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Introduction to Probability and Statistics for Engineering and the Sciences I

by: Alison Vandervort

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Introduction to Probability and Statistics for Engineering and the Sciences I STAT 427

Marketplace > Ohio State University > Statistics > STAT 427 > Introduction to Probability and Statistics for Engineering and the Sciences I
Alison Vandervort
OSU
GPA 3.58

Staff

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This 7 page Class Notes was uploaded by Alison Vandervort on Monday September 21, 2015. The Class Notes belongs to STAT 427 at Ohio State University taught by Staff in Fall. Since its upload, it has received 85 views. For similar materials see /class/210003/stat-427-ohio-state-university in Statistics at Ohio State University.

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Date Created: 09/21/15
STAT 427 lVIidterm 11 Winter 2004 Prof Goel MWF 1030 1120 NOTES This is a closed book examination Please write your name on each sheet You can use your own calculator and a single sided 85 x 11 crib sheet There are four problems worth a total of 50 points You must explain each step in your answer to get partial credits Good Luck Problem 1 7 points The average thickness of plastic casings for magnetic disks is normally distributed with a mean of 15 mm and a standard deviation of 01 mm What is probability that the average thickness of a randomly selected plastic casing is between 135 mm and 17 mm P135Xsr7Ps ZS P 15SZ 20 ltIgt20 ltIgt 15 where I is the cdf of the standard normal distribution Using Normal Table the above expression is equal to 9772 0668 9104 Problem 2 9 points A trial has just resulted in a hung jury because eight members of the jury were in favor of a guilty verdict and the other four members were for acquittal The jurors leave the jury room in random order and each of the first three leaving the room is interviewed Let X the number of jurors favoring acquittal among those interviewed i 6 points What is the probability distribution of the random variable X In a total of N 12 jurors M 4 are in favor of acquittal N M 8 are in favor of a guilty verdict and n 3 jurors are randomly selected for interview The pmf of X is PXxwx0123 1 a 5 3 This is a Hypergeometric Distribution Ifyou have time you can calculate the value of these expressions The pmf is given by 3 n 2 ii 3 points Find the mean of the random variable X From the formula the mean of the Hypergeometric MNn 4123 1 1 of 4 pages Problem 3 16 points A manufacturer of ashlight batteries wishes to control quality of its product by rejecting any lot in which the proportion of batteries having an unacceptable voltage appears to be too high For this purpose out of each large lot 100000 batteries a certain number of batteries are selected at random and tested a 5 points If at least 5 of the 25 tested generate an unacceptable voltage the entire lot will be rejected What is the probability that a lot will be rejected if 20 of the batteries in the lot have unacceptable voltage Let X of batteries among 11 25 selected at random that generate unacceptable voltage Since the lot is very large one can assume that each draw is an independent Bernoulli trial with probability of Success 02 Hence X has a binomial distribution with n25 and p02 and ProbLot is rejected PX z 5 1 P X s 4 1 Bi4 25 2 1 0421 0579 b 5 points If at least 15 out of 1000 tested generate an unacceptable voltage the entire lot will be rejected What is the probability that a lot will not be rejected if 1 of the batteries in the lot have unacceptable voltage In this part X has a binomial distribution with n1000 and p001 However since 11 is large and p is small we can use Poisson approximation ie X has Poisson distribution with I np10000110 Using the Table for the cd we have ProbLot is not rejected PX S 14 F14 10 917 c 6 points If 20 of the batteries in the lot have unacceptable voltage and 900 batteries are tested nd the 67111 percentile for the number of batteries that have unacceptable voltage among those tested In this part X has a binomial distribution with n900 and p02 thus 11 is large but p is not small so we can approximate the Binomial by a NORMAL distribution with p up 90002 180 and variance npl p 18008 144 Thus the standard deviation 039 12 Now from the normal table the 67 11 percentile of a standard normal distribution is Z57 044 Hence the 67 h percentile of the X distribution is given by p O39 z57 180 1244 18528 2 of 4 pages Problem 4 18 points A small accounting firm does not own computing facilities It leases these facilities from Arthur Enron Inc AE1 on an hourly usage basis The firm must plan its computing budget carefully and hence has studied its weekly usage of AEI s systems Suppose the weekly usage X in hours has the following probability density function pdf x 0Sxlt10 100 1 x 20 x 10Sxlt20 f 100 0 otherwise a 5 points Find the probability that the firm s usage of AEI s computers next week will be at most 15 hours 15 10 x 1520x 2020x PXS15fxdxmdx 100 dx10 100 dx 005 0 0 10 15 1 area 0f the triangle to the right 1 125050875 15 20 b 5 points Find the probability that the firm s usage of AEI s computers next week will be between 8 and 12 hours dx 12 10 x 1220x P8sxs12fxdx dx 8 8100 w 100 Since the pdf is symmetric around 10 this area is equal to 10 x P8 s X s122 dx 036 8 100 c 8 points Suppose that AEI charges the firm 200 per hour for its system s use but the minimum amount billed in a week is 1000 irrespective of the time used Express the weekly bill amount for computer usage as a function of X Find the expected amount of weekly expenses for leasing computing facilities from AEI Note that when X is less than or equal to 5 hours the amount billed to the firm is 1000 and when the X is greater than 5 the amount billed is 200 X Thus the amount billed is a function of x given by 1000 x s 5 1100 200x xgt 5 3 of 4 pages Now Expected amount billed EhX j hxfxdx I 1000 fxdx j 200xfxdx 10 20 1000PX s 5 j 20094114 I 200x xdx 5 100 10 100 2 31 2 3 2 8000 4000 10000125 x 20x2 x 125 3 5 3 10 3 3 145833 4 of 4 pages Stat 427 Example of Unauthorized Entrance to a Secure Area Consider the situation where a bank ATM cash withdrawal security system involves keying in a valid K digit account number which if accepted is followed by electronic scanning and recognition of the customer s right thumb print In order for an unauthorized individual to fraudulently withdraw cash they would have to A guess a valid K digit ID number let PrID denote the probability of this event occurring B successfully pass the electronic thumb print scanning process let PTPdenote the probability of this event Assuming that these two events are independent then the probability of fraudulently withdrawing cash FW denoted by PFWwould be calculated as PFW PIDXPTP Calculation of PfFW assuming K is known and N attempts are made to guess the ID Under the assumption that the individual knows the ID length K and walks away after they obtain a valid ID but fail to pass the thumb print scan the PFW can be calculated exactly as follows Let N be the number of attempts allowed by an unauthorized user to randomly select a valid ID of length K Consider the following sequence of mutually exclusive events A1 unauthorized user randomly selects a valid ID of length K on the first attempt A2 unauthorized user fails to select a valid ID on the first attempt and randomly selects a valid ID on the second attempt AN unauthorized user fails to select a valid ID on the first Nl attempts and successfully selects a valid ID on the Nth attempt The probability of successfully selecting a valid ID given N possible attempts is given by N PID 2 HA i1 Let CSID denote the event of correctly selecting a valid ID at random on a single attempt Let PCSID denote the probability of correctly selecting a valid ID at random on a single attempt Assuming sampling from the population of all possible ID s with replacement ie ignoring the negligible finite correction factor that would be required if the sampling was done without replacement then 1 fraudcashwdexampledoc PCSID 1 l where M is the number of valid ID s in the bank s database Let CSID denote the complimentary event of failing to select a valid ID at random on a single attempt then PCSID 1 M 10K It follows that fori l 2 N M M P A 1 quot1 X 10K 10K Therefore N M N M i PIDZPA KZ1 Kl1 i1 10 i1 10 The unconditional probability of fraudulent withdrawal is therefore given by M 10K lHXPTP PFW10 KZ1 Notice that this equation involves two probabilities PID and PTP The probability PID involves three parameters M K and N which are known and in fact K can be set by bank management This probability equation can be used to perform sensitivity analyses to determine how sensitive PFW is to the values of M K and PTP This equation can be studied to determine the appropriate values for K to reduce the probability of fraudulent withdrawals to acceptable levels Figures 1 and 2 are examples of this type of analysis for the evaluation of the effect of the model parameters on PID fraudcashwdexampledoc Scatterplot of PID vs K N12345678 N123456789m Iii 111 1000 Customers M M Scatterplot of PID vs K 10000 Customers n 08 6 04 02 08 0 EU fraudcashwdexampledoc

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