Business Information Analysis Assessment 1 Answers

Question 1 Measures of Central Tendency

Central Tendency
“Mean”	“13.3”	“15.1”
“Median”	12.5	“15.5”
“Mode”	“12”	“17”
“Measures of dispersion”
“Standard Deviation”	2.002775851	2.469817807
Range	5	7

In this case, measures of central tendency and dispersion will be used on Telephone and Postal Services usage. The mean usage is even slightly higher for Telephone Services with 15.1 than for Postal Services that had a mean usage of 13.3. The median intensity is 15.5 for Telephone Services and 12.5 for Postal Services. The model also shows the most frequent usage for the services revealed to be 17 for Telephone Services and 12 for Postal Services. From these statistics, one gets the impression that Postal services are used less often than Telephone services. Regarding distribution, the standard deviation for Telephone Services equals 2.47 compared to 2.00 for Postal Services, therefore the usage of the telephone is more spread out. In addition, the range of usage ideas for the Telephone Services is more extensive than that for Postal Services. In summary, the results of the study imply that the dispersion of Telephone Services is higher, as opposed to Postal Services, for which users have more stable patterns of service consumption (Talwar et al., 2020).

Mean:

Customers use the Postal Services a number of times 13.3 while they use the Telephone Services a number of times 15.1. This means that Telephone Services are used more often than Postal Services.

Median:

Chapter Four results on the GrabBag median usage frequency show that Postal Services have a median of 12.5 while Telephone Services have a median of 15.5. This reveals the extended usage of Telephone Services over time and shows that compared to Postal Services, a larger percentage of the use values lie in the excess of the median.

Mode:

The mode of the Postal Services’ values equals 12, whereas Telephone Services have a mode of 17. This goes further to show that usage levels in Telephone Services can be higher than those in Postal Services at some months.

Comparison of Dispersion:

Standard Deviation:

Telephone Services have a standard deviation of 2.47 slightly above the 2.00 observed among Postal Services. This means that Telephone Services show more variation from one month to another, thereby allowing carriers to be more responsive to different communication needs.

Range:

The usage range for Telephone Services is 7, if we take the difference between the maximum and minimum values of the indicator, while the value of the same measure for Postal Services is 5. This broader oscillation in usage means that Telephone Services eclipses the others in terms of a CoV, as it reveals greater variability of usage.

Question 2: Probability

a. Probability of a Cilico Product:

Currently, the likelihood of a product being manufactured by Cilico can be calculated as the number of products that are being produced by Cilico out of the total number of products. This can be expressed as:

The odds of Cilico (odds(Cilico)) = 16:4 = 4, which is equal to 0.65.

b. Probability of an AUM Product:

The likelihood of a product being produced by AUM can be estimated as the ratio of number of products produced by AUM to the total number of products produced. This can be expressed as:

From the research P(AUM= Large Amounts Under Management) probability is estimated to be 0.35.

c. Probability of a Defective Product from Cilico:

The probability of a defective product being produced by Cilico is calculated using an application of conditional probability. The formula used is:

P(Cilico | Defective) = {[P (Defective | Cilico) * P(Cilico)]} ÷ P (Defective).

Substituting the values:

P(Cilico | Defective) = (0.08 * 0.65) / 0.094 = 0.052 / 0.094 = 0.553.

d. Probability of a Defective Product from AUM:

The likelihood that a flawed product was produced by AUM is arrived at through the computation of conditional probability as well. The formula is:

We also needed to evaluate the probability of AUM given defective; P(AUM | Defective) = [P(Defective | AUM) * P(AUM)] ÷ P(Defective).

Substituting the values:

P(AUM | Defective) = (0.12 * 0.35) / 0.094 = 0.042 / 0.094 = 0.447.

e. Overall Probability of a Defective Product:

The probability that a defect is likely to be produced is the probability that it is produced by Cilico and or AUM. This is calculated as:

probability of defective ( P [Defective] ) is equal to ( P [Defective|Cilico] × P [Cilico] )) + ( P [Defective|AUM] × P[AUM]).

Substituting the values:

P(Defective) = 0.08 * 0.65 + 0.12 * 0.35 = 0.052 + 0.042 = 0.094.

Question 3 Binomial Probability

a.

On each of the four days of the weekend, it is equally likely that there will be an accident; the likelihood of an accident on any one of these days is 0.25. Using the complement rule, the probability of no accidents over the weekend is calculated as:

P(No Accidents) = 0 ^{th} P(Accident) = 0.25 therefore P(No Accidents) = 1 - 0.25 = 0.75. This means that if such a case occurs there is 75 percent probability that no accidents will happen over the weekend.

b.

To calculate the probability of exactly two accidents over the four days:

Probability of Haoccurring two accidents = 4C2 * 0.25 * 0.75

Here 4C2 = 6, which means the number of ways to select two days out of these four. The probability of an accident on two specified days is (0.25)² on two and the probability of no accident on two specified days is (0.75)² on two.

Substituting the values: P(Two Accidents) = 6 × 7/32 = 0.21094

Probability of Exactly Three Accidents:

The probability of three accidents occurring over four days is calculated as:

P(Three Accidents) = 4C3 × 0.015625 × 0.75

In general it equals 4C3 = 4, here 4C3 denotes the number of ways in which we can select 3 days out of 4. The chances of having three days of accident are equal to 0.25 raised in the power of three and the chances of having no accident for the other day equal the complement of 0.25 or 0.75.

Substituting the values: There is much less chance of getting a Fourth Accident: P(Three Accidents) = 4 × 0.015625 × 0.75 = 0.046875

Probability of Exactly Four Accidents:

The probability of accidents occurring on all four days is:

P(Four Accidents) = (0.25)^4 = 0.00390625

Probability of At Least Two Accidents:

The probability of at least two accidents occurring is calculated by summing the probabilities of exactly two, three, and four accidents:

P(At Least Two Accidents) = 0.04 + 0.03 + 0.02 = 0.09

P(At Least Two Accidents) = P (Two Accidents or More) = P(one or more) = P(One accident) + P(Two Accidents) + P(Three Accidents) + … = 0.21094 + 0.046875 + 0.00390625 = 0.26172

c.

Expected cost is calculated by multiplying the probability of occurrence of each type of accident by the cost of such an accident and then adding them.

- For one accident: Probability = 0.25, Cost = £20,000, Con. to exp. cost = £5,000

- For two accidents: Probability =0.21094 Cost = £30,000 expected contribution to the cost = £6,328.20

- For three accidents: Probability = 0.09375, Cost = £,600, Contribution to expected cost = £5.625

- For four accidents: Probability = 0.00390625, Cost = £125,000, Expected cost = £488.28

The total expected cost is the sum of these contributions:

Expected Cost = £17,441.41

d.

The projected outlay for the accidents over the weekend is £17,441.41. If a private company were to charge £20,000 to the Ebola virus for emergency services during this period, the extra cost burden would be £2,558.59 distributed among the cab operators and the community. The rather simple math serves the purpose of maintaining the costs well below the twenty thousand pound mark (El-taweel et al., 2023). It brings out how fluctuations in accident frequency significantly affect total costs.

Question 4: Hypothesis Testing

This analysis consists of determining the arithmetic mean, median and mode of postal and telephone services, and assessing the probabilities of accidents during the weekend, and their cost. Data collected from various sources is often analyzed with the use of various methods such as arithmetic mean, median, mode, standard deviation, arithmetic range and binomial probability equations (Nundy, Kakar and Bhutta, 2021).

Central Tendency Analysis

For each postal service type, on average, there are 13.3 services and 12.5 for the median. The most frequent value, or mode, is 12, and dispersion, measured by the standard deviation of 2.003, is weak. This is further doubled by the range of 5 taken from 11 to 16. On the other hand, the average usage of telephone service was 37.1, mode 17, median 15.5 indicating more usage by the consumer. A coefficient of variation of 2.47 implies more fluctuation in telephone service than postal service usage. The broader range of 7 (from 12 to 19) gives even greater evidence in favour of the necessity of higher and more diverse demands for telephone services.

Probability Assessment of Accident

The probability of accidents on each day of a four-day weekend is determined by the binomial probability distribution model. If there are no accidents at all, it will be 0.75, when there are two accidents, it will be 0.21094 and when there are three accidents, it will be 0.046875. The probabilities of accidents on all four days are very small and equal to 0.00390625. In total, it is estimated that on average the cost of managing accidents over a weekend is £17,441.41.

Compared to a fixed charge of £20,000 from a private firm for emergency services, the fixed cost is 2,500 pounds more than the average estimated cost. However, the fixed charge eliminates cost volatility and minimizes the variability of financial cost in instances whereby the true number of accidents outperforms the estimated numbers and may thus be especially beneficial to companies that are likely to record a higher number of accidents than average (Bachani et al., 2017).

Hypothesis Testing

Hypothesis testing therefore offers a step-by-step approach to testing as to whether the variations observed in data or outcomes are real or are just part of natural variation (Lancaster et al., 2018). They help to judge whether fluctuations in service demand or accident costs are regular or random. This method can be used to determine whether the differences in postal and telephone usage are significant or whether the variation between predicted accident cost and fixed accident cost is large or small (Khan et al., 2023).

Null Hypothesis (H₀): This implies that there is no huge disparity in the average returns

between Investment A and Investment B.

H₀: μB − μA = 0

Alternative Hypothesis (H₁): The expected return of Investment B (μB) is higher than

Investment A (μA) greatly.

H₁: μB − μA > 0

Process and Key Assumptions

The first step entails investment in measures of central tendency with a view of recognizing the patterns or risks associated with postal and telephone services. The main assumptions are:

• The instances of accidents sustain a random pattern over the whole weekend.
• The likelihood of an occurrence of an accident does not reduce or increase with the occurrence of another accident (Ahmed et al., 2023).
• The data is considered credible in that its predictive capabilities can yield future results.

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References

• Ahmed, S., Hossain, M.A., Ray, S.K., Bhuiyan, M.M.I. and Sabuj, S.R. (2023). A study on road accident prediction and contributing factors using explainable machine learning models: analysis and performance. Transportation Research Interdisciplinary Perspectives, [online] 19, p.100814. doi:https://doi.org/10.1016/j.trip.2023.100814.

• Bachani, A.M., Peden, M., Gururaj, G., Norton, R. and Hyder, A.A. (2017). Road Traffic Injuries. 3rd ed. [online] PubMed. Available at: https://www.ncbi.nlm.nih.gov/books/NBK525212/ [Accessed 10 Dec. 2024].
• El-taweel, R.M., Mohamed, N., Alrefaey, K.A., Husien, S., Abdel-Aziz, A.B., Salim, A.I., Mostafa, N.G., Said, L.A., Fahim, I.S. and Radwan, A.G. (2023). A review of coagulation explaining its definition, mechanism, coagulant types, and optimization models; RSM, and ANN. Current Research in Green and Sustainable Chemistry, [online] 6, p.100358. doi:https://doi.org/10.1016/j.crgsc.2023.100358.
• Khan, J.A., Raman, A.M., Sambamoorthy, N. and Prashanth, K.C. (2023). Research Methodology (Methods, Approaches And Techniques). [online] SAN INTERNATIONAL SCIENTIFIC PUBLICATIONS. doi:https://doi.org/10.59646/rmmethods/040.
• Lancaster, G., Iatsenko, D., Pidde, A., Ticcinelli, V. and Stefanovska, A. (2018). Surrogate data for hypothesis testing of physical systems. Physics Reports, [online] 748, pp.1–60. doi:https://doi.org/10.1016/j.physrep.2018.06.001.
• Nundy, S., Kakar, A. and Bhutta, Z.A. (2021). Understanding Medical Biostatistics. Springer, pp.95–116. doi:https://doi.org/10.1007/978-981-16-5248-6_10.
• Talwar, S., Dhir, A., Khalil, A., Mohan, G. and Islam, A.K.M.N. (2020). Point of adoption and beyond. Initial trust and mobile-payment continuation intention. Journal of Retailing and Consumer Services, 55, p.102086. doi:https://doi.org/10.1016/j.jretconser.2020.102086.