Solved problems on finance and statistics

This document contains solved problems on finance and statistics. It includes problems related to annuity, interest, temperature, freezing point, and more. The solutions are provided step-by-step. Desklib is an online library that provides solved assignments, essays, dissertations, and study material for various subjects, courses, and colleges/universities.

1 a) Down Payment
20
100 × 450000=¿
=$90,000
b) Mortgage Amount (Principal)
=House price – Down payment
= $ 450,000 - $ 90,000
= $ 360,000
c) Monthly Payment
M=
p ( r
n )
1−(1+ r
n )
−nt
M= Monthly Payment
P= Principal
R=rate
N= number of payments
M=
360,000( 0.0375
12 )
1−(1+ 0.0375
12 )
−240
= 1125
1−0.4729
= 1125
0.5271 =$ 2134.32
= $ 2134.32
d) Total amount paid
= 2134.32 × (12 ×20 ¿
= $ 512,236.80
Loan Amount = 360,000
Interest = Amount paid – Loan amount
= $ 512,236.80 – 360,000
= $ 152,236.80

2) Cost price = $ 400
Down payment = (0.1 × 4000 ¿=$ 400
Principal = Cost price – Down payment
(4000-400) = $ 3600
Option one
Monthly payment
M=
p ( r
n )
1−(1+ r
n )
−nt
=
3600 ( 0.09
12 )
1−(1+ 0.09
12 )
−12(2)
= 27
0.1642 =$ 164.43
Total Instalments amount
= 104.43 × (12 ×2 )
= 104.43 ×24
=$ 3946
Option two:
Monthly payment
M=
p ( r
n )
1−(1+ r
n )
−nt
M=
4000( 0.095
12 )
1−(1+ 0.095
12 )
−12 ( 3 )

= 31.67
0.2467 = $ 128.37
Total instalment price
= 128.37 × (12 ×3 )
= $ 4621.32
a) Option 1 because it has a total installment price of $ 3,946, which is lower than that of
option 2, which is $ 4,621.32.
b) Choose option 2 because its monthly payment of $ 128.37 is lower than $ 164.43 of
option 1.
3) Present Value = $ 15,000
N=10 years
Option 1
3.25% Compounded Semi-annually
FV=PV (1+i ¿¿n
I = 3.25 %
2 =0.01625
N= (10 ×2 ¿ 20
FV= $ 15,000(1+0.01625 ¿ ¿20
= $ 20,700
Option 2
3% Compounded Monthly
FV=PV (1+i ¿¿n
I= 3 %
12 = 0.0025
N= (10 ×12 ¿ 120
FV=15,000(1+0.0025)
= $ 20,235
Option 1 will result to the highest future value of $ 20,700 compared to $ 20,235 of option 2.
b) Interest

Option 1
3.25 % Compounded semi-annually.
Investment amount 10,000
A=P (1+r ¿n
r = 3.25 %
2 =0.0025
n = (2 ×10 ¿
= 20
A=10,000 (1+0.01625 ¿20
= $ 13,800
Interest = 13,800 – 10,000
= $3,800
Option 2
3% Compounded annually
A=P (1+r ¿n
R = 3 %
12 = 0.0025
N= (10 ×12 ¿
= 120
A=10000(1+0.0025 ¿120
= $ 13,490
Interest = 13,490- 1000
= $ 3,490
Option 1 will result into the largest amount of interest of $3,800 compared to $ 3,490 of
option 2.
4a) Annuity
2% Compounded monthly
Deposit $ 200 each month
n = 30 years

FV=C¿)
C=$ 200
T = (30 ×12 ¿
= 360
r = 2%
12 =0.00167
FV=2000¿)
= $ 1,205,988.024
b) 2.5% Compounded Semi annually
FV=C¿)
C = $ 1000
R= 2.5 %
2 = 0.0125
T = (30 ×2 ¿
= 60
FV=1000¿)
= 88,560
5)
Jan
Column
1
Column
2 Column3
Column
4
High
frequenc
y Low
Frequenc
y Average
16 1 3 2
21.9 1 3.9 2
23 1 5 1
25 1 6.1 2
26.1 2 10.9 1
27 1 12 1
32 1 15.1 2
33.1 1 18 1
35.1 2 19 1
37 1 19.9 1

39 1 23 1
44.1 1 24.1 2
45 2 25 1
46.9 1 27 2
48.9 1 28.9 1
50 2 30 2
51.1 1 30.9 2
52 1 36 1
54 1 41 1
59 1 42.1 1
60.1 1 48 1
61 1
64 2
64.9 1
August Column1
Column
2 Column3
Column
4
High
Frequenc
y Low
Frequenc
y Average
68 1 53.1 1
70 1 54 1
73 2 60.1 5
75 1 61 2
75.9 3 62.1 3
77 2 63 2
78.1 4 64 5
79 5 64.9 8
80.1 3 66 4
81 2
82 4
82.9 1
84 2
a) January
Highest temperature 64.9
Lowest temperature 3
Range 64.9 – 3
= 61.9
August

Highest temperature 84.0
Lowest temperature 53.1
Range = 31.1
January has a larger total range of temperature.
b) January mean high temperature
16+21.9+23+25+26.1+27+32+33.1+35.1+37+39+44.1+45+46.9+48.9+50+51.1+52+54+59+
60.1+61+64+64.9=1015.2
1015.2
29 =35.01
January mean low temperature
3.0+3.9+5.0+6.1+10.9+12.0+15.1+18.0+19.0+19.9+23.0+24.1+25.0+27.0+28.9+30.0+30.9+
36.0+41.0+42.1+48.0=468.9
468.9
29 =16.16
Difference= 35.01- 16.16
=18.85
August
High temperature mean
68.0+70.0+73.0+75.0+75.9+77.0+78.1+79.0+80.1+81.0+82.0+82.9+84.0=1006
1006
31 =32.45
Low temperature mean
53.1+54.0+60.1+61.0+62.1+63.0+64.0+64.9+66.0=548.2
548.2
31 =17.68
Difference= 32.45-17.68
= 14.77
August has a smaller difference between mean high and mean low temperature.

5c) January
Median Low temperature
Median= n+1
2
= 29+1
2
= 15TH
Median is 45
Median high temperature
Median= n+1
2
= 29+1
2
= 15TH
Median is 23
Difference 45-23
= 22
August
Median high
Median= n+1
2
= 31+ 1
2
= 16th
Median is 79
Median low
Median= n+1
2
= 31+ 1
2
= 16th

Median is 64
Difference 79-64
= 15
August has a smaller difference between median high temperature and low median
temperature.
5d)
Freezing point = 0℃ / 32℉
Jan
Column
1
Column
2 Column3
Column
4
High
frequenc
y Low
Frequenc
y average
16 1 3 2 9.5
21.9 1 3.9 2 12.9
23 1 5 1 14
25 1 6.1 2 15.55
26.1 2 10.9 1 18.5
27 1 12 1 19.5
32 1 15.1 2 23.55
33.1 1 18 1 25.55
35.1 2 19 1 27.05
37 1 19.9 1 28.45
39 1 23 1 31
44.1 1 24.1 2 34.1
45 2 25 1 35
46.9 1 27 2 36.95

48.9 1 28.9 1 38.9
50 2 30 2 40
51.1 1 30.9 2 41
52 1 36 1 44
54 1 41 1 47.5
59 1 42.1 1 50.55
60.1 1 48 1 54.05
61 1 61
64 2 64
64.9 1 64.9
Days in January below freezing point.
11
29 ×100
= 37.93%
5e)
Augu
st
Colum
n1
Colum
n2
Colum
n3
Colum
n4
High Frequen
cy Low Frequen
cy
Averag
e
68 1 53.1 1 60.55
70 1 54 1 62
73 2 60.1 5 66.55
75 1 61 2 68
75.9 3 62.1 3 69
77 2 63 2 70
78.1 4 64 5 71.05
79 5 64.9 8 71.95
80.1 3 66 4 73.05
81 2 81
82 4 82
82.9 1 82.9
84 2 84

Days In August with above 80℉
4
31 ×100
= 12.90%