1. PART B
QUESTION 1 (a)
X ~ Bin (9, 0.45)
9
C ( 0.45 ) .(0.55)5
4
i) P( X = 4 ) =
4
= 0.26
ii) P( X ≥7 ) = P ( X = 7 ) +P ( X = 8 ) +P ( X = 9 )
9 9 9
C ( 0.45) .(0.55)2 + C ( 0.45) .(0.55)1 + C ( 0.45) .(0.55)0
7 8 9
=
7 8 9
= 0.0498
iii) P (5 ≤ X ≤ 7 ) = P (X = 5 ) + P ( X = 6 ) + P ( X = 7 )
9 9 9
C ( 0.45) .(0.55)4 + C ( 0.45) .(0.55)3 + C ( 0.45) .(0.55)2
5 6 7
=
5 6 7
= 0.3695
QUESTION 1 (b)
X ~ N (25, 6 2 )
16 − 25
i) P (X < 16) = P (z < )
6
= P ( z < - 1.5 )
= 0.5 – 0.4332
= 0.0668
22 − 25 29 − 25
ii) P ( 22 ≤ X ≤ 29 ) = P ( ≤z≤ )
6 6
= P ( - 0.5 ≤ z ≤ 0.67 )
= 0.2486 + 0.1915
= 0.4401
iii) P ( X ≥ a ) =0.1075
P ( X < a ) =0.8925
 x − 75 a − 75 
P <  = 0.8925
 10 10 
 a − 75 
P z <  = 0.8925
 10 
From the table P ( z < 1.24 ) = 0.8925

2. a − 75
= 1.24
10
a = 87.4
QUESTION 2 (a)
 3.5 
X ~N  40, 
 49 
3.5
i) σX = = 0.5
49
41 − 40
P( X ≥ 41 ) = P( z ≥ )
0.5
= P(z ≥ 2)
= 0.5 – 0.4772
= 0.0228
ii) P(40.5 < X < 40.9)
40.5 − 41 40.9 − 41
= P( <z< )
0.5 0.5
= P(-1 < z < -0.2)
= 0.3413 – 0.0793
= 0.262
QUESTION 2 (b)
30
n = 150, p =
150
30
i) p=
ˆ = 0.2
150
(0.2)(0.8)
σp =
ˆ = 0.0327
150
margin of error
= ± 1.96σ p
ˆ

3. = ±1.96(0.0327)
= ±0.0641
ii) From table, the z-value for 95% CI = 1.96
The 95% Confidence Interval for population mean,
= p ± zσ p
ˆ ˆ
= 0.2 ± 1.96(0.0327)
= 0.072 ± 0.0641
= (0.1359, 0.2641)
95% of confidence interval for the percentage of cakes being effect with fungus is lies between 0.1359 to
0.2641
iii) From table, the z-value for 90% CI = 1.65
E = 0.05
2
 zs 
n = p.q.  
E
2
 1.65 
= 0.2 × 0.8 ×  
 0.05 
= 174.24
The sample size needed is 174 .
QUESTION 3(a)
σ = 1.75 n = 35 , x = 7.32
i. H0 : µ ≥ 8
H1 : µ < 8
ii. Since n is large, 35 >30, we use normal distribution to make the test.
x−µ 7.32 − 8
=
Test statistic, z = σ 1.75 = - 2.2988
n 35
iii. At α = 0.05 , we reject H0 if z < -1.65
Since - 1.65 > -2.29 , we reject H0
We conclude that there is no evidence that the effective rate of broadband networking for High-Speed Company
production of electronic is at least 8.

4. QUESTION 3(b)
24
n= 505 , p=
ˆ = 0.0475
505
24
i. p=
ˆ = 0.0475
505
ii. H0 : p = 0.35
H1 : p ≠ 0.35
iii. Since np > 5 and nq > 5 , we use normal distribution to make the test .
p − p 0.0475 − 0.35
ˆ
=
Test statistic , z = pq 0.35 × 0.65 = -14.25
n 505
At α = 0.01 , we reject H0 if z < -2.58 or z > 2.58
Since -2.58 > -14.25 , we reject H0. We conclude that the proportion of the safety rate of the building is not equal
0.35
QUESTION 4(a)
i. Independent variable: Driving Experience (years)
ii. Scatter diagram
Label x -axis
Label y -axis
Title
Diagram
iii.

5. ∑ x = 1396
2
∑ x = 90
n=8
∑ y = 29642
2
∑ y = 474
∑ xy = 4739
n∑ xy − ∑ x ∑ y
r=
n x2 − ( x ) 2  n y 2 − (
 ∑ ∑  ∑ ∑ y) 
2
  

8 ( 4739 ) − 90 ( 474 )
=
8 ( 1396 ) − 902  8 ( 29642 ) − 4742 
  
= −0.7679
There is a negative correlation between the variables.
QUESTION 4(b)
i. There is a strong positive correlation between cost and age.
ii.
n∑ xy − ∑ x ∑ y
b=
n∑ x 2 − ( ∑ x )
2
6 ( 3100 ) − 49 ( 325 )
=
6 ( 481) − 492
= 5.515
a = y − bx
= 54.17 − 5.515 ( 8.17 )
= 9.112
Regression line: y = 9.112 + 5.515 x
iii. 0.96862 = 0.9382 .
93.82% of the total variation in Cost is explained by Age and another 6.18% is
explained by another factors.
iv.
y = 9.112 + 5.515 x
ˆ
= 9.112 + 5.515 ( 16 )
= 97.35