YFB 3102 SMK SACRED HEART, math m p2 2015Section A [45 marks]

Answer all questions in this section.

1. The frequency distribution of the marks obtained by the 5S1 students in the first school examination is as follows.

Marks Number of students60 – 6970 – 7980 – 8990 - 99

28715

(a) Calculate the median value, first quartile and third quartile values and hence draw the box plot.

[7 marks](b) The school decided to give the top scorers some prizes. The number of recipients only limited to the top five percent of the students. Find the percentile of the prize recipient. [2 marks]

2. A game is played with a set of ten cards, which are numbered 11, 12, 13, 14, … , 20 respectively. The cards are placed in a bag and a player takes a card at random from the bag. If the number on this card is prime (i.e. 11, 13, 17 or 19 ) , the player wins the game, and if the number is even the player loses the game. If the number on the card is 15 the player takes a second card at random from the nine remaining cards in the bag. The player wins the game if the number on the second card is prime and loses the game otherwise.

Find the probability that in a particular game,

(i) The player wins, [2 marks]

(ii) Only one card is taken from the big given that the player loses. [3 marks]

3. (a) The random variable X has a normal distribution with mean µ and standard deviation σ. Given that P ( X < 30) = 0.14 and that P ( X< 60 ) = 0.79, Calculate the values of µ and σ, giving your answer correct to three significant figures. [5 marks]

(b) The random variable Y has a normal distribution with mean 10 and standard deviation 2. Find the value of the constant a such that P (10−a<Y <10+a )=0.95. [5 marks]

4 For ten pairs of married couples from the Sibujaya town, the height of the husband ( x cm ) and the height of the wife ( y cm ) are given as

∑ x=1680 ,∑ x2=282 280 ,∑ y=1620 ,∑ y2=262 450 ,∑ xy=272 179.

Calculate the coefficient of correlation and comment on the result. [ 3 marks ]

5. The table below shows the prices of fish and the quantities of it bought by a housewife at the Sibu Central wet market in January 2013 and January 2015.

Fish January 2013 January 2015Price

(RM per kg)Quantity

(kg)Price

(RM per kg)Quantity

(kg)Selar KuningKembungBawal PutihTenggiriParang

4.008.0010.0012.0011.00

53222

5.0010.0014.0013.0012.00

62121

( 2013 = 100 )

(a) Calculate the Laspeyres quantity index, and comment on the housewife’s change in expenditure on fish. [ 3 marks ]

(b) Calculate the Passche price index, and comment on the housewife’s change in expenditure on fish. [ 3 marks ]

6. The following table shows the sale of a housing developer in Sibu for each quarter from 2010 to 2012.

Year 1st Quarter 2nd Quarter 3rd Quarter 4th Quarter2010 23 32 68 332011 27 40 73 372012 30 43 77 43

(a) Plot a time series, and comment on the appropriateness of a linear trend. [ 4 marks ]

(b) Calculate the centred four-quarter moving averages for this time series. [ 4 marks ]

(c) Calculate the seasonal indices using an additive model. [ 4 marks ]

Section B [15 marks]

7. The dates of birth in a large population are distributed throughout the year so that the probability of a

randomly chosen person’s date of birth being in any particular month may be taken as 1

12 .

(i) Find the probability that, of 6 people chosen at random, exactly two will have birthdays in January. [3 marks]

(ii) Find the probability that, of 8 people chosen at random, at least one will have a birthday in January. [3 marks]

(iii) N people are chosen at random. Find the least value of N so that the probability that at least one will have a birthday in January exceeds 0.9. [4 marks]

(iv) Find the probability that, of 100 people chosen at random, at least 40 will have birthday in May, June July or August. [5 marks]

8. The masses and height of eight students from U6A4 are shown in the following table,

Student A B C D E F G HMass ( x kg) 52 55 51 58 58 54 49 48Height ( y cm ) 165 173 169 175 176 169 167 159

(a) Plot a scatter diagram for the above data. [ 2 marks ]

(b) Calculate the Pearson correlation coefficient r between x and y. [ 3 marks ]

(c) Comment on the value of r with respect to the scatter diagram in (a). [ 1 mark ]

(d) Find the equation of the least squares regression line in the form y = a + bx , where a and b are constants. [ 4 marks ]

(e) Estimate the height of a student whose mass is 45 kg. Comment on your estimation. [ 2 marks ]

(f) Calculate the Spearman rank correlation coefficient rs between x and y . [ 3 marks ]

“ The fear of the Lord is the beginning of knowledge , but fools despise wisdom and discipline . “

[ Proverbs 1 : 7 ]

TEH 2015 STPM 950/2 Mathematics M 1st Trial Exam.

Marking scheme ( 2015 1 st Trial Examination 950/2 Mathematics M Paper 2 ) SMK Sacred Heart, Sibu .

Q1 (a) Median class is 79.5 – 89.5

median =79.5+( 12

(32 )−10

7 ) (89.5−79.5 )=88.07 [ M1A1 ]

First quartile class is 69.5-79.5

fIrst quartile , Q1 = 69.5 + (14

(32 )−2

8¿ (79.5−69.5 )=77 [ M1 ]

Third quartile class : 89.5-99.5

third quartile, Q3 = 89.5+( 34

(32 )−17

15 ) (99.5−89.5 )=94.17 [ M1 ]

Box plot : [ Scale 1, D2 ]

Q1(b) Top five percent 95 percentile = 89.5+( 95100

(32 )−17

15 ) (99.5−89.5 )=98.43 [ M1A1]

Q 2 .(i) Probability that the player wins is 4

10+ 1

10 ( 49 )=4

9 [M1A1]

(ii) P ( only one card is taken / the player loses )

= P (only one card is taken∧the player loses)

P ( the player loses ) =

510

1−49

= 910 [ M2A1]

Q3 (a) Given that X~N(µ,σ2 )

P(X<30) = 0.14 →P(Z<30−μσ )=0.14 →1−P(Z< μ−30

σ )=0.14 → P(Z< μ−30σ )=0.86

From table, μ−30

σ=1.08 -------(1)

P(X<60)=0.79 → P(Z<60−μσ )=0.79

From table, 60−μ

σ=0.806 ---------(2)

Solving (1) and (2), 30σ

=1.886 →σ=15.9

µ= 30 + 1.08σ →µ = 47.2 [ M3A2]

(b) Given Y ~ N( 10,22 ) P (10−a<Y <10+a )=0.95.

P( 10−a−102

<Z< 10+a−102

)=0.95.→P(−a2

<Z< a2 )=0.95→P(|Z|< a

2 )=0.95

→a2=1.96 → a = 3.92 [M2A1]

Q4 ∑ x=1680 ,∑ x2=282 280 ,∑ y=1620 ,∑ y2=262 450 ,∑ xy=272 179.

The coefficient of Correlation between x and y is

r=10 (272179 )−(1680)(1620)

√ [10 (282280 )− (1680 )2 ] [10 (262450 )− (1620 )2]=0.95 [ M1A1]

There is a strong positive linear correlation between x ( Height of the husband ) and y ( height of the wife ) [ B1 ]

Q5. (a) Laspeyres quantity index,

Lq =

= [ M1A1 ] The quantity of the fish spent by the housewife in expenditure in 2015 is decreased by

22.73 % compare to the year of 2013. [ B1 ]

(b) Passche price index, Pp =

= [ M1A1 ]The expenditure of the housewife in fish is increased by 20 % in 2015 compare to 2013.

[ B1 ]

Q6 Axis : [ M1 ]

(a) points & Graph: [ D1D1 ]

The data is increasing linearly showing an upward trend and seasoanally variation has

maximum value at third quarter and minimum value at first quarter.It’s an additive model. [ B1 ]

Q6 (b)

Year Quarter

Sales, y

4 points mov. Ave.

Centred 4 points m.a. Y Deviation, y -Y

20101 23 - - -2 32 - - -3 68 39 39.50 28.54 33 40 41.00 -8

20111 27 42 42.625 -15.6252 40 43.25 43.75 - 3.753 73 44.25 44.625 28.3754 37 45 45.375 -8.375

20121 30 45.75 46.25 -16.252 43 46.75 47.50 -4.53 77 48.25 - -4 43 - - -

[ M3 A1]Q6 (c)

Year / Quarter 1st 2nd 3rd 4th

2010 - - 28.5 -82011 -15.625 -3.75 28.375 -8.3752012 -16.25 -4.5 - -

Average seaonal variation

-15.9375 -4.125 28.4375 -8.1875

Adjusting factor 0.046875 0.046875 0.046875 0.046875Seasonal variation -15.984375 -4.171875 28.390625 -8.234375

The seasonal variation index is as follow ( 4 s.f. )Seasonal variation Index

-15.98 -4.172 28.39 -8.234

[ M3 A1 ]

Q7

(i) Let X be the number having birthday in January in a sample of size 6. X ~ B ( 6, 1

12 )

P(X=2) = (62)( 112

)2

( 1112

)4

=0.07355(0.0736) [M2A1]

(ii) Let Y be the number of people having birthday in January in sample size 8. Y~B(8,112

¿

P (Y ≥ 1 )=1−P (Y =0 )=1−¿ [M2A1]

(iii) Let w be the number of people having birthday in January in a sample of size N.

W ~ B ( N, 1

12 ) ,

P (W ≥1 )>0.9→ 1−P (W=0 )>0.9 →1−(1112 )

N

>0.9 →( 1112 )

N

<0.1→ N >26.46

Therefore, the minimum N 1s 27. [M3A1] (iv) Let T be the number of people having birthday in May, June, July and August in a

sample size 100. T ~ B( 100, 412) [B1]

np = 100(13¿=100

3>5 , nq = 100( 2

3 )=2003

>5 →Using Normal Approximation,

T ~ N ( 100

3, 200

9¿ [B1]

P (T ≥ 40 ) → P (T>39.5 )=P(Z>39.5−

1003

√ 2009

)=P (Z>1.308 )=0.0955 [M2A1]

8 (a) [ D1D1]

8 (b) [ B1]

[ M1A1]

8 (c) r = 0.9036, it shows there is a strong positive linear correlation between the mass (x) and the height (y) of students. [ B1 ]

8(d) y = a + b x

[ M2A1]

: . The equation of least square regression line is y = 98.10 + 1.337 x [ B1]

8(e) When x = 45, y = 98.10 + 1.337 ( 45 ) = 158.27 cm ( 2d.p. ) [ M1]

: . The estimate height of the a student when the mass is 45 kg is 158.27 cm. This estimation is not valid as the mass, 45 kg is not in the range of data collected. [ B1 ]

8(f) Student

sA B C D E F G H

x 4 6 3 7.5 7.5 5 2 1y 2 6 4.5 7 8 4.5 3 1d2 4 0 2.25 0.25 0.25 0.25 1 0

[ M1 ]

Spearman rank correlation coefficient, [ M1A1 ]