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# Engr Statistics IME 2610

WMU

GPA 3.69

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This 25 page Class Notes was uploaded by Kaya Schultz on Wednesday September 30, 2015. The Class Notes belongs to IME 2610 at Western Michigan University taught by Steven Butt in Fall. Since its upload, it has received 38 views. For similar materials see /class/216909/ime-2610-western-michigan-university in Industrial & Manufacturing Engineering at Western Michigan University.

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Date Created: 09/30/15

Continuous Random Variables For a continuous random variable X PXajfxdx 0 A random variable X is said to be continuous if for any value of a 1 it can take on an infinite number of possible values associated with intervals of real numbers and Does this rule out a as a possible value of X 2 if there is a function fx such that a fx 2 0 for all x WWW Recall that the cumulative distribution function for a random b ofxyx Iowaloifx x 1 variable X is de ned as Fb PX s b b 0 Pa S X S b Ifxdx IfX is a continuous random variable with pdf x then b F02 Jfxdx 1 lb ag x fx is called the probability densityfunction pdf of X 11le 2610 Chapter 5 page 1 11le 2610 Chapter 5 page 2 Expectation of a Continuous Random Variable The expectation of a continuous random variable X having pdffx is given by 00 nppzrboxmd Eco foxdx WW u 400 low mm If gx is any realvalued function of a continuous random variable X then Egon Jgxfxdx Jgltxfxdx 400 low mm For constants a and b EaX b aEX In Same with discrete random variables 11er 2610 Chapter 5 page 3 Variance of a Continuous Random Variable The variance of a continuous random variable X cg VX EX pf EX EX2 xwmxmx Iowar bound Alternatively c VXEX24 EX2 1302 4 2 For constants a and b VaX b a2VX Same with discrete random variables 11er 2610 Chapter 5 page 4 Continuous Variable Examples Example 1 Suppose a random variable X has a probability density function given by kai x 03x31 fx 0 elsewhere Find the value of k that makes this a probability density function Example 2 An accounting firm that does not have its own computing facilities rents time from a consulting company The firm must plan its computing budget carefully and hence has studied the weekly use of CPU time quite thoroughly The weekly use of CPU time approximately follows the probability density function given below measurement in hours xR4 x 03x34 0 elsewhere fx a Find the probability that CPU time used by the firm will exceed 2 hours for a selected week 11er 2610 Chapter 5 page 5 Example 2 continued bFind the cumulative distribution function Fx for weekly CPU time X c The current budget of the firm covers only 3 hours of CPU time per week How often will the budgeted figure be exceeded dHow much CPU time should be budgeted per week if this figure is to be exceeded only 10 of the time e Find the expected value and variance of weekly CPU time f The CPU time costs the firm 200 per hour Find the expected value and variance of the weekly cost for CPU time 11er 2610 Chapter 5 page 6 Normal Distribution The most important probability distribution you will ever encounter The Normal distribution is the basis for most statistical tools N0rmalpa f egomews m lt x lt 00 1 x f 6 271 Expected Value EX p Variance VX 62 3 std dev 6 The shape of the Normal probability density function is defined by its two parameters p and 6 11er 2610 Chapter 5 page 7 A very important property of the Normal distribution is that any linear function of a Normally distributed random variable is also normally distributed For example if X has a Normal distribution with EltXgt Hx and VX 63 Y aX b where a and b are constant then Y is also Normally distributed with EY aux b and VY 6125 Standard Normal Distribution Consider the random variable Z where Z N Np 0 6 1 Z is called a Standard Normal Random Variable Let39s verify the Expected Value and Variance of the Standard Normal Distribution mm 2610 Chapter 5 page 8 Standard Normal Distribution Standard Normal Transformations 4 00 1 AzzZ ExPeCted Value39 Ea 4 AJ Z 271 e dz For any random variable X where X Np 6 00 AX p 1 J39ZegzzZdz Z4 6 271 400 N N0 1 Note that his can be rew1itten as X Z6 p Which implies that EX EZc p 6EZ p 0 p p Variance VZ EZ2 EZ2 and VX VZc p xZ 02 IE 00 EZ2 L jzze ZZZdz zze zzzdz 0 0 ca cu ate r0 a 1 1t1est e tan ar 011na rans 011nat10n V2 T 11 pbb39lquot hS ddN lT f 39 is much easier to use than the pdf 0r cdf If we let u Z2 and du szz then 1 00 1271 0 L V27 1 m VZ EZ2 EZ2 1 0 1 ulZeAuZdu EZ2 rem 232r 1 mm 2610 Chapter 5 page 9 mm 2610 Chapter 5 page 10 Calculating Normal Probabilities Determine the following probabilities P0 Z 12 Assume thatXNNCu 2 039 5 P2 X s 7 HVJIE 2610 Chapter 5 page 11 Statistics amp Sampling In an earlier lecture we de ned the following Population Sample Random Sample a subset of a population chosen at random Independent random variables X1 X2 Xquot with the same distribution make up a random sample How closely does a random sample describe the population it is drawn from Numerical descriptive measures Population gt parameters mean variance Sample gt statistics sample mean sample variance A statistic is a function of a random sample HVJIE 2610 Chapter 5 page 12 Sample Mean and Variance Sample Mean and Variance For a sample of size n the sample mean is We have used mmu i 2 1 denote the opulation mean value of a random variable X xi now represents the 111 sample Observation 4 4 2 VX T EX H T 6 Through examples we can show that the sample mean tends to center around the population mean to denote the opulation variance This implies that the sample mean is a good estimate of the l t39 Let x denote the im possible population value of X pap a on mean f is a good estimate of p Then if the random variable can take on only N values with probability lN then the mean becomes 1 Earm n N il and the variance becomes wmimem2 Ni1 I HVJIE 2610 Chapter 5 page 13 HVJIE 2610 Chapter 5 page 14 Sampling Distribution of the Sample Mean The probability distribution of a statistic is called its sampling distribution Ile X2 Xquot are random variables with E u and V 62 and the random variable Y is Y 01X1 CZXZ chn then EY 01111 02112 anyquot And if X 1 X2 Xquot are independent random variables then VY 012612 022622 cnzcnz Furthermore if X 1 X2 Xquot are independent Normal random variables then Y is a Normal random variable HVJIE 2610 Chapter 5 page 15 LetXi denote the random variable corresponding to the 1m sample observation From our de nitions the sample mean X is a random variable since it is the sum of n random variables Since this is a linear function the mean of Y is then 1 quot 1 quot EX H7 ZEX ZHX HX quot11 quot11 where E 11X Now if the sampled observations were selected independently then 1 n 1 n 1 quot n62 62 VX 2erXi7ZVX 262 X X n Ki1 J n 11 n 11 n 739 HVJIE 2610 Chapter 5 page 16 Sampling Distribution of the Sample Mean Finally the standard deviation denoted 6y is 6XW Central Limit Theorem Ile X2 Xn is a random sample of size n taken from a population with mean p and variance 62 and if X is the sample mean then the limiting form of the distribution of Zii 0X A as n gt 00 is the Standard Normal Distribution Alternative de nition If a random sample of size n is drawn from a population with mean ux and variance 62X then the sample mean Y has approximately a Normal distribution with mean My 11x and 2 o varrance 62 X X n Note The approximation improves as the sample size increases HVJIE 2610 Chapter 5 page 17 Random Sampling and Finite Populations The properties of the sample mean and the sample variance hold only for random samples Random Sampling 3 each observation in the sample is selected independently from the same probability distribution For finite populations true independence of sample observations is not achieved For example if a card is drawn from a deck of 52 cards and it is not returned before the next card is drawn then the probability distribution for the second card is different from the first If the finite population is large compared to the sample size then the series of random selections are nearly independent In this last case we will still call the sample a random sample HVJIE 2610 Chapter 5 page 18 Normal Approximation of a Binomial Variable Due to the Central Limit Theorem to be discussed later it can be shown that the Normal distribution can be used to approximate the Binomial and Poisson distributions If X is a Binomial random variable with EX rip and VX np1 p then Z X np an1 p is approximately a Standard Normal random variable The Normal approximation to the Binomial is best when n is large relative to p Rule of thumb we can be confident in the approximation if the variance is at least 10 ie npl p Z 10 another common rule ifnpgt5andn1 pgt5 mm 2610 Chapter 5 page 19 Sampling Distributions Population Proportion Collect a random sample of n observations from a population and let X be the number of observation which belong to the same class X XP4 is a point estimator of the proportion of the n population It belonging to this class For 11 large P is approximately Normal with mean It and variance 7il7in Why Recall that Normal approximation works well if p is not too close to 0 or 1 and ifnp and n1 p are 2 5 or np1 p Z 10 This implies thatP N N7i7il7in and Z0 N N0 l where Z0 10 nil nl n mm 2610 Chapter 5 page 20 Probability A probability is a numeric value representing the chance likelihood or possibility that a particular event will occur Set Notation vs Probability Notation A set S sample space 5 consisting of points individual outcomes labeled 2 5 7 9 will be written as Let sets A and B be defined as then A and B are subsets events of S Note that 2 is a point in S This is equivalent to saying that 2 is an element elementary event of S mm 2610 Chapter 4 page 1 Notation The union of A and B is the set event consisting of all points that are either in A or in B or in both In set notation this is The intersection of A and B is the set event consisting of all points that are in both A and B In set notation this is The complement of B with respect to S is the set event of all points in S that are not in B Two sets events are said to be mutually exclusive or disjoint if they have no points in common eg suppose C 2 9 then 1 A and C are mutually exclusive 2 B and C are not mutually exclusive they have point 9 in common M2610 Chapter 4 page 2 Venn Diagram HVJIE 2610 Chapter 4 Sets Sample Space page 3 Probability An experiment is the process of making an observation of an operation from the sample space that has more than one outcome Random experiment an experiment that can result in different outcomes even though it is repeated in the same manner every time The sample space S of an experiment is a set consisting of all possible experimental outcomes state space Example A company sells computer chips in boxes of500 chips each chip can be classified as either satisfactory or defective The number of defective chips in a particular box is uncertain and the sample space is the set S where An event is any subset of a sample space Let Event C number of defective chips is less than 3 Probability Rules mzsio Chapter 4 page 4 Rule 1 A probability is a number between 0 and 1 that is assigned to an event or outcome of some process or experiment An event that has a probability of 0 is called a null event and has no chance of occuning Rule 2 The event that A does not occur is called A complement or simply notA and is given the symbol A or A If PA represents the probability of event A s occurring then 1 PA represents the probability that event A will not occur Rule 3 If two events A and B are mutually exclusive then the probability that both event A and event B will occur is 0 ie PA and B 0 If two events are mutually exclusive they cannot occur at the same time Rule 4 If two events A and B are mutually exclusive then the probability that either event A or event B will occur is the sum of their separate probabilities PA or B PA PB mm 2610 Chapter 4 page 5 Rule 5 If events in a set are mutually exclusive and collectively exhaustive the sum of the probabilities must add to 10 Rule 6 If two events A and B are not mutually exclusive then the probability of either event A or event B or both will occur is the sum of their separate probabilities minus the joint probability of their simultaneous occurrence PA or B PA PB 4 PA and B Rule 7 Conditional Probabilit In some situations the occurrence of one event affects the probability of occurrence of another event The conditional probability of B occuning given that A has occurred is equal to the joint probability that both A and B will occur divided by the probability of A occurring PA and B 1303 A PA mzsio Chapter 4 page 6 Rule 8 If two events A and B are independent then the probability that both event A and event B will occur is equal to the product of their respective probabilities PA and B PAPB Two events are independent if the occurrence of one event in no way affects the probability of the second event That is A and B are independent if and only if PA PA l B and PB PB l A Example Suppose we toss a fair coin twice so that S HH HT TH TT Let A the event that at least one head occurs HH HT TH B the event that at least one tail occurs HT TH TT C the event of a head occurring on the first toss HH HT D the event of a head occurring on the second toss HH TH a Assuming equally likely outcomes are events A and B independent b Assuming that outcomes are equally likely are events C and D independent mm 2610 Chapter 4 page 7 Rule 9 If two events A and B are not independent then the probability that both A and B will occur is the product of the probability of event B occuning given that event A has occurred PA and B PAPB A PBPA B Probability Example 1 A set of 15 marbles contains 4 red and 11 green marbles If two marbles are taken from this set one at a time and without replacement a What is the probability that the two marbles are red b The first is green and the second is red c What are the probabilities associated with parts a and b if the rst marble is replaced prior to the second being taken mzsio Chapter 4 page 8 Probability Example 2 A manufacturing company has two retail outlets It is known that 30 of the potential customers buy products from outlet I alone 50 buy from outlet II alone 10 buy from both I and II and 10 buy from neither Let A denote the event that a potential customer randomly chosen buys from I and B denote the event that the customer buys from II Find the following probabilities a PA b PA u B 0 NE d A PAB PA m B e PA u E A f PA B PAUB A g mm 2610 Chapter 4 page 9 Probability Example 3 A manager supervises the operation of three power plants Plant X Plant Y Plant Z At any given time each of the plants can be classi ed as either generating electricity 1 or as being idle 0 We will use the notation 0 l 0 to represent the situation where Plant Y is generating electricity but Plants X and Z are both idle The sample space for the status of the three plants at a particular point in time is 000 100 007 016 001 101 004 018 010 110 003 021 011 111 018 013 The probability values for each plant combination are also given in the figure mzsio Chapter 4 page 10 Assume l Eventl is the event that plant X is idle 2 EventB is the event that at least two out of the three plants are generating electricity a What is the probability of A b What is the probability of B c What is PA 1 B HVJIE 2610 Chapter 4 page 11 Distributions of Discrete Random Variables To define the probability distribution of a discrete random variable X a probability is assigned to each value that X can take on PXx i12 n The probability distribution of a discrete random variable is called its probability mass function and is written as xi PX xi i l 2 cain ip fair 00in Cumulative Distribution Function 0 d f Fb PX s 13 return to cain ip mzsm Chapter 4 page 12 Expected Value mean average value of a random variable in EX ire x where n is the number of possible values the random variable X can be assigned return to coinflip Variance ofX 62 V00 EX 102 i x Hzfx i xffx uz i1 i1 return to coinflip HVJIE 2610 Chapter 4 page 13 Discrete Random Variable Example An engineer in charge of the maintenance of a particular machine notices that its breakdowns can be characterized as being due to either an electrical failure within the machine a mechanical failure of some component of the machine or operator misuse When the machine is running the engineer is uncertain what will be the cause of the next breakdown and it can be thought of as an experiment with a sample space S electrical mechanical misuse Through a controlled study it was determined that the probability of each kind offailure is Pelectrical 02 Pmechanical 05 Pmisuse 03 Each of these failures may be associated with a repair cost For example suppose that electrical failures generally cost an average of 200 to repair with mechanical failures having an average repair cost of 350 and operator misuse failures having an average repair cost of only 50 1 De ne a random variable with respect to repair cost Identify all possible values this random variable can be assigned 6139 Define the probability mass function of this random variable to Define the cumulative distribution for this random variable 4 Determine the mean variance and standard deviation of this random variable mzsm Chapter 4 page 14 Discrete Random Variable Distributions Certain basic probability distributions can be developed that will serve as models for a large number of practical problems Bernoulli random variable X Takes one of two possible values 0 or 1 which can represent yes or no true or false success or failure etc Let p denote the probability that X l and l p denote the probability that X 0 Bernoulli Probability Mass Function x p p1p x 0 1 for 0 Sp 3 1 1300 P VX p1 P mzsio Chapter 4 page 15 Binomial Distribution Consider n independent Bernoulli trials Y i l n p remains constant across each trial Yi 1 corresponds to a success on the i3911 experiment Let X correspond to the total number of successes in the n experiments X is a random variable whose probability distribution is termed the Binomial distribution This probability mass function is defined as x in n quotjinnp x 0 1 n X Random variables that fit the Binomial distribution must have the following characteristics M2610 Chapter 4 page 16 1 The experiment consists of a xed number of identical trials n i Each trial can result in one of only two possible outcomes a success Yi l or a failure Yi 0 S The probability of success p is constant from trial to trial ltr39 The trials are independent un39 X is defined to be the number of successes among the n trials Notation X N Bn p Expected Value MEXnp Variance 62 VX npap mm 2610 Chapter 4 page 17 Binomial Example Milk Container Volume A machinefilled milk container is labeled as containing 2 liters However the actual amount of milk deposited into the container by the filling machine varies between 195 and 220 liters There is a probability of 0261 that a milk container is underweight Containers are shipped to retail outlets in boxes of 20 containers What is the distribution of the number of underweight containers in a box Define the random variable including parameters What is the probability that a box contains exactly 7 underweight containers What is the probability that a box contains no more than 2 underweight containers On average how many underweight containers are in a box mzsio Chapter 4 page 18 Geometric Distribution Consider a sequence of independent Bernoulli trials where Xi 1 corresponds to a success on the i3911 trial 1 Assume that the probability of success on each Bernoulli trial is the same p 2 Let Y correspond to the number of trials until the first success occurs including the successful trial EgX10 X20 X30 X4l the number of trials until the first success is 4 Y4 Given 1 and 2 the probability distribution of Y is called the Geometric distribution What is the upper bound on the maximum number of trials until the first success mm 2610 Chapter 4 page 19 Geometric Distribution The Geometric distribution is summarized as follows Pxlp1p391p x123 Expected Value EX Variance 62 VX Only one parameter is necessary to define the Geometric distribution What is it mzsro Chapter 4 page 20 Negative Binomial Distribution Consider a sequence of independent Bernoulli trials where Z 1 corresponds to a success on the i3911 trial 1 Assume that the probability of success on each Bernoulli trial is the same p 2 Let Y correspond to the number of trials that occur until the rm success including the rm successful trial eg suppose x 2 Z10 Z21 Z30 Z40Z51 Here the number of trials until the second ie xm success is 5 3Y5 Given 1 and 2 the probability distribution of Y is called the Negative Binomial distribution M2610 Chapter 4 page 21 Negative Binomial Distribution What is the upper bound on the maximum number of trials until the xLh success What is the lower bound on the minimum number of trials until the xLh success The Negative Binomial distribution is summarized as follows ygl x x Pyxp 1ip1p yxx1x2 V Eg Assuming x 2 what is PY 5 P first 4 trials contain 1 success and the 5 11 trial is a success Independent eventsU P first 4 trials contain 1 success P 5 11 trial is a success T T Binomial B4 p p 4 4 1 p11p3p 1 p21p3 M2610 Chapter 4 page 22 Negative Binomial Distribution Expected Value Variance EY f p c VY x1p N Two parameters are necessary to define the Negative Binomial distribution What are they HVJIE 2610 Chapter 4 page 23 Geometric and Negative Binomial Examples Telephone Ticket Sales A bank of telephone salespersons who start accepting calls at a speci ed time handles telephone ticket sales for a popular event In order to get through to an operator a caller has to be lucky enough to place a call at just the time When a salesperson has become free from a previous client Suppose that the chance of this is 01 a What is the distribution of the number of calls that a person needs to make until a salesperson is reached b A What is the probability that a caller gets through on the fifth attempt c A What is the average number of calls that are needed to get through to a salesperson d A What is the probability that 15 or more calls are needed to get through to a salesperson M2610 Chapter 4 page 24 Personnel Recruitment Suppose that a company wishes to hire three new workers and that each applicant interviewed has a probability of 06 of being found acceptable Furthermore if an applicant is found acceptable the person is hired on the spot a What is the distribution of the total number of applicants that the company needs to interview b A State the value of the parameters of the distribution identified in a c A What is the probability that six applicants need to be interviewed to fill the three positions d The current budget allows for up to six interviews what is the probability that the budget is suf cient to nd 3 suitable applicants What is the mean and standard deviation of the number of interviews that must be undertaken for three suitable applicants to be found e A mm 2610 Chapter 4 page 25 Hypergeometric Distribution So far the distributions we have discussed have involved independent Bernoulli trials The Hypergeometric distribution involves dependent Bernoulli trial H ypergeometric Illustration 0 We have a box with N colored beads in it 0 Among the N beads there are A red beads successes and N A white beads failures 0 Suppose n beads are sampled randomly and sequentially from the box without replacement Yi 1 on the i3911 draw if a red bead is drawn and Yi 0 if a white bead is drawn 0 0 Let X the number of successes red beads among the n beads in the sample 0 X is a called a Hypergeometric random variable mzsio Chapter 4 page 26 Hypergeometric Distribution Why is X a sum of dependent Bernoulli trials The H ypergeometric distribution is summarized as follows A N A PXlnNA forx0lA x Kn x N b with Oifagtb a Expected Value EX Variance 62 VX E l N n N N N l Three parameters are necessary to de ne the Hypergeometric distribution What are they mzsio Chapter 4 page 27 Hypergeometric Distribution If the size of the population N is much larger than the sample size n then sampling without replacement is very similar to sampling with replacement In this case the Hypergeometric distribution can be approximated with the Binomial distribution with Y Bn AN M2610 Chapter 4 page 28 H ypergeom etric Example Milk Container Contents Revisited Milk is shipped to retail outlets in boxes that hold 16 milk containers One particular box which is known by the distributor to contain six underweight containers is opened for inspection and five containers are randomly taken from the box a What is the distribution of the number of underweight milk containers in the sample chosen by the inspector b A State the value of the parameters of the distribution identified in a mm 2610 Chapter 4 page 29 Milk Container Contents Revisited continued c What is the probability that the inspector chooses a sample with 2 underweight containers d What is the expected number of underweight containers chosen by the inspector M2610 Chapter 4 page 30

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