Business Decision Making

This document provides information on business decision making, including topics such as mean, standard deviation, coefficient of variance, relationships, and networking. It includes calculations and explanations for each topic. The document also mentions the challenges faced and improvements that can be made. The subject is business decision making.

BUSINESS DECISION
MAKING

Table of Contents
Task 1...............................................................................................................................................3
a) The mean.................................................................................................................................3
b) Standard Deviation..................................................................................................................3
c) Coefficient of variance............................................................................................................3
Task 2...............................................................................................................................................3
1...................................................................................................................................................3
2...................................................................................................................................................4
3...................................................................................................................................................5
Task 3: Relationships.......................................................................................................................5
1...................................................................................................................................................5
2...................................................................................................................................................6
3...................................................................................................................................................6
4...................................................................................................................................................6
Task 4: Networking.........................................................................................................................6
1...................................................................................................................................................6
2...................................................................................................................................................7
3...................................................................................................................................................7
Task 5...............................................................................................................................................7
1...................................................................................................................................................7
2...................................................................................................................................................7
3...................................................................................................................................................7

Task 1
a) The mean
Sum = 4972
Total sample size = 113
Mean = Sum / sample size
= 4972 / 113
= 44
The result shows that average age of Netflix viewer is around 44 years.
b) Standard Deviation
Standard deviation = √ variance
= √166.7
= 12.91
The standard deviation result shows that total variance between viewers age varies by 12.91.
c) Coefficient of variance
Coefficient of variance = standard deviation / mean
= (12.91 / 44) × 100
= 29.34%
The result shows that on an average the age of viewers are deviated from mean by 29.34%. This
indicates that most of the viewers have age around 44 years.
Task 2
1.
a)
Age
Category Frequency C.f. % C.f.
Under 25 5 5 4.42%
25 to 34 26 31 27.43%
35 to 44 30 61 53.98%
45 to 54 26 87 76.99%

55 and over 26 113
100.00
%
Total: 113
b)
Age Category Mid-term (x) Frequency (f) xf
16 to 25 20.5 5 102.5
25 to 34 29.5 26 767
35 to 44 39.5 30 1185
45 to 54 49.5 26 1287
55 to 64 59.5 26 1547
Total: 113 4888.5
Mean (x¯) = ∑xf / ∑f
4888.5 / 113
43.26106195
Age Category Mid-term (x) Frequency (f) xf
Mean -
x
(Mean -
x)2 f(Mean -x)2
16 to 25 20.5 5 102.5 22.76 518.0176 2590.088
25 to 34 29.5 26 767 13.76 189.3376 4922.7776
35 to 44 39.5 30 1185 3.76 14.1376 424.128
45 to 54 49.5 26 1287 -6.24 38.9376 1012.3776
55 to 64 59.5 26 1547 -16.24 263.7376 6857.1776
Total: 113 4888.5 15806.5488
Standard deviation =
=
= 11.82712826
2.
When calculating the means of grouped and ungrouped data, there will be a variation. The mean
of grouped data is preferred because it is more accurate as compared to the mean of ungrouped
data. The mean of ungrouped data may lead to wrong manipulation of the median therefore it is

considered inefficient in most cases. Thus, findings from grouped data are more accurate than
ungrouped data.
3.
25 34 44 54 64
0.00%
10.00%
20.00%
30.00%
40.00%
50.00%
60.00%
C.f. Curve
C.f. Curve
The minimum age requires to reach top 25% of viewers is 34 years. In the graph; on X axis
cumulative percentage is given and Y axis age of viewer is given.
Task 3: Relationships
1.
10 20 30 40 50 60 70
0
0.1
0.2
0.3
0.4
0.5
0.6
f(x) = − 0.00665618618218745 x + 0.551808975592273
R² = 0.724087835181154
Job Separation*
(Probability)
Linear (Job Separation*
(Probability))
Age
Probability

2.
The best predictor variable chosen for regression analysis is Age factor. The reason behind
choosing this variable is perfect negative relationship between both Job Separation Probabilities
and age. The relation clearly shows that with the increase in age; probability of Job Separation
reduces. While other variables doesn’t show any proper negative or positive relationship with
Job separation.
3.
The coefficient of determination or r2 is 0.724 which is equivalent to correlation coefficient of
0.85. The value of correlation is near 1; this indicates that there is nearly perfect relationship
between age and job separation probabilities.
4.
The regression equation identified in scatter diagram is ‘y = -0.006x + 0.551’.
The above equation shows that both Age and Job separation probabilities have negative relation.
That means if age increases; probability of Job Separation decreases.
In the above equation; the value of gradient is -0.006x and value of intercept is 0.551. Here,
independent factor is Age (x). So, to find that probability of Job Separation at specific age; this
equation is best fit.
Task 4: Networking
1.

2.
The total duration of the project is 16 weeks; which is minimum time taken by the project to get
complete.
There are two critical paths:
1. S→A→C→E→F→H
2. S→B→E→F→H
3.
Critical activities are those for which the start and finish times are strictly defined. They are
critical in the sense that their delay results in the delay of the whole project. The start time for
non-critical activities can, to a certain extent, be freely selected. In above diagram; D→G→H is
non-critical activities and can put on hold if necessary.
Task 5
1.
I think I have done all parts with perfection except Task 3; because in this task, scatter diagram
and identification of equation has perfectly done but the reason for choosing Age factor has weak
evidence.
2.
The most challenging aspect was Scatter diagram because I have less knowledge about how to
use excel for data presentation and also am not use to with various features of MS excel.
3.
I would try to present data and diagram in more professional way with proper labeling and
interpretation. I will try to gain more knowledge about excel to present my data in effective way.