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Building paragraphs for paper 1

Building paragraphs for paper 1

Thesis paragraph—will work on in conference

Paragraph 1

Summary of article

Paragraph 2

Rhetorical paragraph—this paragraph does not need to be in this order but have as much information as possible to show how effective it was Rhetorically.

Writer

What do they use to help construct the argument to show Ethos—types of examples and evidence

Audience

Purpose—overall intention of the article

Exigence

Pathos—how does the author(s) get an emotional response from the audience

Logos—body of paper

Paragraph 3

logical steps and how they construct the article.

Evidence that is most important to you

How does this evidence relate to the larger argument

Paragraphs 4-7/8

Start making connections of your small piece of evidence to the larger article

Management Controls

Management Controls

Task 1

The main advantage of a fixed budget is cutting down the cost of doing business. It is well known that a fixed budget aids the management of a business in terms of maintain strict controls on the developed budget. This implies that the management does not have to worry about the outlook of the company in future. It has become problematic for many businesses to get their act together in terms of maintaining a specified budget.

The disadvantage of fixed budgets is that the business cannot make adjustments when the market changes. The problem with a fixed budget is that no adjustments can be made until the end objective has been attained. This means that a business with a fixed budget system cannot make good progress in terms of dealing with market changes. The business has to be in a good position in terms of countering changes in the market.

The first merit of a variable budget is that the business can effect changes within the course of doing business. It is evident that the modern business platform keeps on changing. The changes seen in the market system are reflective of choices made by stakeholders in the market. The good aspect with a variable budget is that changes can be made from time to time when there is need.

The main disadvantage of a variable budget is execution of policies. When it comes to the variable budget, the mode of implementation is considered challenging. This is because various issues have to be integrated in terms of making the latter work. For example, the management of the business ought to understand how to use all controls in implementation.

Task 2

City On time Deliver Performance

Quarter 1 (Ending march 31st) Quarter 2 (Ending June 30th) Quarter 3 (Ending September 30th) Quarter 4 (Ending December 31st)

Luton 10,000 10,000 0 0

Glassgow 0 0 0 10,000

Table 2.1

Legends:

Bonus of 10,000/quarter will only be paid to employees if and only if the Deliver Performance will be 98% or more then that

City Profitability Bonus Per Month

Bonus for Moths of Jan, Feb and March Bonus for Month of April, May and June Bonus for Month of July, Aug and Sep Bonus for Month of Oct, Nov and Dec

Luton 5333.33/ Month 5,666.66/Month 4,666.66/Month 6,000/Month

Glassgow 10,666.66/Month 10,000/Month 12,000/Month 12,666.66/Month

Table 2.2

Legend

The formula for calculating month bonus is [(quarterly profit * 0.02) / 3] Where 3 represent number of months in one quarter.

City Total Deduction from Bonus per Quarter

Quarter 1 Quarter 2 Quarter 3 Quarter 4

Luton 9,000/Quarter 13,000/Quarter 5,000/Quarter 12,500/Quarter

Glasgow 17,500/Quarter 17,000/Quarter 14,000/Quarter 11,000/Quarter

Table 2.3

Legend

The formula for the deduction is [(total cost of return from sales * 0.5)]

City Bonus after Deduction

Quarter 1 Quarter 2 Quarter 3 Quarter 4

Luton 2,333.33/Month 1,333.33/Month 3,000/Month 1,833.33/Month

Glasgow 4,833.33/Month 4,333.33/Month 7,333.33/Month 9,000/Month

Table 2.4

Legend

The formula for total bonus per month after deduction is {[(Total Bonus per Month before Deduction * 3) – (Total Deduction per Quarter)] / 3} Where 3 represent number of months in one Quarter.

Description

As seen from the above calculations, the need for subtracting total bonus from the quarterly income is to gain a net value of total business. This is normally done in order to get an overview of the financial system of the business. The financial system ought to be in a good position in terms of indicating all calculations.

The bonus scheme operated by Quality Tyres has been instrumental in terms of affecting the behavior of managers positively. A bonus scheme improves the output of the manager by motivating the manager to work hard in terms of meeting objectives. Every company has its own mechanism of motivating managers. Despite of the approach used in terms of motivating employees, the objective of a bonus scheme should be improving the output of employees. The percentage bonus scheme aids managers in terms of getting focused about reaching the required targets. The other positive aspect associated with the bonus scheme approach is the idea of making management effective and easy. As long employees are treated well, they tend to respond by being part of running the organization. This provides an ample time for the management in terms of rolling out management policies. A good bonus scheme ensures that employees are rewarded on the basis of their hard work. This implies that a situation where some employees who underperform gain more than others. In other words, the bonus scheme strategy creates a level playing field for all stakeholders in the organization.

Task 3

The best strategy of improving traditional budgeting is through incorporating the latest trends in terms of budgeting and cost controls. It is evident that businesses are skeptical about the relevance of traditional budgeting mechanism in terms of implementing their cost control objectives. This is an indication that traditional budgeting requires further additions in relation to making it better. The approach of using traditional budgeting hit a snag after it became apparent those poor and outdated budgeting tools cannot counter changes experienced in the market system. The current challenges facing the industry are a manifestation of the need to incorporate new measures in terms of budgeting and cost controls.

The first benefit of using return on capital employed as a performance measure is to understand the growth path of the business. Return on capital employed aids the business in quantifying various components of the business such as revenues and cost components. The use of return on capital employed in terms of tracking the movement of the business has become instrumental in terms of bolstering the efforts of the business in terms of understanding the dynamics of the business. Return on capital employed plays a role in relation to guiding the management of the business in terms of making good decisions. Prudent decision-making is an aspect of building a good business and should not be underestimated.

The first non-financial measure that can be taken by the company is changing the budgeting approach in the long term. In a bid to make the right projection, the company has to keep on introducing the new principles of budgeting in phases. This ensures that the right transition is made in terms of having good cost control measures. The second non-financial measure that can be used in terms of budgeting is to understand short and long-term market needs.

Midterm

Midterm

Morgan Holland

5/22/2020

Instructions

You will need alcohol. RData dataset to complete the Midterm.

The dataset includes data on 9,822 individuals. Each individual report various demographic and health characteristics, as well as a variable that equals 1 if the person reports abusing alcohol.

Question 1.

Load the alcohol.RData dataset. Convert it to a table named “alcohol.” Use print(head(alcohol)) to show me the first six rows of the dataset. Copy and paste your code and output into the answer box.

Print(head(alcohol))

Question 2

unemrate is the unemployment rate in a respondent’s State. At what level of measurement is this variable? (Nominal Scale, Ordinal Scale, Interval Scale, or Ratio Scale)?

Ratio scale since it has a defined zero point.

Question 3-6

The famsize variable measures how many people are in the respondent’s family. Fill in the missing values in the following relative frequency table:

Family Size Number of Respondents Relative Frequency Cumulative Relative Frequency

1 2595 0.264 0.264

2 2311 0.235 0.499

3 1766 0.180 0.679

4 2000 0.204 0.883

5 740 0.075 0.958

6 257 0.026 0.984

7 99 0.010 0.995

8 32 0.003 0.998

9 9 0.001 0.999

10 6 0.001 0.999

11 5 0.001 1.000

12 1 0.000 1.000

13 1 0.000 1.000

Total number of respondents = 9822

Family size 2: 2311/9822 = 0.235

Family size 4: 2000/9822 = 0.204

Cumulative Relative Frequency Family size 3: 0.499+0.180 = 0.679

Cumulative Relative Frequency Family size 3: 0.679+0.204 = 0.883

Question 7-8f

Create a barplot of the famsize variable Copy and paste your code in the answer box. In the next question, upload your barplot.

# Simple Bar Plotcounts <- table(mtfamily$size)barplot (counts, main=”Number of Respondents”,   xlab=”Family Size”

Question 9

Does family size appear to be right-skewed, left-skewed, or not skewed at all?

Family size appears to be left-skewed

Question 10-11

educ is the respondent’s level of education, in years. Calculate the sample mean and sample standard deviation of educ.

Sample mean = average (E2:E9823) = 13.30961

Sample standard deviation =std. s (E2:E9823) = 2.898751

Question 12

Suppose we assume that unemrate is normally distributed with mean 5.57% and standard deviation 1.51%. What is the probability of picking a person at random who faces an unemployment rate less than 4%?

µ = 0.0557 and σ = 0.0151

P(x<0.04) = P(x-µ<0.04-0.0557) = P (x-µσ<0.04-0.05570.0151)0.04-0.05570.0151= -1.04P(z<-1.04) = 0.1492

Question 13

What is the probability of picking a person who faces an unemployment rate greater than 7%, assuming the unemployment rate is normally distributed as in question 12?

µ = 0.0557 and σ = 0.0151

P(x>0.07) = P(x-µ>0.07-0.0557) = P (x-µσ>0.07-0.05570.0151)0.07-0.05570.0151= 0.95P(z>0.95) = 0.1711

Question 14

What is the 98th percentile of unemrate, assuming it is normally distributed as in question 8?

0.98 to z score

Percentile to z score

P(z<?) = 0.98

Z = 2.05

2.05=x-0.05570.0151X = 2.05(0.0151) + 0.0557

X = 0.0867 = 8.67%

Question 15-16

Now let’s informally test our assumption that the unemployment rate is normally distributed.

plot a histogram of unemrate, using a binwidth of 0.1. Copy and paste the code you used in the answer box. In the next question, upload your histogram as a .png or .jpeg

Question 17

Based on your histogram, do you think the unemployment rate is normally distributed? Justify your reasoning in 2-3 sentences.

The graph is not normally distributed.

The graph is nowhere close to the bell shape. It peaks at somewhere in the middle of the histogram and also further to the right.

Question 18

Suppose you are rolling a six-sided die. Let event A= {2,4,6} and B = {1,3,5}. True or False, these events are mutually exclusive.

False. The two events can happen at the same time.

Question 19

Suppose you are rolling a six-sided die. Let event A= {2,3,4} and event B= {1,2,3}. What is the intersection of these two events? That is, what is A∩B?

A∩B = {2,3}

Question 20

Suppose you are rolling a six-sided die. Let event A= {2,3,4} and event B= {1,2,3}. What is the union of these two events? That is, what is A∪B?

A∪B = {1,2,3,4}

Question 21

Suppose you flip a biased coin two times. The probability of getting heads in this coin is p = 0.2. What is the probability of getting two heads in the two flips?

Hint: There are two ways to solve this. You can use R’s built-in Binomial distribution (type help(dbinom)) for more information) or you can write out each event and calculate the probability using the probability rules.

Since the probability of getting heads in one toss/flip = 0.2

The probability of getting two heads on two-coin tosses/flips = 0.2 x 0.2 (because these are independent events) = 0.04

Thus, the answer is: 0.04

Question 22

Suppose X is Bernoulli distributed with parameter p=0.8. What is the mean of X?

Bernoulli distribution is a type of discrete probability distribution which have two possible outcomes where probability of x = 0(failure) is 1-p and probability of x = 1(success) is p.

Mean = Summation xp(x) = 0 * (1-p) + 1 * (p) = p = 0.8

Therefore, mean(x) = 0.8

Question 23

Suppose you know that the number of customers that arrive at a grocery store in an hour is a Poisson random variable with λ=200 That is, you know that on average 200 customers enter the store every hour. On average, how many customers can you expect to arrive in the next ten minutes?

Here λ = 200/hr

Or, λ = 2003 min=103Expected number of customers in the next minute = λt

= 10310=1003=33 customers