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Applied Quantitative Methods

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c) Inter quartile Range (IQR) of selling of chocolate bars and analysation of usefulness of IQR over standard deviation

The IQR Range is obtained by finding the difference between the thirds and the first quartile.

The Inter quartile range helps to depict the central tendency and spread and has a few more benefits too which include:

       Skewness

It gives a comparison to the quartiles to confirm whether it is a skewed data or not. According to Puth et al. (2014, p.186), a data consisting of high values tilts the median to lie in third quartile whereas the reverse of the above scenario tilts it closer to first quartile.

       Outlier identification

It eliminates the outliers, which are more than 1.5 times the IQR distance, and it is slightly above the third Quartile and just below the first Quartile.

d) Calculation and interpretation of the correlation coefficient

The bonding between two sets of values gives Correlation Coefficient. A positive correlation gives linear graph, which signifies highly correlation between the given sets of values.

a) Calculation and interpretation of the Regression Equation

A regression equation is written as y = a + bx, where “a” and “b” represents the following formula

The below table shows, the equation of regression and the above parameters “a”, “b” and “x” are predicted henceforth:

b) Calculation and interpretation of Determination Coefficient

The determination coefficient reflects the value of r^2, which is known as the correlation coefficient. The calculation is shown in Question two.

Q4: Probability Distribution

a) Determination of the probability for random selection of the player who have been receiving the training of Grassroots or from the Holmes

Probability of the given condition

P (Grassroots training + Holmes) = (35 + 92 + 12)/ (35 + 92 + 54 + 12) = 139/ 193= 0.7202 (Approximately).

b) Identification of the probability for random selection of the player who have been receiving the scientific training or from the External

Probability of the given condition

P (External training + Scientific training) = 54/ (35 + 92 + 54 + 12) = 54/ 193= 0.2798 (Approximately).

c) Finding of the probability of the player from Holmes to be in the scientific training

Probability of the given condition

P (Holmes recruitment + Scientific training) = 35/ (35 + 92 + 54 + 12)= 35/ 193= 0.1813 (Approximately).

d) Identification of the level of dependency between training and recruitment

There are two distinct events -training and recruitment, both of them are independent.

Probabilities of the players to be recruited from Holmes have been receiving scientific training.

That is, P (Holmes + Scientific training) = 35/ (35 + 92 + 54 + 12)= 35/ 193 = 0.1813 (Approximately).

Probabilities of the players to be recruited from Holmes have been receiving Grassroots training.

That is, P (Holmes + Grassroots training) = 92/ (35 + 92 + 54 + 12)= 92/ 193 = 0.4767 (Approximately).

Probabilities of the players from external recruitment have been receiving scientific training.

That is, P (External recruitment + Scientific training) = 54/ (35 + 92 + 54 + 12) = 54/ 193 = 0.2798 (Approximately).

Probabilities of the players from external recruitment have been receiving Grassroots training.

This implies that, P (External recruitment + Grassroots training) = 12/ (35 + 92 + 54 + 12)= 12/ 193 = 0.0622 (Approximately).

Therefore, the events can be said independent, as both the incidents are not affecting the other (Dopkins et al. 2017, p.1660). Several events, which are not affecting each other by the supplementary knowledge concerning the occurrences of the number of the remaining events, are known as independent events.

Q5: Bayes’ Rule

a) Determination of the probability for the customer from Segment A prefers the product X over Y o Z

Bayes’ Theorem states that:

Probability (person from the Segment A) = P (A) = 55/ 100 = 0.55

Probability (consumer choosing product X) = P (B) = 20/ 100 = 0.2

Probability (consumer choosing other products) = P (BA) = 80/ 100 = 0.8

Therefore, the probability of the given scenario = P (AB) = (0.8 * 0.55)/ 0.2

P (AB) = 0.44/ 0.2 = 2.2

b) Probability of random customer to prefer product X

P (Consumer from any Segment) = 1/ (0.55 + 0.3 + 0.1 + 0.05) = 1/ 1 = 1

P (Consumer to choosing product X) = 1/ (0.2 + 0.35 + 0.6 + 0.9) = 1/ 2.05 = 0.4878 (Approximately).

Q6: Poisson distribution

a)  Determination of the probability of only 2 people or less of the 8 people to buy anything

Probability is 0.1

Mean (x bar) = (8 * 2)/ 2 = 8

Variance (s2) = (82 * 2)/ 2 = 64

P (2 or less than the respective 8 people will purchase) = P(2 < 8) = (x bar * Euler’s constant)/ x!

That is, P (2 < 8) = (8 * 2.71828)/ 8! (Euler’s constant = 2.71828)

Hence, P (2 < 8) = 21.74624/ (8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) = 21.74624/ 40320

b) Determination of the probability for 9 people to enter the store in next 2 minutes

People (x) = 9

E(x) = (9 * 2)/ 2 = 9

Probability of 9 people

P (9 people within next 2 minutes) = (1 - (9 * 2.71828))/ 9! = (1 - 24.46452)/ 362880 - 23.46452/ 362880

Q7: Normal distribution

a) By Normal distribution, Probability of apartment selling at greater than $2 million

Standard Deviation (as per last 12 months sale) = $385000

Normal Distribution Formula for more than one type is:

f (x) = (number of events - mean of the data)/ standard deviation.

Therefore, probability of selling the apartment is

f (selling the apartment at more than $2 million) = f(z > 200000)

That is, f (z > 200000) = (200000 - 110000)/ 385000 = 90000/ 385000 = 0.2338 (Approximately).

b) Probability of selling the apartment at less than $1.1 million but more than $1 million

Standard Deviation (as per last 12 months sale) = $385000

Normal distribution formula of less than type is

f (x) = (upper case - lower case)/ standard deviation.

Probability of the given condition:

f (selling the apartment at less than $110000 but more than $100000) = f(100000 < z <110000)

f (100000 < z <110000) = (110000 - 10000)/ 385000

Therefore, f (100000 < z <110000) = 10000/ 385000 = 0.0260 (Approximately).

Q8: Z-Distribution

a) Reasons for using or not using the Z-distribution

Standard Deviation (as per last 12 months sale) = $385000

The probability of selling at greater than $ 1 million but less than $ 1.1 million is

f (selling the apartment at less than $110000 but more than $100000) = f(100000 < z <110000)

f (100000 < z <110000) = (110000 - 10000)/ 385000 = 10000/ 385000 = 0.0260 (Approximately).

Since it is not a normal distribution, hence, it is not possible to apply Z-distribution over here.

b) Determining the probability of 30% investors

Total number of investors = 2000

45 of the total investors will invest $100000 each.

Hence, the sum will become = $4500000

Therefore, the amount will be of $1.1 million.

30% of the 45 investors = 45 * (30/ 100) = 45 * 0.3 = 13.5.

P (30% of 45 investors will commit to investing more than or equal to $1 million) = 13.5/ 45

Therefore, P (30% of 45 investors will commit to investing more than or equal to $1 million) = 0.3.

Bibliography

Cohen, P., West, S.G. and Aiken, L.S., (2014). Applied multiple regression/correlation analysis for the behavioral sciences, pp. 14

Coleman, J., (2017). The mathematics of collective action. Abingdon: Routledge, pp. 19.

Cox, D.R. and Wermuth, N., (2014). Multivariate dependencies: Models, analysis and interpretation. Chapman and Hall/CRC, pp. 22.

Darlington, R.B. and Hayes, A.F., (2016). Regression analysis and linear models: Concepts, applications, and implementation. New York: Guilford Publications, pp. 23

Dopkins, S., Varner, K. and Hoyer, D., (2017). Variation in the standard deviation of the lure rating distribution: Implications for estimates of recollection probability. Psychonomic bulletin & review24(5), pp.1658-1664.

Keith, T.Z., (2014). Multiple regression and beyond: An introduction to multiple regression and structural equation modeling. Abingdon: Routle

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