2019-3-KED-KTEK-MARKING SCHEME

Question Marking Scheme Trial Exam 2019 954/3 Mark

1a
1351 B1
Mean = = 56.29= 56.3
24

Standard deviation = √ ❑2 M1

= 17.475 A1
52+54
Median = = 53 B1
2
1b
40+ 48 62+ 64
Q1 = = 44 , Q3 = = 63
2 2 M1 A1

IQR = 63 – 44 = 19

Lower fence = 44-1.5(63 - 44)= 15.5

Upper fence= 63 + 1.5 ( 63 - 44) = 91.5
Outlier= 99
D1 (Quartiles)

D1 (all correct
with outlier)

B1
B1
[ 10 marks]

2a P(A∩ B ¿=P ( A ) . P ( B ) , A∧Bareindependent . B1

= 0.4 P(B)
P(A ∪ B ¿=P ( A ) + P ( B ) −P( A ∩ B)
0.88 = 0.4 + P(B) – 0.4 P(B)
0.48 = 0.6 P(B) M1
P(B)= 0.8 A1

2b P(A∩ B ¿’ = 1- P(A ∩ B ¿ B1
= 1- 0.4 P(B)
A1
= 1- 0.4 (0.8)
= 0.68
[5 marks]

3 (a) 1 1
P(X ≤ - )=
2 16
−1 M1
2
1
∫ ax +b dx= 16
−1
 -6a+8b = 1 ……..(1)

71 M1
P(X≥ ) =
162
c
7
∫ ax+ b dx= 16
1
2

M1
ac 2 a b 7
( + bc) – ( + ) = ……..(2)
2 8 2 16
c
A1 A1 A1
3(b) f(x)= ∫ ax +b dx=1
−1
2
ac a M1
( +bc) – ( - b )= 1 ………(3)
2 2

From (1), (2) and (3) A1

1 1
a= ,b= c= 1 (prove) A1 (all
2 2
correct)
x
1 1 [ 9 marks]
for -1≤ x ≤ 1 , F (x)= ∫ t+ dt
−1 2 2
1 2
= ( x +2x+1)
4

1 2
F(x) = {0 , x <−1 ( x +2 x+ 1 ) ,−1 ≤ x ≤1 1, x >1
4

4a 96% = z= 2.054 B1 (2.054)

96% confidence interval for the proportion of pens that fail to write
immediately = B1 (0.12)
24 24 M1 ¿)
( −2.054 √❑ , +2.054 √❑ ) M1(correct
200 200
formulae)
A1
= (0.0728, 0.167)
M1
4b 3 A1
Proportion= = 0.06 [7 marks]
50
Since 0.06 is not lies in the interval, so reject supplier’s claim.

5 The 95% confidence interval is given by

σ
( x ± 1.96
√❑ B1 for
z0.025=1.96
σ
x +1.96 ………..(1)
√❑
B1(either (1)
σ or (2)
x −1.96 ………..(2)
√❑
M1A1
 (4 Marks)
x = 1062.5 , 𝛔= 120

6a Unbiased estimate of mean

735.68
^μ = M1
88
= 8.36 A1

Unbiased estimate of variance M1

88 6636.5 735.68 2
σ^ 2= [ -( ) ] A1
88−1 88 88
6b
= 5.589
B1
H0 : μ = 8

H1 : μ ¿ 8

Significance level , α = 0.05 B1

Critical region, Z0.05= 1.645

Since n=88¿ 30, by CLT

5.589
X N (8, )
88

Test statistic:
M1
8.36−8.0
Z= A1
√❑

= 1.428

Reject H0 if zcalc > 1.645 A1

Since z=1.428 < 1.645, do not reject Ho.

There is not enough evidence at 5% significance level to reject A1

the manager’s claim that each employee in the firm works for [ 10 Marks]
an average of 8 hours per day
7a P(T≥ 250 ¿ B1(for ≥ 250)
=e
−21
5500
×250 M1

A1
=0.385 B1
1 −21
= 1- e 5500 t
7b 2
M1
21 1
t= A1
5500 2
 t=181.54 hours
7c d M1
f(t) = ¿]
dt
21 −21 t
= 5500
5500 e
A1
−21
21 5500 t ,t ≥0
f ( t )={ e 0 , otherwise
5500

D1( e− x
shaped curve
in first quad)
D1 (all
correct)
B1

∞ M1
21 −21 t (integration by
E(T) = ∫ t × 5500
0
5500 e parts)
7d
−21
t A1
21 e 5500 ∞ 5500 −21 t
= [ t × −21 - ∫ 5500
5500 0 21 e [ 1 ] dt M1
5500 A1
[15 marks]
5500 5500 t ∞
−21 −21
t
= [−t × e 5500 − e ]
21 0

5500
= [ -0-0] – [-0- ]
21
5500
=
21

8a Let y = x-200
❑ ❑
∑ (x−200)=41.56 ∑ y=41.56
❑ ❑
❑ ❑
∑ (x−200) 2
= 107.4673 ∑ y=107.4673
❑ ❑

❑
∑ y 41.56 M1
The mean, y = =
❑
40
❑ A1

41.56
x = 200 +
40
M1
= 201.039
A1
The unbiased variance
A1
 (41.56)
8b 2 (107.4673− ) B1( 1.96)
σ^ =
y 40
B1(std error)
39
M1 (correct
= 1.6484 formulae)
σ^ 2x =σ^ 2y=1.6484 A1

B1
8c 95% confidence interval for μ
= ( 201.039± 1.96 √ ❑ )

= (200.64, 201.44)
M1
Ho : μ=200 A1
H1 : μ ≠ 200

95% confidence interval for μ= ( 200.64, 201.44)

M1
Since 200 does not lie inside the 95% confidence interval. M1
8d Hence there is enough evidence to reject Ho at the 5%
significance level.
A1

2.326 √ ❑ ≤ 0.5
n≥ 35.673

The least sample size is 36