AJC JC 2 H2 Maths 2011 Mid Year Exam Solutions Paper 2

Solutions to MYCT H2 Math Paper 2 with marker’s comments Qn

Solutions

1

In a systematic sampling, all the users are assigned a unique number from 1 and 18500. A random number between 1 and 185 inclusive is chosen initially. This user is chosen and the next 185th resident is chosen until 100 residents are chosen.

A stratified sampling gives a representative sample of the population as the opinions of the game may differ according to age groups. 2

Let X be the random variable denoting the number of marks scored by a candidate. If X  N (78.1, 15.22 ) , then

P( X > 100) = 0.0748 i.e. around 7.5% of the student cohort will score above 100 marks if the random variable X is following a Normal distribution of mean 78.1 and std dev 15.2. This is unrealistic as students cannot score more than 100 marks.



Since n is large, by CLT, X  N  78.1,



15.22   approximately. 60 

P(80.0 < X < 85.0) = 0.166 3(i)

p = 35 P (a randomly chosen component is faulty) = ( 0.35 )( 0.08 ) + ( 0.65 )( 0.05 ) = 0.0605

3(ii)

f(p) = P( component supplied by A / faulty ) = P(faulty component supplied by A) P(faulty) p ( 0.08) 100 = 100 − p p ( 0.08) + ( 0.05) 100 100

=

0.08 p 0.08 p (shown) = 0.08 p + 5 − 0.05 p 0.03 p + 5

f ' ( p) = =

( 0.03 p + 5) 0.08 − 0.08 p ( 0.03) 2 ( 0.03 p + 5) 0.4

( 0.03 p + 5)

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Qn

Solutions

For 0 ≤ p ≤ 100 , f ' ( p ) > 0 ⇒ as p increases, f ( p ) increases. ⇒ f ( p ) is a strictly increasing function. i.e. when the company buys a larger percentage of its electronic components from supplier A, the probability of a randomly chosen faulty component is supplied by A increases. 4(i)

Using the symmetrical properties of the normal distribution, The mean of the distribution =

150 + 250 = 200 2

Let X be the random variable denoting the mass of an orange sold in the supermarket.

X  N (200, σ 2 ) P(X > 250) = 0.1 50 P(Z > ) = 0.1

σ

50 σ 39.0 = 1.28155 ⇒ =

σ

4(ii)

Let Y be the random variable denoting the total mass of a ‘small’ bag. Let W be the random variable denoting the total mass of a ‘large’ bag.

Y  N (5(200),5(392 )) ⇒ Y  N (1000, 7605) Y1 + Y2  N (2(1000), 2(7605)) ⇒ Y1 + Y2  N (2000,15210) W  N (10(200),10(392 )) ⇒ W  N (2000,15210) Y1 + Y2 − W  N (0,30420) P(| Y1 + Y2 − W |< 300) = P(−300 < Y1 + Y2 − W < 300) = 0.915 4(iii)

0.001(Y1 + Y2 )  N (2, 0.01521) 0.0012(W )  N (2.4, 0.021902) 0.001(Y1 + Y2 ) − 0.0012(W )  N (−0.4, 0.037112) P(0.001(Y1 + Y2 ) − 0.0012(W ) < 0) = 0.981

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(a)(i)

no. of arrangements = 5! X 7! = 604800

(ii)

no. of arrangements = 9P 3 x 8! = 20321280

2

Qn

Solutions (iii) No of ways of dividing the people into 3 groups of 3 and 1 group of 2

C3 • 8C3 • 5C3 • 2C2 = 15400 3!

11

=

(b) Treat man and his wife as 1 unit. ⇒ arrange 4 units + 3 elderly man + 1 empty seat in a circle. No of arrangements = (8-1)! X (2!)4 = 80640 6(i)

Let X be the mass of the component of the electrical product and µ be the mean mass of the component. To test H0 : µ = 12 vs

H1 : µ ≠ 12

Since population variance is known and X follows a normal distribution, we use a Ztest. Test statistic , Z =

X − 12

σ

~ N (0,1)

n Use a 2-tailed test at 5% level, ie reject Ho if p < 0.05 Using G.C. with x = 12.28 , n = 60 ; σ = 1.005 , p = 0.0309 < 0.05 . We reject Ho and conclude that there is sufficient evidence at 5% level to dispute the factory owner’s claim. To test H0 : µ = 12 6(ii)

vs

H1 : µ < 12

Since sample size is small and population variance is unknown, we use a t-test. Test statistic, T =

X − 12 ~ t (19) s n

From the sample, n = 20 , s2 =

20 (1.03) = 1.0842 19

For the owner’s claim to be accepted ⇒ reject H0 ie reject Ho if T < -1.729

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Qn

Solutions

⇒

x − 12 < −1.729 ⇒ x < 11.6 ( 3 s.f.) 1.0842 20

For part (ii), the conclusion is to reject H 0 i.e. T cal will lie in the rejection region.

6(iii)

Since the population variance is said to be known and equals to s2. A Z-test will be carried out and Z cal = T cal (since σ = s ) N(0,1)

t(19)

From the diagram, we can conclude that there is still sufficient evidence to support the owner’s claim that the mean mass is less than 12 grams at 5% level of significance. Alternatively, Z-test is to be used since population variance is known. From part (ii), Tcal =

Since σ = s , Z cal =

x − 12 < −1.729 (to reject Ho) s 20

x − 12 = Tcal < −1.729

σ

20 At 5% significance level, Zc = −1.6448 ∴ Z cal < −1.729 < −1.6448 ( to reject Ho)

There is still sufficient evidence to support the owner’s claim that the mean mass is less than 12 grams at 5% level of signifance. 7(i)

Let X be the number of telephone calls received by a call centre in a 4-min interval, ie X ~ Po(12). P(X > 10 ) = 1 – P(X ≤ 10) = 0.653

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Qn

Solutions

7(ii)

Let Y be no of calls received at the call centre in n seconds,

 3   n  n  , ie Y ~ Po    60   20 

Y ~ Po 

Given P ( Y ≤ 1) < 0.1

n   1 +  < 0.1  20  x − x   Sketch y = e 20  1 +  − 0.1 .  20 

⇒ e

−

n 20

From G.C., for y < 0 ⇒ x >77.8 ⇒ x = 78 The shortest length of time required = n = 78 secs. 7(iii)

Let W be the no of calls received at the call centre in a 10-hr interval , ie W ~ Po ( 1800 ) Since λ = 1800 > 10 ⇒ W ~ N ( 1800, 1800 ) approx P( W > 1850 ) = P ( W > 1850.5 ) by continuity correction = 0.11696 = 0.117

7(iv)

P(manpower shortage | ‘busy’ day) = 0.15 P(manpower shortage) = 0.15 (0.11696) = 0.017544 Let S be the no of days with manpower shortage in 30 randomly chosen working days of 10 hours , S ~ B( 30, 0.017544) P( S < 2 ) = P(S ≤ 1) = 0.903

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