Probability Distribution and Statistics Management

University: University of Cambridge

Unit No: N/A

Level: Undergraduate/College

Pages: 2 / Words 599

Paper Type: Assignment

Course Code: MAT274

Downloads: 32

Organization Selected : Rowlinson Knitwear

PROBLEM 1

A. Analysing mean value of hotel room rate

Mean

159.05

B. Analysing median of hotel room rate

Median

161

C. Analysing mode of hotel room rate

Mode

167

D. Analysing First quartile of hotel room rate along with its concept

1st Quartile

120

This is the 25% of the data set or variables which is lower half of the same.

E. Analysing Third quartile of hotel room rate along with its concept

3rd Quartile

120

This is the 75% of variable which is upper half of database.

PROBLEM 2

A. Determining the probability of selecting lowest or second lowest cost fund

FUND	Undergraduates	MBA	Combined
Lowest cost fund	1/5	1/5	1/5
Second lowest cost fund	3/8	2/5	2/5

B. Probability of analysing the lowest cost funds and is an undergraduate

FUND	Undergraduates
Lowest cost fund	1/5
Undergraduate	1/2
Probability of outcomes	2/3

C. Probability of lowest cost fund or is an undergraduate

FUND	Undergraduates
Lowest cost fund	1/5
Undergraduate	1/2

D. Analysing the probability for highest cost fund in undergraduate students

FUND	Undergraduates
Highest cost fund	1/4

PROBLEM 3

A. Probability workers take public transportation daily

Considering the Binomial random variable concept

P(X =3)

= 10!/ 3!(10-3)! 0.3^3(1-0.3)^10-3

=0.267

Moreover, X is a Binomial random variable

P= 0.3 and N = 10.

B. Probability of at least 3 workers take public transportation daily

P(X ≥ 3)	=1- P(X = 0) – P(X = 1) – P(X = 2)
P(X = 0)	= 10!/0! (10 – 0)!
0.3^0(1 – 0.3)^10-0
=0.0282
P(X = 1)	= 10!/1! (10 – 1)!
0.3^1(1 – 0.3)^10-1
=0.1211
P(X = 2)	= 10!/2! (10 – 2)!
0.3^2(1 – 0.3)^10-2
=0.2335
P(X = 0)+ P(X = 1)+ P(X = 2)	0.0282+ 0.1211+ 0.2335
=0.3828
=1- 0.3828
= 0.6172

PROBLEM 4

A. Measuring the mean and standard deviation

Mean	199.69
standard deviation	25.02

B. Probability of early morning trading will be less than 180 million

Z= (X - µ)/ ðž¼

n	13
Mean	199.69
standard deviation	25.02
Z=	-7.461630695

C. Probability of early morning trading more than 230 million

P(230<x)

= P (1.1639 < Z)

= 0.122

PROBLEM 5

FVN = PV (1+r) N

Particulars	Figures
Present Value (C0 )	500000
Rate per Period (r)	7%
Periods (n)	10
Future value	983575.68

PROBLEM 6

FVN = PV (1+rsm) mN

= $100000(1+0.074) 4(1)

= $107,185.90

PROBLEM 7

FVN=A [(1+r) N−1r]

=$20,000[(1+0.07)20−10.07]

=$819,909.85

PROBLEM 8

Particulars	Figures
Future value	75000
Interest rate	6%
Periods (n)	5
Present value	56044.36

PROBLEM 9

A. Payback period

Year	Project A	Cumulative cash inflow of project A	Project B	Cumulative cash inflow of project B
1	1000	1000	200	200
2	800	1800	600	800
3	600	2400	800	1600
4	200	2600	1200	2800

Project A: 2 + 200 / 600

= 2.3 or 2 years and 3 months

Project B: 3 + 400 / 1200

= 3.3 or 3 years and 3 months

B & C. NPV

Year	Project A	PV factor @ 10%	Discounted cash flow of project A	Project B	Discounted cash flow of project B
0	-2000	-2000
1	1000	0.909	909	200	182
2	800	0.826	661	600	496
3	600	0.751	451	800	601
4	200	0.683	137	1200	820
Total discounted cash inflow	2158	2098
Less: initial investment	2000	2000
NPV	158	98

D. Justification

As per making investment in both projects on which Project A needed to be selected as it has presented the favourable or most profitable outcomes. NPV value of Project A is 158 while Project B has 98.

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