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Introduction to Statistics I

by: Sydney Rutherford

Introduction to Statistics I STA 2122

Marketplace > Florida International University > Statistics > STA 2122 > Introduction to Statistics I
Sydney Rutherford
GPA 3.96

Dane McGuckian

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Dane McGuckian
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This 20 page Study Guide was uploaded by Sydney Rutherford on Monday October 12, 2015. The Study Guide belongs to STA 2122 at Florida International University taught by Dane McGuckian in Fall. Since its upload, it has received 56 views. For similar materials see /class/221817/sta-2122-florida-international-university in Statistics at Florida International University.


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Date Created: 10/12/15
Statistics Exam Two Notes Events Sample Spaces and Probability This section introduces the basic concept of the probability of an event Three different methods for nding probability values will be presented The most important objective of this section is to learn how to interpret probability values The Rare Event Rule for Inferential Statistics If under a given assumption the probability of a particular observed event is extremely small we conclude that the assumption is probably not correct An example of the Rare Event Rule would be as follows Say that you assume that a college graduate will have a starting salary of 75k or more but a random survey of 32 recent graduates indicates that the starting salaries were around 35k If your assumption is actually true the probability that a sample of 32 recent grads would have an average salary of only 35k would be extremely small so we must conclude your assumption was wrong in actuality we need to know what the standard deviation is before we could decide how probable the above sample results would be but we will get to that later Before we get to probability there are some terms we need to discuss An experiment is an act of observation that leads to a single outcome that cannot be predicted with certainty An event is a specific collection of sample points For example EventA Observe an even number 1 03 A sam le oint or sirn le event is the most basic outcome of an experiment For example getting a four on a single roll of a die 55 S Sample point N outcome A sample space denoted S is the collection of all possible outcomes of an experiment For example a roll ofa single die 3 1 2 3 4 5 6 J 1 List the different possible families that can occur when a couple has three hildren L I w List the possible outcomes for three ips ofa fair coin Solutlon HHH HHT HTH HTT THH THT TTH TTT Here are some commonly used notational conventions P denotes a probability A B and C denote specific events P A denotes the probability of eventA occurring The probability of an event A is calculated by summing the probabilities of the sample points in the sample space for A In other words PA n of times you observed event A number of ways A can happen N number of observations Number of total possibilities You might have noticed that the statements in the parenthesis in the above definition seem to define a second definition of probability That is because there are two ways to think about probability I have those two ways defined below Rule 1 Relative Frequency Approximation of Probability Conduct or observe a procedure and count the number of times eventA actually occurs Based on these actual results PA is estimated as follows number of times A occurred number of times trial was repeated Rule 2 Classical Approach to Probability Requires Equally Likely Outcomes Assume that a given procedure has 71 different simple events and that each of those simple events has an equal chance of occurring If event A can occur in s of these 71 ways then PA number of ways A can occur number of total possible outcomes View the video example here to see how the different versions of the formula above are used 39 r J Using the sample space created above for having three children nd the probability of having at least two girls Law of Large Numbers As a procedure is repeated over and over again the relative frequency of an event tends to approach the true probability for that event Probability Rules for Sample Points 1 All sample point probabilities must be between 0 and 1 That is 0 S PA S 1 0 indicates an impossible outcome anal 1 a certain outcome 2 The sum of the individual sample point probabilities must be equal to 1 That is 2 p 1 ifyou have listed all of the possible outcomes of an experiment 11 then you must have100 of the probability t Use the data summarized below to determine the probability that a subject did not use marijuana 397 Find the probability of a false 7 positive The box below gives an important recap of basic probability Notice that basic robability is a single fraction and deals with a single event Basic Probabilng Key Words Find the probability I randomly selected F011nulas PA oftimes A happened OR PA ofways A can happened Total of observations Total of poss1ble outcomes A bag contains 6 red marbles 3 blue marbles and 7 green marbles If a marble is randomly selected from the bag what is the probability that it is blue A 13 B 17 C 316 D 113 C 316 A bag contains 8 red marbles 4 blue marbles and 1 green marble Find Pnot blue A 913 B 9 C 139 D 413 A 913 Unusualness We will de ne arbitrarily an unusual event as one that only occurs 5 or less of the time Based on that idea answer the following question ipl f Assume that one student in a class of 27 students is randomly selected to win a prize Would it be unusual for you to win Assume unusual is a probability less than or equal to 005 A Yes B No Solution A Yes Okay let s take a little break from probability to look into a method of counting I know you know how to count but this is no ordinary countingiwe are talking Combinatorics here In many probability problems the big obstacle is nding the total number of outcomes and this section presents several methods for nding such numbers without directly listing and counting the possibilities The fundamental Counting Rule For a sequence of two events in which the rst event can occur m ways and the second event can occur 71 ways the events together can occur a total of MD ways 39 l If you can choose three different shirts to go with a pair of pants and four different pairs of shoes how many unique out ts can you create 34 12 different out ts Before we introduce the next method we need to cover some notation The factorial symbol 1 denotes the product of decreasing positive whole numbers Factorial nnxn lxn ZXgtlt1forexample 4 4X3 X2 X l 24 note 0 l by de nition The Factorial Counting Method A collection of n different items can be arranged in order n different ways This factorial rule re ects the fact that the first item may be selected from n different items the second item may be selected from n 7 1 items and so on In how many different orders can a UPS driver drop off five packages The final counting rule we will discuss is called the Combinations Rule Combinations Rule If a sample of r elements is to be drawn from a set of n elements then the number of n lv 3 If a sample of 5 elements is drawn from a set of 20 elements How many different samples are there 20 20 20gtltl9gtlt18gtltgtlt3gtlt2gtltl Solutlon n different samples possible is denoted by nCr J r 2 15 504 520 5 5X4X3X2Xl15gtltl4gtltl3gtltgtlt3gtlt2gtltl L W How many different sets of numbers can we form by choosing 6 numbers from 49 numbers M U 10C A 80640 B 40320 C 45 D 5 Solution C 45 Please note that the branch of mathematics called Combinatorics is quite vast and we have only just scratched the surface The recap box below indicates that these questions will ask for the number of ways something can be done Look for that in problems you face on the test The majority of the exam questions should say find the probability but not these questions Counting Key Words In how many ways or How many Formulas Fundamental Counting Rule of choices for 1st selection gtlt choices for 2nd selection gtlt gtlt choices for last selection Factorial Rule 71 n order does not matter Combinations C rn r n r Unions and Intersections In this section we discuss some very basic ideas from set theory The Union of two events is the event that A or B or Both occur on a single performance of an experiment We use the notation A UB Think of a union as the merging of two sets We simply create a larger set that contains all the elements of the two sets however we discard any repeats That is if both sets have the number 2 we only put one number 2 into the union 1391 Find the union A UB given that A 1 23 and B 1 4 5 AUB 12345 The Intersection of two events A and B is the event that occurs if both A and B occur on a single performance of the experiment We use the notation A MB When thinking of an intersection think of where the sets overlap or what the sets have in common Find the intersection A MB given that A 123 and B 1 4 5 Mal 6 W AUB AnB Complementary Events The Complement of an event A is the event that A does not occuriAll the sample points in the sample space not in event A Notation A A A39 A complement A Toss an even number AC Toss an odd number B Toss a number s 3 BC Toss a number 2 4 r 395 5 Ar w B 12345 AM me 5 Ne illne rAiror 5 occul Rule of Complements PAPA l P A 17P AC Using a small bit of Algebra we have PAC17PA 4 If the probability of losing money on a business adventure is 20 the probability of making money is 35 and the probability of breaking even is 45 what is the probability of not losing money An important rule of probability is the probability of at least one The rule is as follows P at least one liP none 1 What is the probability when ipping a coin twice that at least one head turns up De ne event A as A At least one head on two coin ips and event A complement as Ac No heads then PA 17PAC Now create a list of all of the equally likely outcomes for the experiment of ipping a coin twice HH HT TH and TT Since there are 4 outcomes and only TT that belongs to A complement we can say PA17PAC17 Another way to solve the problem would be to use the direct approach sinceA HHHTTH PA Please understand in these simple problems it is feasible to work the problem with either approach however that is not always the case It is often necessary to apply the probability of at least one rule given above Liz Find the probability of awaking in the morning when using three alarm clocks feach alarm clock has a 99 chance ofwaking you Recap box Probabilim ofAtLeast One Key Words Sometimes it will say Find the probability that at least one F011nula P At Least 1 l P None The Additive Rule and Mutually Exclusive Events The main objective of this section is to present the addition rule as a device for nding probabilities that can be expressed as PA or B the probability that either event A occurs or event B occurs or they both occur as the single outcome of the procedure Addition Rule of Probability PA UB PAPB P A n3 where PA MB PA and B which denotes the probability that A and B both occur at the same time as an outcome in a trial or procedure w Find the probability of randomly selecting one card from a deck of 52 cards and getting either a face card or a heart Two events are mutually exclusive if they cannot occur together at the same time If A and B are mutually exclusive eventsPAmB 0 When to events are mutually exclusive we say Events A and B are disjoint That is disjoint events do not overlap The Venn Diagrams below show the difference between sets that overlap and those that do not To l al Area l PM 775 PM and B Tofu Area 7 PM PlB Find the probability of randomly selecting one card from a deck of 52 cards ither a face card or a three L r quot Use the table below Find the probability of her having a false positive or a false negative exam result Recap Box Addition Rule Key Words Find the probability I randomly selected or F0rmula PA0 B PAPB PA NB of subjects in group A of subjecm in group B of subjecm belonging to both A amp B PA orB Total Total Total Conditional Probability Additional information or other events occurring may have an impact on the probability of an event For example if I ask you what is the chance I rolled a six on a die you would say PRolling a 6 is onesixth but if I told you an even number was rolled the probability of a 6 goes up to onethird Conditional Probability Formula To nd the probability that event A occurs given event B occurs we divide the probability that both event A and B occur by the probability that B occurs P A B P A I B ilt 133 Notation for Conditional Probability PBLA represents the probability of event B occurring after it is assumed that eventA has already occurred readBlA as B givenA j Find the probability of a subjecting having a false positive result given that the subject has a positive result 39w u 1 I Find the probability of a subject getting a positive result given that the subject does smoke marijuana Do the problem again given the person does not smoke marij uana Find the probability of a person having a positive result on a pregnancy test given that they are not pregnant Then nd the probability of a person not being pregnant given that they have a positive pregnancy test Are these probabilities the same The following table contains data from a study of two airlines which y to Small Town USA If one of the 87 ights is randomly selected nd the probability that the ight selected arrived on time given that it was an Upstate Airlines ight A 4387 B 1176 C 4348 D None of the above is correct C 4348 r y In a study 55 of sampled executives had cheated at golf The same study revealed that 20 of sampled executives had cheated at golf and lied in business What is the probability that an executive had lied in business given they had cheated in golf Independent Events Two events A and B are independent if the occurrence of one does not affect the probability of the occurrence of the other Several events are similarly independent if the occurrence of any does not affect the probabilities of occurrence of the others If A and B are not independent they are said to be dependent Using the conditional probability formula to con rm independence Two events are independent iff if and only if PABPA and PBAPB A very interesting and counter intuitive result is that two events that are mutually exclusive meaning they do not occur together are as a result dependent This bothers most people but the answer lies in the fact that mutually exclusive events have no intersection Recall Events A and B are mutually exclusive if A n B contains no sample points This leads to the idea that mutually exclusive events are dependent PBA 0 If we know A has occurred there is no chance that B occurred since there is no intersection between the two 6 Example A common error in logic made when playing games of chance like dice is to think that the dice have a memory Statements like it s been odd four rolls in a row an even is due assume that what happened during a previous roll effects what will happen on the next roll Let s look at a simple example when rolling a single die twice Find the probability of rolling an even given that an odd was rolled rst If you rolled ten odds in a row would an even be due 9 PEn0 i l PE O PO 36 32PE 6 This result tells us that the two rolls of the die are independent events actually we also need to show the same holds for P O l E Recap Box Conditional Probability Key Words Find the probability I randomly selected given tha Formula PltA B PAnB Number of Subjects inBoth GroupsTotal PB Number of Subjects in Group BTotal short cut Only look at the numbers in the row or column relating to the given tha condition the total of that row or column is your denominator then just get your numerator The Multiplicative Rule and Independent Events The conditional probability formula can be rearranged into the Multiplicative Rule of Probability to nd joint probability Multiplication Rule for Probability The probability that two events A and B occur is given y PA and B PA mB PA 0PB lA If the two events A and B are independent we may use PA and B PA mB PA0PB If the outcome of the rst event A somehow affects the probability of the second event B it is important to adjust the probability of B to re ect the occurrence of eventA only use the independent formula for events that are independent 39Are A dB Y QZEMW as PA and a pm Pm I 7 N I PAandB PAJ PBIA H Find the probability of a guessing the correct answers on three different multiple choice questions that have 5 answer choices each P and B Multiplication rule 39 If a store has 20 DVD players on a shelf that work perfectly and 3 that are defective nd the probability of randomly selecting two defective DVD players when purchasing a pair of DVD players from the store 39 If a box contains 8 identical red poker chips and 5 identical blue poker chips what is the probability that two randomly selected poker chips taken from the box without replacement are the same color If a sample size is no more than 5 of the size of the population treat the selections as being independent even if the selections are made Without replacement so they are technically dependent H In order for a container of Sony HDTV sets to be of oaded from a truck for sale at a retail store a random selection of three sets is inspected from the container If the three sets are defect free the entire container will be accepted without further inspection If an individual Sony HDTV has a 02 defective rate what is the probability that a container of Sony HDTV sets will require further inspection Recap Box Multiplication Rule Key Words Find the probability more than 1 randomly selected For example 4 randomly selected 7 7 7 7 F0rmula PA r B PAPB Independent with replacement OR PA r B PAPB A Dependent without replacement Random Sampling A random sample is selected in such a way that every different sample of size n has an equal chance of selection i After shuf ing a deck of cards well is selecting ve cards from the deck constitute a random sample of ve cards If so what is the probability that any particular sample of ve cards is chosen Discrete Random Variables This chapter will deal with the construction of discrete probability distributions by combining the methods of descriptive statistics and those of probability Probability Distributions will describe what will probably happen instead of what actually did happen A random variable is a variable that assumes numerical values associated with the random outcomes of an experiment where only one numerical value is assigned to each sample point Example If I take two penalty kicks in a soccer match over the course of a game I can make two one or no goals The sample space is as follows I Goal Goal I Goal Miss I Miss Goal I Miss Miss Let X the number of goals that I make then X can equal 2 l or 0 We can assign these values to the points in our sample space Goal Goal 2 Goal Miss 1 Miss Goal 1 Miss Miss 0 There are two kinds of random variables I A discrete random variable can assume a countable number of values 0 Example Number of steps walked visiting the Eiffel Tower I A continuous random variable can assume any value along a given interval of a number line 0 Example The time a tourist stays at the top once she gets there i Remember Discrete random variables take on a countable number of values More examples of discrete random variables 0 Number of sales 0 Number of calls 0 Shares of stock 0 People in line 0 Mistakes per page Continuous random variables can assume any value contained in one or more intervals More examples of continuous random variables 0 Length 0 Depth 0 Volume 0 Time 0 Weight Label the random variables listed below as discrete or continuous a The length of time customers spend waiting in line at Publix b The number of books purchased last year c The amount of weight gained by students during freshman year d The number of oil spills off the Alaskan coast a Continuous b Discrete c Continuous d Discrete Probability Distributions for Discrete Random Variables This section introduces the important concept of a probability distribution which gives the probability for each value of a variable that is determined by chance The probability distribution of a discrete random variable is a graph table or formula that speci es the probability associated with each possible value the random variable can assume An example of a probability histogram one of the graphs used to represent discrete probability distributions is given below 0013925456757101112 Emma mm or Number qr Mmmn murvcan Jurors Ammg 12 Think of a probability distribution as how the 100 of total probability is divided up among the possible outcomes of an experiment H Assume that having a boy or a girl is equally likely When having a child and derive the probability distribution for the random variable X the number of girls When having two children Requirements of a probability distribution 1 0 S p x 1 2 Zpxl Determine if the following is a probability distribution P x Expected Values of Discrete Random Variables The mean or Emected Value ofa discrete random variable X is y Ex Zxpx Find the mean of the given probability distribution PX 01536 00256 X 0 1 2 3 4 v How much money on average will an insurance company make off of a 1 year life insurance policy worth 10000 if they charge 290 for the policy and you have a 0999 probability of surviving the year Note even if the company has to pay out it keeps the 290 for itself so a payout is only 10000 290 9710 a 17 What is your expected value on the following game You offer your friends 10 if they can roll a six on a die but they pay a dollar amount equal to what they roll on the die if they roll any number other than 6 Should your friends play this game against you 2 A contractor is considering a sale that promises a profit of 38000 with a probability of 07 or a loss due to bad weather strikes and such 0f 16000 with a probability of 03 What is the expected profit A 26600 B 22000 C 37800 D 21800 D 21800 The population variance for a random variable x is Z Z 02 Ex Zx px 2x2 POO 2 The standard 39 39 quot for a random variable x is given by taking the square root of the above mentioned population variance 039 o2 The rules we learned in chapter 2 can be used to describe the distribution of data on the number line within one standard deviation from the mean two standard deviations and so on The table below shows the relative probabilities under the different rules Pyio39ltxlty039 PH72039 ltx lty20 PH73039 ltx lty30 Li39 I v The following table gives the probability that x patients out of five will be cured Calculate the expected value for the rv x calculate the standard deviation and use Chebyshev s rule to interpret the standard deviation IX 0 1 I2 l3 I4 Is The Binomial Distribution This section presents a basic de nition of a binomial distribution along with notation and it presents methods for nding probability values Binomial probability distributions allow us to deal with circumstances in which the outcomes belong to two relevant categories such as acceptabledefective or surviveddied Characteristics of a binomial random variable Experiment consists of n identical trials Flip a coin 3 times 2 There are only two possible outcomes for each trial success or failure omes are Heads or Tails 3 The probability of a success remains the same from trial to trial PHeads 539 PCTails l5 5 4 The trials are independent H on ip 139 doesn t change PH on ip i l 5 The random variable is the number ofsuccesses in n trials Let X number of heads in three ips Is the following a binomial experiment A marketing rm conducts a survey to determine if consumers prefer the appearance of the bottle for Absolute Vodka over the bottle of its two top rivals 1000 people will be asked to pick their favorite bottle and the number of people who select the Absolute bottle will be counted What if we instead recorded the name of the brand that was chosen by each person The rst scenario is a binomial experiment but the second is not Is the following a Binomial experiment Let x represent the number of correct guesses on 5 multiple choice questions where each question has 5 answer options Yes there are two possible outcomes correct or incorrect a xed number of trials 5 they are independent trials the probability for a correct answer is 15 for each guess and x counts the number of successes Derive the probability distribution for the above problem and again let x represent the number of correct guesses on 5 multiple choice questions where each question has 5 answer options IX l0 l1 l2 l3 l4 l5 I Px The Binomial Probability Distribution pxnp q quot forx0 12 n X p probability of a success q 1 p x number of successes n number of trials Some tips when nding binomial probability 39239 Be sure that x and p both refer to the same category being called a success 39239 When sampling without replacement consider events to be independent if n lt 005Nthe population sample size 71 Say 40 of the class is female If I randomly select ten student numbers from the roster what is the probability that exactly 6 will be female Pxquotp qquot x 10 46 61076 6 x 210004096 1296 1115 A Binomial Random Variable has Mean Variance and Standard deviation Mean u n p Variance 0392 npq Standard Deviation 039 ln p q Find the mean and standard deviation for the number of correct guesses on the above mentioned fivequestion multiple choice quiz Would it be unusual to pass a fivequestion quiz by blind guessing Sometimes the calculations can be tedious using the binomial formula so to speed things up we can use a Binomial Table 1 1 Use the binomial table in the appendix of your text to confirm our calculations for the fivequestion multiple choice example above by finding the probability that a person gets 3 or less questions correct by guessing Here is a small part ofa similar table 115 k p015 0 044371 p 020 032768 p 025 023730 the S 5 100000 100000 100000 answer 099328


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