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514 The discussion requires a minimum of 300 words, 3 scholarly sources, including the textbook. Make sure that you use APA style with your references. Under no circumstances use any direct quotes. Any directly quoted or copied material will result in a zero for the assignment. Let’s be sure to write it in own work 100\% and give appropriately when using someone’s else work. Reference for textbook attached: Mirabella, J. (2011). Introduction to Statistics. Nashville, TN: Savant Learning Systems. 1 Define Probability using various sources. Present a scenario where you might collect data and use probability within your current work setting. 1,500 word count and there is a total of 3 questions each (not including in-text citation and references as the word count), a minimum of 4 scholarly sources are required in APA format. For the 4 scholarly sources, one from the textbook that’s posted below and the other two from an outside source . Let’s be sure to write it in own work 100\% and give appropriately when using someone’s else work. Under no circumstances use any direct quotes. Any directly quoted or copied material will result in a zero for the assignment. Reference for textbook attached: Mirabella, J. (2011). Introduction to Statistics. Nashville, TN: Savant Learning Systems. Use the Student_Data which consists of 200 MBA students at Whatsamattu U. It includes variables regarding their age, gender, major, GPA, Bachelors GPA, course load, English speaking status, family, and weekly hours spent studying. 1 Create a pivot table of Gender and Major. Then complete the Joint Probability table so you can answer the following: a) What is the probability of randomly choosing a Female? b) What is the probability of randomly choosing a Male AND Finance major? c) What is the probability of randomly choosing a Female OR Leadership major? d) Given that the student you selected is a Male, what is the probability he has no major? e) Given that the student you selected has no major, what is the probability the student is male? 2 Lets assume that the Student_Data.xls file was the entire population. We know the mean and standard deviation of student ages to be 42.3 and 8.9, respectively. Using the Normal_ Probability.xls file, compute the percentage of students that are older than 50, younger than 40, between 41 and 46, and oldest 10\% are at what age? Then compare to the truth as found in the actual file. mba_514_unit_2.pdf Unformatted Attachment Preview CHAPTER TWO PROBABILITY & THE NORMAL DISTRIBUTION R I What comes to mind when you hear the word probability? Most people think of gambling C and it is understandable since you can never bet on a sure thing. But what is probability A event occurring. Whether you are deciding to really? It is the chance or likelihood of an take an umbrella because you just heard R there would be a 50\% chance of rain or you are making the decision to take another cardDin blackjack based on your total and a dealer’s up card or you are choosing which route, to take home based on likely traffic from past Basics of Probability experience, probability is a part of our daily lives. We unknowingly use probability to make routine decisions, but a better understanding of the concepts could actually help in A your making the best decisions. D Probability is NOT the same as betting odds R (payoffs); Kentucky Derby and Pro Sports sets odds to insure they make money and Ithe odds depend on how the betting goes. Lower odds does not mean more likely to win, just that more people think it is true. Vegas tables E set payoffs lower than true odds so they always come out ahead. N N Mathematically, probability equals the number of events meeting the specified condition E a deck of playing cards, there are 52 possible divided by the number of possibilities. With events. Thus, the probability of drawing an ace from a deck of cards = the number of cards meeting the condition of being an ace (i.e., 1 4) divided by the number of possibilities (i.e., 52) = 4 / 52. 9 0 result of an experiment; an experiment is the An event is just an uncertain outcome, the process of making an observation. If you2flip a coin (experiment), there are two possible events à heads or tails. If you flip a coin T several times, you expect 1/2 Heads, 1/2 Tails. If you flip it 10 times you expect 5 Heads, 5S Tails, but what if you get 10 Tails in a row? Has the probability changed? Not really. The probability is still 50\% of getting Heads or Tails because what happened in the past has nothing to do with the future (when dealing with an independent event like a coin flip). Probability is always between 0 and 1 (0\% - 100\%) inclusive. It is NEVER less than 0\% and NEVER greater than 100\%. Copyright 2011, Savant Learning SystemsTM Introduction to Statistics by Jim Mirabella 2-1 Chapter Two: Probability & the Normal Distribution Solving Probability Problems Probability problems can easily be solved with tables. Setting up the table is usually the tricky part, but once it is done, the rest is easy. With a deck of cards, you would make a table with the values serving as the columns (A, 2, 3, …, J, Q, K) and the suits serving as the rows (diamonds, hearts, clubs, spades). Basically, we just employ rules of Set Theory in which you circle the objects in question and count what is in the relevant circle. Suppose 100 people were asked about their political affiliation and the following shows their responses: RTotal I 40 Male 10 25 5 C 60 Female 30 20 10 A 100 Total 40 45 15 R Now if you were asked the following, you can use Dsimple rules to solve them. , 1. Prob[Female] Dem 2. 3. 4. 5. 6. Rep Ind Prob[Republican] A D Prob[Female and Democrat] R Prob[Female or Democrat] I E of female given Democrat) Prob[Female / Democrat] (i.e., prob N Prob[Democrat / Female] (i.e., prob of Democrat given female) N E Let’s take them one at a time. 1 9 Ind0 52 10T 15S 1. Prob[Female] = total females / total people = 60/100 = 60\% Dem Rep Male Female 10 30 25 20 Total 40 45 Total 40 60 100 2. Prob[Republican] = total republicans / total people = 45/100 = 45\% Dem Rep Ind Total Male Female 10 30 25 20 5 10 40 60 Total 40 45 15 100 Copyright 2011, Savant Learning SystemsTM Introduction to Statistics by Jim Mirabella 2-2 Chapter Two: Probability & the Normal Distribution 3. Prob[Female and Dem] = intersection of females and democrats / total people = 30/100 = 30\% Dem Rep Ind Total Male Female 10 30 25 20 5 10 40 60 Total 40 45 15 100 4. Prob[Female or Dem] = union of females and dem / total people = (10+30+20+10)/100R=70/100 = 70\% Dem Rep Male Female 10 30 25 20 Total 40 45 IndI 5C 10A 15R D , Total 40 60 100 5. Prob[Female/Dem] = percentage of democrats who are female (i.e., prob of female given Dem) = # of female democrats / total democrats = 30/40 = 75\% A D Male 10 R Female 30 I Total 40 E N N who are democrat (i.e., prob of 6. Prob[Dem/Female] = percentage of females Dem given female) = # of female democrats E / total females = 30/60 = 50\% Dem Dem Rep Ind Total 1 9 Female 60 30 20 10 0 2 T S The screenshot on the next page from Joint_Probability.xls shows the results of entering the discussed political affiliation data. Copyright 2011, Savant Learning SystemsTM Introduction to Statistics by Jim Mirabella 2-3 R I C A R D , Excel will solve this for you, but it is important to understand how the answers are derived. Is political A (A given B) = Prob (A). In other words, the probparty independent of gender? To be independent, Prob ability of an event occurring is unchanged by the existence of the other event being known. The Prob D [Female] = 60 / 100 = 60\%. The Prob [ Female givenRDemocrat] = 30 / 40 = 75\%. Knowing the person’s political party alters the probability in regards to gender, so the two variables are dependent. I E An example of independent variables would be the suit and value of a playing card. The probability of Nthat the card drawn is known to be a diamond, the drawing an Ace from a deck of cards is 4 / 52. Given N 4 / 52 = 1 / 13, the two variables must be indepenprobability that the card is an Ace is now 1 / 13. Since dent. Thus, the probability of a card being a specific E value is independent of the suit (i.e., knowing the suit will not help you), AND the probability of a card being a specific suit is independent of its value (it works in both directions). 1 9 0 The Normal Distribution If you were to look at a 5’8” man, you’d probably say2he was short. If you were to look at a 5’7”” woman, you’d probably say she was tall, despite the fact thatTshe’s shorter than the man. Why? It is because we tend to compare each data point to its respective mean. S If we wanted to solve problems regarding heights, you would think we would need two separate formulas (one for men and one for women), but it is not true. Thanks to the Normal Probability Distribution, we can put everything on the same scale. The Normal curve is a bell-shaped curve which peaks in the middle at the mean. Units on the curve are measured in terms of the standard deviation; one standard deviation in both directions from the mean captures 68\% of the data, two standard deviations in both directions captures 95\% of the data, and three standard deviations in both directions captures 99.7\% of the data. As expected, the bulk of the data is close to the mean. Copyright 2011, Savant Learning SystemsTM Introduction to Statistics by Jim Mirabella 2-4 Chapter Two: Probability & the Normal Distribution The Standard Normal Distribution has values mainly between -3 and +3 as measured by a z-score, with the z-score being the number of standard deviations a value is from the mean. The formula is Thus the male height of 5’8” is about one standard deviation below average (z = -1) and the female height of 5’7” is about two standard deviations above average (z = +2). The female would show up to the right of the male on this standard scale. The Standard Normal Distribution is commonly used. You often hear about standardized testing, but this is what it means. The SAT scores, for example, are computed by the comparing one’s raw score with the mean raw score and dividing by the standard deviation (thus giving the z-score); then the z-score is R converted to the SAT scale, which forces the mean to be 500 and the standard deviation to be 100. If the I mean raw score on the math section were 40 and the standard deviation was 10, then a person who scored C and since the standard deviation is 10, that person 60 would have scored 20 points higher than the mean, A = +2.00). On the SAT system, this translates to scored 2 standard deviations above average (z = 60-40/10 R would be 700. Since it is normally distributed, we 200 points above the average of 500, so the SAT score know that about 68\% of students score between 400Dand 600, 95\% score between 300 and 700, and the remaining 5\% score below 300 or above 700. , One of the great features of using the Normal Distribution is that we can compute probabilities very easily. A We only need the mean and standard deviation to completely define the curve. If you know that you scored 400, and you know the mean = 500 and the standard deviation = 100, you can compute that your D score is 1 standard deviation below the mean. R I E N N E 1 9 0 2 T S --------------|--------------|--------------|--------------|--------------|-------------2.5\% 13.5\% 34\% 34\% 13.5\% 2.5\% |------------68\%-----------| |---------------------------95\%--------------------------| |----------------------------------------99.7\%--------------------------------------| Copyright 2011, Savant Learning SystemsTM Introduction to Statistics by Jim Mirabella 2-5 Chapter Two: Probability & the Normal Distribution If you have ever asked a teacher to grade on a curve, be careful. A true grading curve means that any score more than 1 standard deviation about the mean gets an A, less than one standard deviation about the mean gets a B, up to one standard deviation below the mean gets a C, one to two standard deviations below the mean gets a D, and more than two standard deviations below the mean gets an F. This results in approximately 16\% A’s, 34\% B’s, 34\% C’s, 13.5\% D’s, and 2.5\% F’s. Some teachers adjust the curve a bit so that the cutoff for a B is even higher so there are more C’s and fewer B’s, but the concept is the same. And this means one’s grade depends on the grades of others. While a 90\% is an A or A- in most schools, if the class mean were 95\%, then this score is below average, and depending on the standard deviation, a 90\% could be in the C or D range (although we only seem to recall the situations where our poor scores got raised). Standardizing grades mostly benefits those who R performed worst (and rarely those who did very well), and the irony is that the curve killer in a class is not the top student but rather the bottom I one. A student scoring 100\% no doubt raises the class average a bit, but a student scoring 0\% will C deviation an astronomical amount. So in a drop the average a bit more while raising the standard class of 20 students where the average is 80\% andAthe standard deviation is 10\%, the cutoff for an A would be 90\%, a B would be 80\%, a C would R be 70\% and a D would be 60\%. A grade of 100 added to the mix would likely raise the mean by about D one point to 81\% while barely impacting the standard deviation; the cutoffs for A, B, C and D ,might be 91\%, 81\%, 71\% and 61\% respectively. But if a grade of 0 were added to the mix instead, the mean would likely drop to around 76\% while the standard deviation might increase to over 20\%; you might then see a cutoff of 102\% for an A, 76\% for a B, 50\% for a C and 24\% for a D. ThusA an A would be impossible, a B would be lowered D a solid F can now pass the course, courtesy a bit, a C would be lowered a lot, and a student with R getting what they deserve. Rather unfair, but of a poor student who deprived the A students from such is the result of standardized scores for smallI samples (but SATs and other standardized tests are given to over a million students, making this E a non-issue). N N E 1 9 0 2 T S In the Normal_Probability.xls file, you can do multiple computations at once, given just a mean and a standard deviation. For example, we know the mean SAT math score is 500 and the standard deviation is 100. We can then compute the probability of scoring above a number, below a number, or between two numbers (this can also be read as the percent of people who scored in those ranges). We can also do the reverse and determine what score corresponds to a certain percentile (in the case above, to score in the top 5\% and be at the 95th percentile, you would need to score at least 664.49). Copyright 2011, Savant Learning SystemsTM Introduction to Statistics by Jim Mirabella 2-6 Chapter Two: Probability & the Normal Distribution R I C A R D The Normal_Probability.xls file also includes tabs, that show each of the computations graphically. Here we see a computation for the probability of scoring less than 650 on the SAT Math. This corresponds to a z-score of 1.50 (650 – 500 = 150, 150 divided by 100 = 1.50), as 650 is 1.50 standard A deviations above the mean. The area under the standard normal curve = 1 or 100\%, and the area shaded in red corresponds to the area less than aD z of 1.50. The red area accounts for 93.32\% of the total area, which means that 93.32\% of thoseRtaking the SAT score less than 650 on the Math section. I E N The Central Limit Theorem N When a distribution is not Normal, we cannot compute probability as is. However, if we take E samples from the data, the means of the samples will appear normal if the sample size is large enough. Regardless of the look of the original distribution, larger samples will result in the curve 1 of at least 30 guarantees that the distribution starting to appear Normal. As a general rule, a sample of sample means will be normally distributed. The 9 mean of the new distribution is the same as the original mean, but the spread is cut down dramatically and is known as the standard error (which 0 is computed from the standard deviation and the sample size). So instead of solving problems 2 involving individual values, we solve problems involving means, but essentially they are similar. T When solving problems regarding the probabilitySof a sample mean being in a specified range, you only need the mean, the standard deviation AND the sample size. So in the case of the SAT, you might want to know what the probability is of taking a sample of 4 students and getting a mean greater than 600. Copyright 2011, Savant Learning SystemsTM Introduction to Statistics by Jim Mirabella 2-7 Chapter Two: Probability & the Normal Distribution R I C A R Here we see in the Normal-CLT tab of the Normal_Probability.xls file that 2.28\% of the time, D a sample of 4 students will have a sample mean greater than 600. This doesn’t mean that 2.28\% , greater than 4, the probability would get of the students score above 600. If we used a sample smaller because as you take a larger and larger sample, there is a greater likelihood that you will be closer to the true mean. With scores ranging from A 200 to 800 on the exam and only 16\% scoring above 600, it is hard to imagine taking a D large sample and having the mean be greater than 600. R The Central Limit Theorem is quite powerful. With I a measure of central tendency and a measure of dispersion, we can completely define a distribution E and solve any probability problems; the only catch is that the data must be normally distributed, but if it isn’t, there are ways to overcome N this and make the data work for you. As we will see in later units, this theorem will form the N basis for hypothesis testing and we will take advantage of working with data, whether or not it is E normally distributed. 1 9 0 2 T S Copyright 2011, Savant Learning SystemsTM Introduction to Statistics by Jim Mirabella 2-8 CHAPTER TWO KNOWLEDGE ASSESSMENT Probability & the Normal Distribution Discussion Questions DISCUSSION QUESTION 1 SAMPLING VS. RELIABILITY: “The larger the sample, the more reliable the results.” Do you agree or disagree with this statement? Explain. DISCUSSION QUESTION 2 R SEEING THROUGH THE CLAIM: An auto manufacturer advertises that “90\% of the cars we’ve I ever made are still on the road.” Assuming this is literally true, how can it be explained? What C this misleading claim? facts / statistics would you need to know to expose Practice Problems: Real A R D Estate , Solutions are provided to practice problems so you can check your work. A Use the Real_Estate.xls file which consists of 100D homes purchased in 2007 and appraised in 2008. It includes variables regarding the number of bedrooms, number of bathrooms, whether the house R has a pool or garage, the age, size and price of the home, what the house is constructed from, and I the appraisals from two agents. E N PRACTICE PROBLEM 1: Create a pivot table of Pool and Garage. Then N complete the Joint Probability table so you can answer the following: E a) b) c) d) What is the probability of randomly choosing a home that has no pool? 1 What is the probability of randomly choosing a home that has a pool AND no garage? 9 What is the probability of randomly choosing a home that has a pool OR a garage? Given that the home you selected has a0pool, what is the probability it also has a garage? 2 PRACTICE PROBLEM 2: T Let’s assume that the Real_Estate.xls file was the S entire population. We know the mean and standard deviation of home sizes to be 2212 sq. ft. and 235 sq. ft., respectively. Using the Normal_ Probability.xls file, compute the percentage of homes that are a) smaller than 2500 sq. ft.? b) larger than 2300 sq. ft.? c) between 2000 and 2100 sq. ft.? d) The largest 20\% of homes are greater than what size? In each case, compare the computed results to the truth as found in the actual data file. Copyright 2011, Savant Learning SystemsTM Introduction to Statistics by Jim Mirabella 2-9 CHAPTER TWO KNOWLEDGE ASSESSMENT Probability & the Normal Distribution PRACTICE PROBLEM 3: Knowing the mean and standard deviation of home sizes to be 2212 sq. ft. and 235 sq. ft., respectively, if we were to visit 4 homes at random every day and compute the mean size, what is the probability that the mean would be a) smaller than 2500 sq. ft.? b) larger than 2300 sq. ft.? c) between 2000 and 2100 sq. ft.? d) 20\% of the time what would you expect the mean to be above? R I C Assigned Problems: Student Data A R Use the Student_Data.xls file which consists of 200 MBA at Whatsamattu U. It includes variables D regarding their age, gender, major, GPA, Bachelors GPA, course load, English speaking status, , family, weekly hours spent studying. A ASSIGNED PROBLEM 1: Create a pivot table of Gender and Major. ThenDcomplete the Joint Probability table so you can answer the following: R I a) What is the probability of randomly choosing a Female? E b) What is the probability of randomly choosing a Male AND Finance major? N N d) Given that the home you selected is a Male, what is the probability he has no major? E c) What is the probability of randomly choosing a Female OR Leadership major? e) Given that the home you selected has no major, what is the probability the student is male? 1 9 ASSIGNED PROBLEM 2: 0 Let’s assume that the Student_Data.xls file was the entire population. We know the mean and 2 standard deviation of student ages to be 42.3 and 8.9, respectively. Using the Normal_ Probability. T older than 50, younger than 40, between 41 and xls file, compute the percentage of students that are S to the truth as found in the actual file. 46, and oldest 10\% are at what age? Then compare Copyright 2011, Savant Learning SystemsTM Introduction to Statistics by Jim Mirabella 2-10 APPENDIX PIVOT TABLE TOOL TUTORIAL In this tutorial, you will learn to utilize the Pivot Table tool built into Excel. This tool allows you to create Rgrouping the data as you wish, and even analyzing crosstabulations and to dig into a data file very deeply, with statistical options and graphs. I C We will use the Real Estate data from the course, which consists of 100 homes. First, open the file and A select the data with your mouse (to row 101 ... Purchase answer to see full attachment