Applied Quantitative Methods Assignment II

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This document provides solutions for Applied Quantitative Methods Assignment II. It covers topics such as number of passengers at train station, frequency histogram, central tendency measures, correlation coefficient, regression line calculation, probability calculations, Bayesian probability, binomial distribution, Poisson distribution, and more.

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APPLIED QUANTITATIVE METHODS
ASSIGNMENT II
Student Name
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Question 1
(a) Number of passengers at train station (Melbourne)
(b) Frequency histogram
(c) Central tendency measures
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Question 2
(a) The dataset provided is labelled as a sample as it corresponds to selected values
which have been chosen from the population of all weeks in a year. If the given
data corresponded to the population data, then all the 52 weeks contained in a
year would have been represented. Instead only selected values have been
captured which denote a sample.
(b) standard deviation
(c) Inter Quartile Range
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Sorted
In cases where the underlying data is skewed, standard deviation does not provide an
accurate picture and as a result, it is better to rely on IQR to capture the extent of
dispersion in the data, IQR is not impacted by extreme values (Eriksson and
Kovalainen, 2015).
(d) Correlation coefficient
Correlation coefficient (r) = 0.968
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The correlation coefficient indicated above highlights that the underlying relationship
between the given variables is positive. Also, the strength of the relationship between
the variables is quite strong which is apparent from the high magnitude of the
correlation coefficient (Hair et al., 2015).
Question 3
Calculation for regression line
Slope = a =
476181676576 339223425707
{7 (1676576)2 33922} = 1628.689
Intercept = b =
7 23425707 339247618
{7 (1676576)2 33922} = 10.677
y = a + bx; y = 1628.689 + 10.677 x
Chocolate bars sold = 1628.689 + (10.677 Student Weekly Attendance)
The two examples to refers the regression line
Holmes is closed in holidays which implies that the weekly attendance is 0.
Number of chocolate bars sold = 1628.689 + (10.677 weekly attendance)
Number of chocolate bars sold = 1628.689 + (10.677 0.00) = 1628.689 or 1629
If there is an increase in the student attendance by 10, then the chocolates bars sold
would be increased by 1068 units approximately which can also be derived by
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multiplying the slope of regression equation by the change in weekly attendance of
students.
(b) Calculation of R2
R2 = (Corelation coefficient)2 = (0.968)2 = 0.937
The above value highlights that 93.7% of the variation in chocolate bars sale can be
accounted for by corresponding changes in the weekly attendance of students. This
clearly implies that the model is a good fit with high prediction capability (Flick, 2015).
Question 4
(a) Probability (grassroots training or Holmes University)
(b) Probability (scientific training and external).
(c) Probability (scientific training and Holmes University)
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(d) The two events i.e. training (P) and recruitment (Q)will be independent when.
PP× PQ= P(P and Q)
P(P) = P (External) = (54 + 12)/ (35 + 92 + 54 + 12) = 0.3412
P(Q) = P (Scientific training) = (35 + 54)/ (35 + 92 + 54 + 12) = 0.4611
PP× PQ= 0.1573
PP and Q= PExternal and Scientific training= 0.2576
PP× PQ P(P and Q)
Tr
aining (P) and recruitment (Q) will be independent because the condition has not
satisfied.
Question 5
(a) Probability Segment A ( prefer X
Y , Z )
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P Segment A= 0.55, P (Segment B) = 0.30
P Segment C= 0.10, P (Segment D) = 0.05
P(Product X|Segment A) = PXa=A= 0.20
P(Product X|Segment B) = PXa=B= 0.35
P(Product X|Segment C) = P Xa=C= 0.60
P(Product X|Segment D) = P Xa=D= 0.90
P(X) = (0.55 0.2) + (0.35 0.3) + (0.6 0.1) + (0.9 0.05) = 0.32
Bayesian Probability
P Segment Aa=Product X) = P Aa=X= P AP (X|A)
P (X) = 0.55 0.2
0.32 = 0.35375
(b) Preference of customer be product X
P(X) = (0.55 0.2) + (0.35 0.3) + (0.6 0.1) + (0.9 0.05) = 0.32
Question 6
(a) Probability 2 or less customers will buy
Binomial Distribution
P x 2= Px = 0+ Px = 1+ Px = 2
Number of trials = 8
Probability of success = 0.1
P x 2= {8
0 (0.1)00.98} + {8
1 0.1)10.97+ {8
2 0.1)20.96= 0.96
(b) Probability that 9 customers will come for buying in 2 min
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Poisson distribution
γ = 8
x = 9
P x = 9= eγγx
x! = 𝑒889
9! = 0.124
Question 7
(a) Probability (Apartment will sell > $2 Mn)
μ= $1.1 million = $1100000, σ= $385,000, x = $2 million = $2000000
z = x μ
σ =
2000000 1100000
385000
z = 2.34
From Standard Normal Table
P z > 2.34= 0.0097
P x > $2 million= 0.0097
(b) Probability (Apartment will sell b/w $1Mnand $1.1 Mn)
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μ= $1.1 million = $1100000, σ= $385,000
x1 = $1 million = $1000000, x2 = $1.1 million = $1100000
Px1 < x < x2= P 1000000 < x < 1100000
= P
1000000 1100000
385000 < x μ
σ <
1100000 1100000
385000
Px1 < x < x2= P 1000000 < x < 1100000= P 0.26 < z < 0
From Standard Normal Table
P z < 0= 0.5000
P z < 0.26= 0.3975
Px1 < x < x2= P 1000000 < x < 1100000= 0.5000 0.3975
P 1000000 < x < 1100000= 0.1025
Question 8
(a) As per the Central Limit Theorem, it is fair to assume that the underlying sample
distribution is normal provided the sample size is sufficiently large. This sample size has
been defined as 30. It is known that the sample size provided in the given case is 50
which is significantly greater than 30 which implies that despite the original population
not being normal, the given sample data can be assumed to be normal (Hillier, 2016).
(b) Sample proportion p= 11
45 =0.244
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30% investors agreed to participate into fund.
𝑃𝑝 > 0.3= 𝑃 𝑧> 0.3 0.2444
0.064 = 𝑃𝑧> 0.87= 1 𝑃𝑧< 0.87= 1 0.807
= 0.192
S.E.= (1 p)p
n = 0.24441 0.2444
45 = 0.064
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References
Eriksson, P. and Kovalainen, A. (2015) Quantitative methods in business research. 3rd
ed. London: Sage Publications, pp. 115
Flick, U. (2015) Introducing research methodology: A beginner's guide to doing a
research project. 4th ed. New York: Sage Publications, pp. 206
Hair, J. F., Wolfinbarger, M., Money, A. H., Samouel, P., and Page, M. J.
(2015) Essentials of business research methods. 2nd ed. New York: Routledge, pp. 165
Hillier, F. (2016) Introduction to Operations Research.6th ed.New York: McGraw Hill
Publications, pp. 189
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