logo

(Solved) Decision Support Tools- Assignment

   

Added on  2021-04-21

12 Pages862 Words24 Views
Decision Support Tools
Assignment
Student Name
Student Id
Subject Name
Subject Code
Lecturer Name

STUDENT NAME:
STUDENT NUMBER:
Question 1
Probability
(a) Any probability distribution is essentially a statistical function capable of representing the
possible values that a given random variable can assume and also lists down the
corresponding probability for the same (Flick, 2015).
There are various differences between continuous and discrete probability distribution
which are highlighted as follows (Fehr & Grossman, 2013).
The primary difference pertains to the random variable which is discrete for discrete
probability distribution and continuous for a continuous probability distribution.
The representation of a continuous probability distribution is achieved through
probability density function while the same is achieved through probability mass
function in case of discrete probability distribution.
The nature of frequency plot is continuous in case of continuous probability
distributions and the same is discrete when discrete probability distributions are
considered (Hastie, Tibshirani & Friedman, 2011).
Example:
Continuous Probability Distribution – Normal Distribution
Discrete Probability Distribution – Binomial Distribution
(a) Daily Sales with respect to number of days of selling the lead of bread is represented
below:
1

STUDENT NAME:
STUDENT NUMBER:
(i) “The probability to sell 3 or 4 loaves on any one day”
P ( sell 34 loaves ) =P ( sell 3loaves ) + P ( sell 4 loaves )
P ( sell 3 loaves )= 25
100 =0.25
P ( sell 4 loaves ) = 20
100 =0.20
Hence,
P ( sell 34 loaves )=0.25+ 0.20=0.45
Hence, 0.45 is the probability that baker’s top selling would sell 3 or 4 loaves on any one day.
(ii) “The average (mean) daily sales over period (100 days)”
2

STUDENT NAME:
STUDENT NUMBER:
Therefore,
The average (mean) daily sales over period (100 days) ¿ 285
100 =2.850
Hence, 2.850 is the average daily sale over the period of 100 days.
(iii) “The probability to sell 2 or more loaves on any one day”
P( sell 2more loaves)=P( sell 2 loaves)+ P(sell 3 loaves)+ P (sell 4 loaves)+ P(sell 5 loaves)
P ( sell 2loaves ) = 20
100 =0.20
P ( sell 3 loaves )= 25
100 =0.25
P ( sell 4 loaves ) = 20
100 =0.20
P ( sell 5 loaves ) = 15
100 =0.15
Hence,
P ( sell 2more loaves )=0.20+0.25+ 0.20+0.15=0.80
Hence, 0.80 is the probability that baker’s top selling would sell 2 or more loaves on any one
day.
(iv) “The probability to sell 4 or less loaves on any one day”
P ( sell 4less loaves ) =1P (sell 5loaves)
P ( sell 5 loaves ) = 15
100 =0.15
P ( sell 4less loaves )=10.15=0.85
3

End of preview

Want to access all the pages? Upload your documents or become a member.

Related Documents
Statistics for Analytical Decisions PDF
|11
|1749
|71

(PDF) Decision support systems and tools
|9
|1164
|35

Sample assignment on Statistics (pdf)
|8
|1496
|50

Probability and Hypothesis Testing in Employee Health Care Claims
|5
|642
|165

Data Analysis Plan for Age-Related Hearing Loss in Puerto Rico
|6
|810
|453

Probability and Expected Value
|6
|679
|180