COMP 211 DATA AND SYSTEM MODELING (PROB/STAT) Spring 2012 Assignment #2 Due: Monday‚ 5pm‚ 4/16/2012 Total points: 200 (each question 20 points) Please submit a softcopy (in PDF format) of your assignment to WebCT before the deadline. Late penalty: within 24 hours after the deadline: ‐20%; after 24 hours: 0 point. Question 1: [20 points] A film-coating process produces films whose thickness are normally distributed with a mean of 110 microns and a standard deviation of 10 microns. For a certain application
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probably not attributable to chance is: (Points : 1) | Type I error Type II error Statistical significance In the semi-quartile range | 5. A score that is likely to fall into the middle 68% of scores of a normal distribution will fall inside these values: (Points : 1) | . +/- 3 standard deviations +/- 2 standard deviations +/- 1 standard deviation semi-quartile range | 6. It is important to assess the magnitude or strength of
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appraisal based on a forced distribution system: its drawbacks and remedies Rachana Chattopadhayay International Management Institute‚ Kolkata‚ India‚ and Anil Kumar Ghosh Theoretical Statistics and Mathematics Unit‚ Indian Statistical Institute‚ Kolkata‚ India Performance appraisal based on a FDS 881 Received 8 August 2011 Revised 29 January 2012 1 May 2012 Accepted 24 June 2012 Abstract Purpose – Performance appraisal based on a forced distribution system (FDS) is widely used
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(2) Jan 2001 3) A fair six-sided die is rolled. The random variable Y represents the score on the uppermost‚ face. (a) Write down the probability function of Y. (b) State the name of the distribution of Y. (2) (1) Find the value of (c) E(6Y + 2)‚ (d) Var(4Y – 2).
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the five-number summary and the mean and standard deviation of the data. C. Describe the distribution of the data‚ citing both the plots and the summary statistics found in questions 1 and 2. AP Statistics Exam Review Topic II: Normal Distribution [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] [pic] FREE RESPONSE A set of 2‚000 measurements had a symmetric‚ mound-shaped distribution. The mean is 5.3 and the standard deviation is 0.7. Determine an interval that contains
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Special Probability Distributions Chapter 8 Ibrahim Bohari bibrahim@preuni.unimas.my LOGO Binomial Distribution Binomial Distribution In an experiment of n independent trials‚ where p is a the probability of a successful outcome q=1-p is the probability that the outcome is a failure If X is a random variable denoting the number of successful outcome‚ the probability function of X is given P X r nCr p r q nr Where q=1-p r=0‚1‚2‚3‚….. X~B(n‚p) The n trials
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GEMA 5400 | Brunswick Distribution Inc | Case Analysis | | | | Table of Contents Introduction ………………………………………………………………..1 Executive Summary ……………………………………………………2 Application and Analysis ……………………………………………..3 Literature Review………………………………………………………….4 Conclusion……………………………………………………………………..5 Bibliography…………………………………………………………………..6 Appendix…………………………………………………………………………7 INTRODUCTION Brunswick Distribution started as a small distribution company 10 years ago when
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#1 True or false: Even if the sample size is more than 1000‚ we cannot always use the normal approximation to binomial. Solution: If a sample is n>30‚ we can say that sample size is sufficiently large to assume normal approximation to binomial curve. Hence the statement is false. #2 A salesperson goes door-to-door in a residential area to demonstrate the use of a new Household appliance to potential customers. She has found from her years of experience that after demonstration‚ the
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Question 1 The following table gives the classification of the amount paid and the method of payment at a department store. Cash Credit Debit Total < $20 10 8 6 24 $20 - $100 15 25 10 50 Over $100 5 15 6 26 Total 30 48 22 100 a) Find the probability that the amount paid is < $20 Answer: P(<$20) = b) Find the probability that the method of payment is credit Answer: P(Credit) = c) Find the probability that the amount is <$20 and the method of payment
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Chapter 6 Continuous Probability Distributions Case Problem: Specialty Toys 1. Information provided by the forecaster At x = 30‚000‚ [pic] [pic] Normal distribution [pic] [pic] 2. @ 15‚000 [pic] P(stockout) = 1 - .1635 = .8365 @ 18‚000 [pic] P(stockout) = 1 - .3483 = .6517 @ 24‚000 [pic] P(stockout) = 1 - .7823 = .2177 @ 28‚000 [pic]
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