Statistics for Managerial Decisions: Comprehensive Assignment

This document provides a detailed solution to a statistics assignment focused on managerial decision-making. The solution covers various statistical concepts, including stem and leaf plots, relative frequency histograms, bar charts, and price vs. fair value analysis for stock recommendations. It calculates mean, median, quartiles, standard deviation, and range for price earnings ratios, along with box and whisker plots. The assignment also includes probability calculations related to causes of death in Australia, Poisson distribution analysis for rainfall patterns, and normal probability plots for assessing data normality. Furthermore, it constructs confidence intervals and identifies variables with significant differences between float and non-float glass compositions. Desklib is a platform where students can find similar solved assignments and study resources.

STATISTICS FOR MANAGERIAL DECISION
ASSIGNMENT
[Pick the date]
Student Name

Question 1
(a) Quarterly opening price
Stem and leaf plot
1

(b) Relative frequency histogram and frequency polygon
2

(c) Bar chart for total market capitalisation of six companies as on January 1, 2017 is indicated
below.
AMP
Insurance Australia Group
SunCorp Group
QBE Insurance
IAG
Macquarie
0.00 5.00 10.00 15.00 20.00 25.00 30.00
Market Cap
Market Cap (Bn)
Companies Name
(d) The price vs fair value graph has been obtained for both the companies which would provide
an hint as to which stock is currently undervalued based on which a recommendation may be
given.
PPT Stock
MQG Stock
3

Based on the above, it is apparent that PPT stock price is currently lower than the fair value
which MQG is currently trading at little above the fair price. Hence, the PPT stock must be
purchased.
Source: http://www.morningstar.com
Question 2
(a) Mean, median, first quartile and third quartile of average price earnings ratio is shown below:
(b) Standard deviation, mean absolute deviation, range of average price earnings ratio is shown
below:
(c) Box and whisker plot of average price earnings ratio is shown below:
4

(d) Typically for stocks which have high potential of future earnings, the P/E ratio is high. On
the other hand, businesses with lower growth potential or stagnant earnings are likely to trade
at lower P/E. The P/E ratio tends to impact the price of the stock as for the same earnings per
share, two companies can have different prices owing to the future growth, associated risk
and robustness of the business model.
Question 3
(a) Probability that a randomly selected Australian person will die from neoplasms.
Total number of persons =147,678
Number of person who will die from neoplasms = 44,674
p= 44674
147678 =0.3025
Therefore, there is a 0.3025 probability that a randomly selected Australian person will die from
neoplasms.
(b) Probability that a randomly selected Australian person will die from disease of circulatory
system and is female.
Total number of persons =147,678
5

Number of female person who will die from disease of circulatory system = 22,493
p= 22493
147678 =0.1523
Therefore, there is a 0.1523 probability that a randomly selected Australian female person will
die from disease of circulatory system.
(c) Proportions have been calculated as shown below:
Highest proportion of male deaths compared to female deaths = Neoplasms
Highest proportion of female deaths compared to male deaths = Mental and behavioural system
(d) Probability that a randomly selected Australian person is dying from diabetes mellitus.
Total number of persons =147,678
Number of Australian person is dying from diabetes mellitus = 4328
p= 4328
147678 =0.0293 07
6

Therefore, there is a 0.029 probability that a randomly selected Australian person is dying from
diabetes mellitus.
Question 4
(a) Poisson distribution
1st week has started = 2nd January 2017
(i) Probability that on a given week in a year there would be zero rainfall.
Total number of days on which rain was experienced in a year = 134
Therefore, the mean number of days on which rain was experienced per week = 134/52 = 2.5769
No rainfall: x=0
P ( X=0|λ=2.173¿= ( 2.5769 )0 × e−2.5769
0 ! =0.076009
(ii) Probability that there would be 2 days or more than 2 days of rainfall in a week.
No rainfall: x ≥ 2
P ( X ≥2| λ=2.5769 ¿=1− ( 2.5769 ) 2 × e−2.5769
2 ! =1−0.5242=0.4757
(b) Normal distribution
The mean weekly rainfall in a given week = total amount o weekly rainfall / total number of
weeks
Mean=525
52 =10.096 mm per week
Standard deviation = 12.4638 mm
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Standard error ¿ Standard deviation
√ ( 52 ) =1.728
(i) Probability that in a given week, there would be rainfall amount between 5 and
15mm.
Z 1= 5−10.096
1.728 =−2.948
P ( Z 1<−2.948 )=0.001597
Z 2= 15−10.096
1.728 =2.8379
P ( Z 2<2.8379 )=0.9977
Now,
Rainfall amount between 5 and 15mm ¿ P ( 5<x< 15 )
P ( 5< x< 15 )=P ( Z 1<Z < Z 2 ) =0.9977−0.001597=0.9961
(ii) Total amount of rainfall if 10% of weeks have the same or higher amount of rainfall.
Here,
P ( rainfall> x )=0.10
The z value with respect to 10% can be found through NORMSINV () excel function.
z=NORMSINV ( 0.10 ) =−1.2815
Now,
-1.2815 = (X- 10.096)/ 1.728
x=7.88 mm
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Question 5
(a) Normal probability plots for the variables
1.51 1.515 1.52 1.525 1.53 1.535 1.54
-4
-3
-2
-1
0
1
2
3
4
Normal Probability Plot
Refractive Index
Sorted Data (Refractive Index)
z vaue
Comment: The above plot can be approximated as a linear shape and hence thus the given
variable can be approximated as normal distribution.
10 11 12 13 14 15 16 17 18
-4
-3
-2
-1
0
1
2
3
4
Normal Probability Plot
Sodium
Sorted Data (Sodium)
z value
Comment: The above plot can be approximated as a linear shape and hence thus the given
variable can be approximated as normal distribution.
9

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
-4
-3
-2
-1
0
1
2
3
4
Normal Probability Plot
magnesium
Sorted Data (magnesium)
Z value
Comment: The above plot deviates significantly from a linear trend and hence the given variable
cannot be approximated as a normal distribution.
0 0.5 1 1.5 2 2.5 3 3.5 4
-4
-3
-2
-1
0
1
2
3
4
Normal Probability Plot
Aluminimum
Sorted data (Aluminimum)
z value
Comment: The above plot can be approximated as a linear shape and hence thus the given
variable can be approximated as normal distribution.
10

69 70 71 72 73 74 75 76
-4
-3
-2
-1
0
1
2
3
4
Normal Probability Plot
Silicon
Sorted data (Silicon)
z value
Comment: The above plot can be approximated as a linear shape and hence thus the given
variable can be approximated as normal distribution.
0 1 2 3 4 5 6 7
-4
-3
-2
-1
0
1
2
3
4
Normal Probability Plot
Potassium
Sorted data (Potassium)
z value
Comment: The above plot deviates significantly from a linear trend and hence the given variable
cannot be approximated as a normal distribution.
11

4 6 8 10 12 14 16 18
-4
-3
-2
-1
0
1
2
3
4
Normal Probability Plot
Calcium
Sorted data (Calcium)
z value
Comment: The above plot deviates significantly from a linear trend and hence the given variable
cannot be approximated as a normal distribution.
0 0.5 1 1.5 2 2.5 3 3.5
-4
-3
-2
-1
0
1
2
3
4
Normal Probability Plot
Barium
Sorted data (Barium)
z value
Comment: The above plot deviates significantly from a linear trend and hence the given variable
cannot be approximated as a normal distribution.
12

0 0.1 0.2 0.3 0.4 0.5 0.6
-4
-3
-2
-1
0
1
2
3
4
Normal Probability Plot
Iron
Sorted data (Iron)
z value
Comment: The above plot deviates significantly from a linear trend and hence the given variable
cannot be approximated as a normal distribution.
(b) 90% confidence interval for variables for both float and non-float glasses.
For float glass
13

14

15

For non-float glass
16

17

18

19

The variable(s) which are significantly different in composition between float and non-float type of
glasses are as follows.
1) Aluminium
2) Barium
3) Magnesium
For the above mentioned variables, there is no overlapping of the confidence intervals predicted
for float type and non-float type glasses.
20