Q3 =18

thobservation = 70 [1 mark]

Refer graph. [3 marks]

5. (a) Let Xbe a random variable representing the number of free telephones at 11 a.m.

on that Monday.

ThenX~B(8, 0.2)

P(X=x) =8

(0.2)

x(0.8)

8x[1 mark]

= If

then 1 1 [2 marks]

8x

4 (x + 1)

5 x 4x 0.8 [1 mark]Thus, So,X= 1 has the highest probability, i.e. the most likely number of free telephones

is 1. [1 mark]


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5. (b) (i) P(X= 2) =8 (0.2) 2 (0.8) 6 = 0.2936

The probability that exactly 6 telephones are used is 0.294. [2 marks]

(ii) P(X2) = 1P(X1)

= 1P(X= 1)P(X= 0)

= 10.33550.1678= 0.4967

The probability that at least 2 telephones are free is 0.497. [2 marks]

6. Population mean, = 32.5 yearsPopulation standard deviation, = 9.2 yearsFinite population size,N= 400

Random sample size, n = 50

Since sampling is done at random without replacement from finite population, thus

the sample mean will have a sampling distribution with:Mean, = = 32.5

and standard error, = =

[1 mark]

= 1.2186 [1 mark]

Now as sample size n = 50 30, then by Central Limit Theorem, we have = = [1 mark]

P = [1 mark]=

= 10.0202

= 0.9798 [1 mark]


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7. (a) [ ][ ]=

[][]=

= 0.9707 0.971 [3 marks]The coorrelation coefficient r= 0.971 indicates that a strong positive relationship

exists between the sales,y (in thousand RM) and the weekly advertising expense,x

(in thousand RM). [1 mark]

(b) The least squares regression line is

y = bx + a

1 1 1

2

2

1 1

n n n

i i i i

i i i

n n

i i

i i

n X Y X y

b

n X X

=

= 2.2468

2.25 [2 marks]

a y bx

=

= 74.0667 74.1 [2 marks]Thus the regression line isy = 2.25x + 74.1 [1 mark]

b = 2.25 indicates that a one-unit change in x is associated with 2.25 units change in

y. This means that each one thousand ringgit in advertising expense is associatedwith 2250 ringgit increase in sales. [1 mark]


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8. (a) Graph drawn correctly [3 marks]

(b) Median = 1291Semi interquartile

Q1= 121.5, Q3 = 138

Semi interquartile = (Q3Q1)= 8.25 [5 marks](c) Mean =

= 129.7 [2 marks]

Standard Deviation = = 13.1 [2 marks]

9. (a)

[5 marks]

(b) The critical path is ADFH - I [1 mark]

(c) Minimum time to complete the project is 17 weeks [1 mark]

(d)

Task Independent float (in weeks)

A 4 -0 -4 =0B 6-0 -3=3

C 9 -4 -5 = 0

D 642 = 0

E 156 -1 = 8

F 1367 = 0

G 17116= 0

H 15-13-2=0

I 17-15-2=0

[2 marks]

0 01

4 42

6 63

A

4 D

G

C

B

E

S

I

HF

3

2

5

1

7

6

2

2

9 114

13 13

5

15 156

17 17

7


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10. Lasperyres price index for year 2009

=

= 111.35 [2 marks]

Lasperyres price index for year 2010

=

= 119.86 [2 marks]

For the base year 2008, price increased about 11.35% and continuous to increase

19.86% from 2009 up to 2010. [2 marks]

11. (a) Plot time series correctly. [3 marks]

(b) Linear trend is not appropriate for the given time series data because there is no

clear increasing or decreasing trend. [1 mark]

(c)

Year Quarter

Fixed

Deposits

(RM 000)

Moving

average

2007

1 13

2 20

3 35 22.625

4 22 23.250

2008

1 14 23.750

2 24 24.000

3 35 24.500

4 24 25.250

2009

1 16 26.750

2 28 28.750

3 43 30.000

4 32 30.500

2010

1 18 29.500

2 30 28.875

3 50

4 20 [3 marks]


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(d)

Year Quarter

Fixed

Deposits

(RM 000)

Moving

average

2007

1 132 20

3 35 22.625 12.375

4 22 23.250 1.250

2008

1 14 23.750 9.750

2 24 24.000 0.000

3 35 24.500 10.500

4 24 25.250 1.250

2009

1 16 26.750 10.750

2 28 28.750 0.7503 43 30.000 13.000

4 32 30.500 1.500

2010

1 18 29.500 11.500

2 30 28.875 1.125

3 50

4 20

Quarter

Year

1 2 3 4

2007 12.375 1.250

2008 9.750 0.000 10.500 1.250

2009 10.750 0.750 13.000 1.500

2010 11.500 1.125

Average

seasonal

variation

10.667 0.125 11.958 1.000

Adjusting

value 0.104 0.104 0.104 0.104

[Adjusted

seasonal

varation

10.771 0.021 11.854 1.104

[2 marks]


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12. (a) Let x = number of Alpha to produce

y = number of Beta to produce

Maximize profit Z= 5x + 8y

Subject to

16 x + 8y 80004 x + 12y 48006 x + 6y 3600

x [5 marks](b) Graph correctly drawn. [4 marks]

Number of Alpha = 300

Number of Beta = 300Maximum Profit = 5(300) + 8 (300)

= RM 3900 [4 marks]


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solutions math s p2 stpm 2011 trial sabah

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