Applied Quantitative Methods - Desklib

This article covers various topics like frequency distribution, frequency histogram, computation of mean, median, mode, correlation coefficient, probability computation, and more related to Applied Quantitative Methods. It also includes sample calculations and interpretations. The subject is Mathematics, the course code is MATH101, and the college/university is not mentioned.

APPLIED QUANTITATIVE METHODS
STUDENT NAME/ID
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Question 1
(a) Frequency distribution (class size =10) for prices of the parts
(b) Frequency histogram for prices of the parts
(c)Computation of mean
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Computation of median: Sorted data (ascending order)
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Median
Computation of mode
Maximum frequency is reported for data point 140.
Mode = 140
Question 2
Data about number of flights at the airport and the associated sale of baggage for 7 weeks is
highlighted below.
(a) Sample
It can be concluded that the given set of data is a sample data because it is data corresponding to
only7 weeks which has been extracted from a given population (large data set).Further, the data
would be considered as population only if every week data for which the airport has been
operated is provided rather than just for 7 weeks data. Hence, this is a sample data only.
(b) Standard deviation for variable ‘Number of flights at the airport’
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(c) Inter quartile range ‘IQR’ for the variable number baggage sold at the airport
Sorted data (Ascending order)
The inter quartile range does not get affected because of the existence of extreme value and skew
and therefore, it is considered as an appropriate measure while comparing with standard
deviation (Harmon, 2016).
(d) Calculation and interpretation of correlation coefficient
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Interpretation: The positive value of correlation coefficient is indicative of the fact that variables
are positively associated. Further, the strength of the association is termed as moderate because
the value of correlation coefficient is nearby 0.5.
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Question 3
(a) Calculation for least square regression line
Y =a+bX =24.90+0.391 X
Number of baggage sold=24.90+(0.391∗Number of flights at airport )
(b) The square of the correlation of coefficient is termed as coefficient of determination.
Coefficient of determination (R^2) = (0.4535)^2 = 0.2057
The value of coefficient of determination has been derived as 0.2057 which implies that only
20.57% deviation in the number of baggage’s sold would be offered explanation by deviation in
the number of flights at airport. Clearly, this value is not high and hence, the regression model
cannot be assumed to be a good fit (Hillier, 2016).
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Question 4
(a) Let the probability of being a club member and get grassroots training is P(A)
Now,
P ( A )= 100+ 20+40
40+50+100+20 =0.762
(b) Let the probability of being a an external and get scientific training is P(B)
Now,
P ( B ) = 50
40+50+100+20 =0.238
(c) Let the probability of being a club member and get scientific training is P(C)
Now,
P ( B )= 40
40+100 =0.286
(c) Events recruitment and training are independent or dependent
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Since the underlying probabilities of receiving particular form of training tends to vary across the
two recruitment methods, hence it would be correct to conclude that the given events show some
inter-dependency and thereby cannot be termed as independent.
Question 5
Let make the probability table for representing the various scenarios.
The requisite probability is 0.58 which implies that a randomly chosen customer who is from
segment A will select TV over radio.
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Question 6
Computation of mean
Mean=E ( x ) =∑ ¿ ¿
Mean=0.61132
Computation of standard deviation
s= √ ∑ ( x−E ( x ) ) 2
. P ( x )
s= √43.9729=6.63
Therefore, the mean and standard deviation for the probability distribution is 0.61132 and 6.63
respectively.
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Question 7
Probability computation
Mean μ=250 km/hour∧Standard deviation σ=30 km/hour
(a) P(train average < 200 km/hour)
P ( x<200 )=P {( x−μ
σ )<( 200−250
30 ) }
P ( x<200 ) =P( z ←1.67)=0.0475 (Standard normal table) (Flick, 2015)
0.0475 is the probability that
the train average will be
lower than 200 km per hour.
(b) P(train average > 300
km/hour)
P ( x>300 )=P {( x −μ
σ )<( 300−250
30 ) }
P ( x>300 )=P(z >1.67)=0.0475 (Standard normal table)
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0.0475 is the probability that the train average will be higher than 200 km per hour.
(c) P(train average between 210 and 280 km/hour)
P ( 210< x<280 ) =P {( 210−250
30 ) < x−μ
σ < ( 280−250
30 )}
P ( 210< x<280 )=P (−1.3< z <1 )=0.7495 (Standard normal table)
0.7495 is the probability that the train average will be between 210 and 280 km/hour.
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Question 8
Sample size n = 49
Mean μ=448 shoppers
Standard deviation σ =21 shoppers
Now,
P( average will lie between 441 shoppers and 446 shoppers )
P ( 441< x <446 ) =P {( 441−448
21/ √ (49) ) < x−μ
σ < ( 446−448
21/ √ (49) ) }
P ( 441< x <446 )=P (−2.3<z <0.667 ) =0.2415(Standard normal table)
Therefore, there is a 0.2415 probability that average will lie between 441 shoppers and 446
shoppers.
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References
Flick, U. (2015) Introducing research methodology: A beginner's guide to doing a research
project. 4th ed. New York: Sage Publications
Harmon, M. (2016) Hypothesis Testing in Excel - The Excel Statistical Master. 7th ed. Florida:
Mark Harmon.
Hillier, F. (2016) Introduction to Operations Research. 6th ed. New York: McGraw Hill
Publications.
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