MATHEMATICS T (MS)

MARKING SCHEME FOR STPM 2020 SEMESTER 3 TRIAL EXAMINATION

SECTION A

1(a) Median = 1 ( 12 + 13 ) = 12.5 B1
2
1 ( 10 + 11 ) = 1 B1
1 = 10.5 and 3 = 2 ( 16 + 17 ) = 16.5 B1
2

Interquartile range = 16.5 - 10.5 = 6

(b)

32 is an outlier B1,B1
(c ) Skewed to the right or Positively skewed B1
B1

2. 0.01 D Marks
(a) A 0.99 D’
D M2
0.45 0.05 D’
D B1
0.25 B 0.95 M1
0.30 C 0.02 D’ M1
A1
0.98
M1
P(D│A) = 0.01 M1
A1
b(i) P(D) = P(A ∩ D) + P(B ∩ D) + P(C ∩ D) 9
= (0.45 × 0.01) + (0.25 × 0.05) +

(0.30 × 0.02)

= 0.023

(ii) P(A│D) = P(A∩D)

P(D)

= 0.45×0.01
0.023
= 9 or 0.19565 or 0.1957 or 0.196

46

Total

3. M1 E(X) = Var(X) = λ is applied to
(a) Var(X) = E(X2) –[ E(X)]2 Poisson distribution
M1
λ = 20 – λ2
(λ - 4)( λ + 5) = 0 A1
Since λ > 0, therefore λ = 4

(b) P(X ≥ 1) = 1 - P(X = 0) = 1- e−440 M1
A1 Accept only 3 to 5 s.f
0!

= 0.982 or 0.9817 or 0.98168

(c) Let Y : The number of televisions sold in 3 B1
weeks.

Y ~ Po (12) M1

P(Y = 4) = e−12124 A1 Accept only 3 to 5 s.f

4!

= 0.00531 or 0.005309 or 0.0053086

Total 8

4.

(a) ̂ = ̅ = 14050 = 70.25 ……………………………………. B1

200 M1 (his mean)
A1
̂ 2 = (∑ 2 − ̅2)

−1

= 200 (995350 − 70.252)………………………………….

199 200

= 41.897…………………………………………………..

(b) 95% confidence interval for mean weight

̅ ± 1.96 ̂ = 70.25  1.96(√41.897) ……… B1 for value 1.96M1 correct formula

√ √200

= 70.25  0.8971

95% confidence interval for mean weight is (69.353 , 71.147) .. A1 (accept 3,4,5 sf)

(c ) Since 75 kg does not fall in the 95% confidence interval for the mean height,therefore we

can state that the male students in that college are not obese. A1

5.(a) H0 :  = 10

H1 :  > 10 B1
Critical value Z0.1 = 2.326

Rejection region Z > 2.326

147 −10 M1
Z = 12

3.4
12

= 2.294 A1

Since Z = 2.294 < 2.326, H0 is not rejected. M1

Conclusion : There is insufficient evidence to conclude that the actual time spent is

more than 10 minutes on average. A1

(b) Because it is a normal population A1

6. H0 : There is no relationship between the dropout and the income of the family. B1
H1 : There is a relationship between the dropout and the income of the family.

Number of dropouts Total

Male student Female student 18
13
Low 18(27) 18(23) 19
50 = 9.72 50 = 8.28 50

M1 (AInlcl oromwes are coMrreedcitu)m 13(27) 13(23)
of the family 50 = 7.02 50 = 5.98

High 19(27) 19(23)
50 = 10.26 50 = 8.74

Total 27 23

( − )2


13 9.72 1.107

5 7.02 0.581

9 A1 (All col1u0m.2s6are correct) 0.155

5 8.28 1.299

8 5.98 0.682

10 8.74 0.182

2 = 1.107 + 0.581 + 0.155 + 1.299 v = (3 – 1)(2 – 1) 02.1(2)= 4.605 B1B1
=2
+ 0.682 + 0.182 M1

= 4.006 A1
Since 4.006 < 4.605, do not reject H0.

There is sufficient evidence to conclude there is no relationship between the dropout and

the income of the family at the 10% significance level. A1

SECTION B

7. H0 : = 153 B1 both
B1 either
H1 : > 153

̅ = 150 + 154 =155
150
(115500)2]=
̂ 2 = 150 [11000 − 72.8188
149
150

= 0.07, critical region : z > 1.476 B1 1.476
Test statistic, z = 155−153 M1

√721.851088

= 2.870 > 1.476 A1 2.870

At 7% sl, sufficient evidence to reject H0. A1

The publisher has understate the mean number pages in each

book at 7% sl.

A 93% symmetrical CI for is B1 1.811

(155 ± 1.811√72.8188 ) M1

150

= (153.738 , 156.262) A1

H0 : = 0.8 B1 -1.881
H1 : < 0.8 B1 √(0.8)(0.2)
= 0.03, critical region : z < -1.881
150
̂ −0.8 < -1.881
M1
√(0.8)(0.2)
M1
150 A1
A1
̂ < 0.7386

80+ 0,7386
150

< 30.79

Greatest value of is 30

8(a) H0 : This sample of beads provides evidence to support the claim stated in
the list of contents.

H1 : This sample of beads does not provide evidence to support the claim. B1
M1A1
Colour O E (O − E)2 (Ei)
E
Red 37 1505 = 30 M1A1
1.633 (O − E)2
25
E
Yellow 22 24 0.166
B1
Pink 11 18 2.72 B1
B1
Green 15 12 0.75

Orange 25 24 0.0417

Purple 9 12 0.75

Blue 31 30 0.033

Total 150 150 6.094

Level of significance : 5%

Degrees of freedom : 7 – 1 = 6

The critical value,  2 at 5% level of significance = 12.59
6

Since 6.094 < 12.59, do not reject H 0 .

There is sufficient evidence, at 5% level of significance, to conclude that the
sample of beads provides evidence to support the claim stated in the list of
contents.

8