Stpm Trial 2009 Matht2 Q&a (pahang)

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CONFIDENTIAL* PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP Three hours

STPM 2009

954/2

MATHEMATICS T PAPER 2

PEPERIKSAAN PERCUBAAN SIJIL TINGGI PERSEKOLAHAN MALAYSIA NEGERI PAHANG DARUL MAKMUR 2009

Instructions to candidates: Answer all questions. Answers may be written in either English or Malay. All necessary working should be shown clearly. Non-exact numerical answers may be given correct to three significant figures, or one decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. Mathematical tables, a list of mathematical formulae and graph paper are provided.

This question paper consists of 6 printed pages.

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2

CONFIDENTIAL*

Mathematical Formulae for Paper 2 Mathematics T : Logarithms :

Integration :

log a x =

log b x log b a

dv

Series :

∫

n

1 r = n(n + 1) ∑ 2 r =1 n

1 r = n(n + 1)(2n + 1) ∑ 6 r =1 2

n

∑r3 = r =1

du

∫ u dx dx = uv − ∫ v dx dx

1 2 n (n + 1) 2 4

f ' ( x) dx = ln f ( x) + c f ( x) 1 1  x dx = tan −1   + c 2 a +x a

∫a

2

∫

 x dx = sin −1   + c a a2 − x2 1

Series: n n n (a + b) n = a n +  a n −1b +  a n − 2 b 2 +  +  a n − r b r +  + b n , where n ∈ N 1  2 r Coordinate Geometry : The coordinates of the point which divides the line joining (x1 , y1) and (x2 , y2) in the ratio m : n is  nx1 + mx 2 ny1 + my 2  ,   m+n   m+n The distance from ( x1 , y1 ) to ax + by + c = 0 is ax1 + by1 + c

a2 + b2 Maclaurin expansions (1 + x) n = 1 + nx +

n(n − 1) 2 n(n − 1)  (n − r + 1) r x ++ x +  , where x < 1 r! 2!

x2 xr e = 1+ x + + ... + + ... 2! r! x

ln (1 + x ) = x −

(− 1) x r + ...,−1 < x ≤ 1 x2 x3 + − ... + 2 3 r r +1

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3

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Mathematical Formulae for Paper 2 Mathematics T : Numerical Methods :

Correlation and regression :

Newton-Raphson iteration for f ( x) = 0 : f ( xn ) x n +1 = x n − f ' ( xn )

Pearson correlation coefficient: ∑ xi − x y i − y r= 2 2 ∑ xi − x ∑ y i − y

(

(

)(

)

) (

Trapezium rule : 1 h[ y 0 + 2( y1 + y 2 +  + y n −1 ) + y n ] ∫a 2 b−a where yr = f (a + rh) and h = n b

f ( x)dx ≈

Regression line of y on x : y=a+bx where b =

∑ (x − x )( y − y ) ∑ (x − x ) i

i

2

i

a = y − bx

Trigonometry

sin( A ± B) = sin A cos A ± cos A sin B cos( A ± B) = cos A cos B  sin A sin B

tan( A ± B) =

tan A ± tan B 1  tan A tan B

cos 2 A = cos 2 A − sin 2 A = 2 cos 2 A − 1 = 1 − 2 sin 2 A

sin 3 A = 3 sin A − 4 sin 3 A cos 3 A = 4 cos 3 A − 3 cos A

 A+ B  A− B sin A + sin B = 2 sin  cos   2   2 

 A+ B  A− B sin A − sin B = 2 cos  sin   2   2   A+ B  A− B cos A + cos B = 2 cos  cos   2   2   A+ B  A− B cos A − cos B = −2 sin  sin   2   2 

)

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CONFIDENTIAL*

4

1. A passenger in an aeroplane flying at a certain height , H , sees two towns , Pekan town and Kuantan town, directly to the left of the aeroplane. The angles of depression to the Pekan towns and Kuantan town , are α° and β° respectively where α° < β° . If the two towns are d meter apart, show

H = d sin α sin β cos ec (β − α ) .

[3]

2.

The points A , B and C lie on the circumference of a circle as shown in the diagram above with ∠ABC= 80° and ∠ACD=50°. The tangent to the circle at point A meets the chord CD produced at point T. (a) Show that AC = AT. [4] (b) Show that the length of the chord CD is equal to the radius of the circle.

[4]

3. The forces F1 = (5 i + 3 j )N , F2 = (4 i − 6 j )N and F3 = (− 2 i + 7 j )N act at a point. (a) Calculate the magnitude of the resultant force. [2] (b) Using the “scalar product”, calculate the angle between the resultant force and force, F4 = (5 i + 3 j )N . [4] 4. (a) A certain substance evaporates at a rate which is proportional to the amount of substance left. Given that the initial amount of the substance is A and the amount which has evaporated at time t is x , write a differential equation to show the rate of evaporation. [1] (b) Solve the differential equation and sketch the graph of x against t.

[6]

(c) Given that it took ln 2 seconds for half the amount to be evaporated, find how 3 long it takes for of the initial amount to be evaporated. [4] 4 5. The following table show the mean and standard deviation of the marks of the male and female students who sat for a semester test. Student Number of students Mean Standard deviation Male 80 52. 9 5. 3 Female 100 61. 4 4. 1 Calculate the mean and standard deviation of the marks of all the students. [6]

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CONFIDENTIAL*

5

6. Two bowlers , Adam Lambert and Kris Allen , take turns to throw, with bowler Adam Lambert taking the first throw. The probabilities of bowlers Adam Lambert and 2 4 Kris Allen scoring a strike are and respectively. Find the probabilities of bowler 3 5 Adam Lambert scoring a strike first. [3] 7. There are one or two flowers on the faces of 50 cents stamps. 90% of all these 50 cents stamps have two flowers while the rest of the stamps have single flower. From the stamps which have single flower, 95% of these stamps have a flower at the centre of the stamps while the rest have a flower on the left side of the stamps. (a) By using a suitable approximation, determine the probability that between 5 and 15 stamps inclusive have one flower , out of a random sample of 100 pieces of 50 cents stamps. [5] (b) By using a suitable approximation, determine the probability that less than 3 stamps have only one flower on the left side of the stamp out of a random sample of 100 pieces of 50 cents stamps. [4] 8. X is a continuous random variable with a probability density function, f defined as

 1 2x  f (x) =  4 e  0

,0 ≤ x < k otherwise

(a) Find the value of k.

[2]

(b) Obtain the cumulative distribution function F ( x ) for X.

[3]

(c) Find the mode and E(X).

[4]

9. The following stemplot shows the masses, in gm of mangoes harvested in a particular day. Stem Leaf 3 2 3 4 5 7 4 0 8 8 9 5 2 2 2 4 7 9 6 1 1 2 4 7 7 9 8 0 6 9 9 (a) Find the mean and standard deviation of the masses of the mangoes . [3] (b) Find the median and semi-interquartile range of the masses of the mangoes .

[3]

(c) Draw a boxplot to represent the data and identify possible outliers . Comment on the shape of the distribution and give a reason for your answer.

[5]

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CONFIDENTIAL*

6

2 sin 4θ − sin 6θ − sin 2θ = tan 2 θ . 2 sin 4θ + sin 6θ + sin 2θ Hence, find the value of tan 2 15° , leaving your answer in surd form.

10. (a) Prove that

(b) Solve the equation

2 sin 4θ − sin 6θ − sin 2θ = cos 2θ for 0° < θ < 360° . 2 sin 4θ + sin 6θ + sin 2θ

[5] [5]

11. (a) At noon, a man A is travelling at a speed of 20 km/h in the direction of N 25° W. Another man , B is 2 km to the north of A and travelling at the speed of 15 km/h in the direction of N 60° W. Find the magnitude and the direction of the velocity of A relative to B. Hence, find the shortest distance between man A and man B . [9] (b) Find the course man A must travel in order to intercept B if man A maintains his speed but changes its course. [4]

12. (a) A machine is used to fill up bottles with a mean volume of 550 ml. Suppose that the volume of water delivered by the machine follows a normal distribution with mean , µ ml and standard deviation 4 ml. Find the range of values of mean , µ , if it is required that not more than 1% of the bottles contain less than 550 ml. [4] (b) The mass of a box of chocolate cereal is distributed normally with mean 100 g and standard deviation 2 g. (i) Calculate the probability that three boxes of chocolate cereal chosen at random, each has a mass less than 98 g. [2] (ii) Calculate the probability that three boxes of chocolate cereal chosen at random, have a total mass exceeding 305 g.

[2]

(iii) Calculate the probability that out of three boxes of chocolate cereal chosen at random, exactly two have a mass greater than 98 g while the mass of the remaining box is greater than 105 g. [3]

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1 Marking Scheme PEPERIKSAAN PERCUBAAN STPM NEGERI PAHANG 2009 Mathematics T Paper 2 ( 954 / 2 ) Question

Scheme

Marks

1

∆ABK , sin β =

H AK

⇒

AK =

H sin β

M1

∆AKP , Sine Rule : AK d = sin α sin(β − α )

,

 H   sin β sin α

  =

d

sin(β − α )

H = d sin α sin β cos ec (β − α )

Question 2. (a)

Scheme ∠ADC = 180° − 80° ( Opposite angles of a cyclic = 100° quadrilateral are supplementary) ∠TDA = 180° − 100° ( Angles on a straight line ) = 80° ∠TAD = ∠ACD = 50° ( Angles in the alternate segment) ∆ATC , ∠DTA or ∠CTA = 180° − ( 80° + 50°) = 50° { The sum of angles in a triangle is 180° ) Since ∠CTA = ∠ACT = 50°, ∆ATC is an isosceles triangle because the base angles are the same. ∴AC = AT

M1

A1

Marks M1

M1

M1 A1

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2 2. (b)

Join A and C to the centre , then OA=OC=radius

M1 ∠AOC = 2 ×∠ABC = 160° ( Angle at the center = 2 × angle at the circumference.) ------------------------------------------------------------------1 ∠OCA = (180° − 160° ) = 10° 2 ∆OAC is an isosceles triangle ( OA = OC = radius ) ------------------------------------------------------------------∆ODC, ∠ OCD =∠OCA + ∠ACD = 50°+10° = 60° ∠ OCD =∠ODC = 60° ∆OCD is an isosceles triangle ( OA = OD = radius )

M1

M1

∴∠COD= 180° − ( 60° + 60° ) = 60° Since ∠ OCD =∠ODC = ∠COD = 60° , therefore ∆OCD is an equilateral triangle and CD = OC = OD = radius of circle.

Question 3. (a)

Scheme Resultant force, FR = (5 i + 3 j ) + (4 i − 6 j ) + (− 2 i + 7 j ) = (7 i + 4 j ) Magnitude of the resultant force, FR = 7 2 + 4 2 = 65 = 8.062 N

3. (b)

cos θ =

FR • F4 FR • F4 cos θ =

, cos θ =

(7 i + 4 j ) • (5 i + 3 j )

7 +4 • 5 +3 47 cos θ = 65 • 34 θ = 1°13' 2

Marks B1 B1 M1

7 i + 4 j • 5i + 3 j

(7 × 5 ) + (4 × 3 ) 2

A1

2

M1 2

M1 A1

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3 Question 4. (a)

4. (b)

Scheme Amount which has evaporated at time t = x Amount which has not evaporated at time t = A− x dx dx ∝ (A − x) , = k(A − x) dt dt Interpret initial condition t =0 , x = 0 x t 1 separate variables dx = ∫ 0 (A − x) ∫ 0 k dt

Marks

B1

M1

[− ln( A − x )] 0x = k [t ] 0t

correct integration Substitute upper limits & lower limits  A− x  A− x ⇒ ln  = −k t   = e−k t  A   A 

(

)

x = A 1 − e− k t

M1 M1 M1 A1

Graph

D1

4. (c)

(

x = A 1 − e− k ∴k=2

t

) ,Substitute , t = ln

(

) , Substitute

x = A 1 − e− 2 t

x=

2 , x=

A , 2

A1 M1

3 A 4

∴ t = ln 2

Question 5

Mean =

∑x

Mean =

A1

M

+ ∑ xF

n M + nF

Scheme , Mean =

(80 × 52.9 ) + (100 × 61.4 )

[

]

∑ ( x ) = 80 × [5.3 + 52.9 ] = 226 120 Female : ∑ ( x ) = 100 × [4.1 + 61.4 ]= 378 677  ∑ x + ∑ x   10372  − Standard deviation =   2

2

2

2

2

F

2

2

M

 

F

180

 604797   10372   −   180   180 

  180  

A1

B1 B1

M

2

Marks M1

80 + 100

4232 + 6140 10372 , Mean = = 57.622 180 180 2 Male : ∑ ( x M ) = n × σ 2 + µ 2

=

M1

2

M1

2

= 6.2978

A1

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4 Question

6

Question 7. (a)

7. (b)

Scheme P(A) + P(A′ K′ A ) + P(A′ K′ A′ K′ A ) + … 2 1 1 2 1 1 1 1 2 =   +  × ×  +  × × × ×  +…  3  3 5 3  3 5 3 5 3 2   3 =   Infinite Geometric Series 1   1 −  15   5 = 7

Marks

Scheme X represents the number of stamps with one flower X ∼ B ( 100 , 0.1 ) Normal Approximation : mean=10 , variance = 9 15.5 − 10   4.5 − 10 ≤Z≤ P (5 ≤ X ≤ 15 ) = P   3 3   = P (− 1.833 ≤ Z ≤ 1.833 ) = 1 − 2 × (0.0334 ) = 0. 9332 Y represents the number of stamps with only one flower on the left side. p = 0.1 × 0.05 = 0.005 Y ∼ B ( 100 , 0. 005 ) Poisson Approximation : mean, λ = 0.5 P( X < 3 ) = P( X = 0 ) + P( X = 1 ) + P( X = 2 )

Marks B1

M1

M1

A1

B1 M1 Standardize M1 Continuity correction

A1 B1

B1

 0.5 2  = e − 0.5 × 1 + 0.5 +  2   = 1.625 e −0.5 = 0.9856 Question 8. (a)

8. (b)

M1

A1

Scheme

Marks

k

x  1 2x  1 2 ∫0 4 e dx = 1 ,  2 e  = 1  0 k = 2 ln 3 k

0 < x < 2 ln 3 , F ( x ) =

∫

x

0

Integration

M1 A1

x 2

1 e dx 4

Integration

M1

x

x 1 x   1 , F( x ) =  e 2  F ( x ) = e 2 − 1 2  2 0 The cumulative distribution function is  0 ,x < 0  1  x  F ( x ) =  e 2 − 1 ,0 ≤ x < 2 ln 3  2   1 x > 2 ln 3

A1 (values 0 and 1) A1 (second interval)

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5 8. (c)

Mode = 2 ln 3 ,

E( X ) = ∫

2 ln 3 0

f ( x ) is an increasing function x

x 2 e dx 4 2 ln 3

x x =  e2  2 0

−

2 ln 3

∫

2 ln 3 0

M1 x M1 1 2 e dx ,Integration by parts 2

2 ln 3

 x 2x   2x  − e  = e   2 0  0 = 3 ln 3 − 2

Question 9. (a)

A1

Scheme Mean =

Marks B1

1351 = 56.29 24 2

83379  1351  Standard deviation = −  =17.475 24  24  9.(b)

9.(c)

B1

52 + 54 = 53 2 x + x7 40 + 48 Q1= 6 ⇒ Q1 = = 44 2 2 x + x 19 62 + 64 Q 3 = 18 ⇒ Q3 = = 63 2 2 1 Semi-interquartile Range = × [63 − 44 ] = 9.5 2 Lower boundary = 44 − 1.5 × ( 63 − 44 ) =15. 5 Upper boundary = 63 + 1.5 × ( 63 − 44 ) =91. 5

Median =

B1 for 83379 M1 A1 B1

M1 A1

B1 l.b.&u.b. B1 outlier

∴Outlier is 99. D1 Box D1 Whiskers

, Q 2 − Q1 = 9 Q 3 − Q 2 = 10 Since , Q 3 − Q 2 > Q 2 − Q1 the distribution is skewed to the right.

B1

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6 Question 10. (a)

Scheme 2 sin 4θ − sin 6θ − sin 2θ 2 sin 4θ − (sin 6θ + sin 2θ ) LHS, = 2 sin 4θ + sin 6θ + sin 2θ 2 sin 4θ + (sin 6θ + sin 2θ ) 2 sin 4θ − (2 sin 4θ cos 2θ ) = 2 sin 4θ + (2 sin 4θ cos 2θ ) 2 sin 4θ (1 − cos 2θ ) = 2 sin 4θ (1 + cos 2θ )

[ [

] ]

1 − 1 − 2 sin 2 θ = 1 + 2 cos 2 θ − 1 2 sin 2 θ = 2 cos 2 θ Q.E.D = tan 2 θ 2 sin 4θ − sin 6θ − sin 2θ Substitute θ = 15° into tan 2 θ = 2 sin 4θ + sin 6θ + sin 2θ 2 sin 60° − sin 90° − sin 30° tan 2 15° = 2 sin 60° + sin 90° + sin 30°  3 −1− 1 2 2 2  =   3 + 1+ 1 2  2  2  2 3−3 = 2 3+ 3 2 3−3 2 3−3 × = 2 3+3 2 3−3

( (

) ( ) (

Factor formulae M1

Double Angle M1

A1

Values of sinθ and cosθ M1

) )

=7 −4 3 10. (b)

Marks

2 sin 4θ − sin 6θ − sin 2θ = cos 2θ , tan 2 θ = cos 2θ 2 sin 4θ + sin 6θ + sin 2θ sin 2 θ = 2 cos 2 θ − 1 2 cos θ sin 2 θ = 2 cos 4 θ − cos 2 θ

A1 M1 M1 sin θ tan θ = cos θ

1 − cos 2 θ = 2 cos 4 θ − cos 2 θ

M1 Basic Identity

1 , cos θ = ±0.8409 4 θ = 32.8° ,147.2° ,212.8°327.2°

M1

cos 4 θ =

A1

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7 Question 11. (a)

Scheme

Marks

D1 arrows D1 directions/angles

VB

A

2

= 20 2 + 15 2 − 2(20 )(15 ) cos 35°

VB =

A

M1

20 2 + 15 2 − 2(20 )(15 ) cos 35°

A1 V B =11.5546 km/h The magnitude of the velocity of A relative to B is 11.6 km/h . A

15 11.5546 = sin θ sin 35°

θ = 48°8' ,

⇒ sin θ =

15 sin 35° 11.5546

M1

α = 48°8' −25° = 23°8' or 23.13°

The direction of the velocity of A relative to B is N 23° 8′ E.

A1 D1 diagram

d = 2 sin 23°8'

M1

d = 0.7857 km ∴The shortest distance between A and B is 0.786 km

A1

11. (b) 15 20 = sin α sin 120°

sin α =

15 sin 120° 20

α = 40°30'

θ = 180° − 120° − 40°30' θ = 19°30' In order to intercept B , man A must travel in the direction of N 40° 30′ W . or (bearing 319° 30′)

D1 arrows D1 directions/angles

M1

A1

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8 Question 12 (a)

Question 12(b) (i)

Scheme

P ( X < 550 ) < 0.01 ,

Marks

550 − µ   P Z <  < 0.01 4  

M1

 550 − µ    < −2.326 4  

B1 for −2.326

µ > 559.3

A1

B1 for “<”

Scheme X represents the mass of a box of chocolate cereal X ∼ N 100 ,2 2

(

Marks

)

  98 − 100   P ( X < 98 ) × P ( X < 98 ) × P ( X < 98 ) =  P  Z <  2   

3

M1 Standardize

= [P ( Z < −1)]

3

= (0.1587 ) = 0.003997 (ii)

3

X 1 + X 2 + X 3 ∼ N (300 ,12 )  305 − 300   P ( X 1 + X 2 + X 3 > 305 ) = P  Z > 12   = P ( Z > 1.443 ) = 0.0745

(iii)

A1

M1 Standardize

A1

3 × P ( X > 98 ) × P ( X > 98 ) × P ( X > 105 ) 2

  98 − 100   105 − 100   = 3 ×  P Z >  × P  Z >  2 2     

M1 Standardize

= 3 × [P ( Z > −1)] × P ( Z > 2.5 ) 2

= 3 × [0.8413] = 0.0132

2

× 0.00621

M1 A1