Marks: 1 Assume that X has a normal distribution‚ and find the indicated probability. The mean is μ = 60.0 and the standard deviation is σ = 4.0. Find the probability that X is less than 53.0. Choose one answer. a. 0.5589 b. 0.0401 c. 0.9599 d. 0.0802 Question2 Marks: 1 Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation. Assume that the population has a normal distribution. Weights of eggs: 95% confidence;
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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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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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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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April 2013. SPECIAL DISTRIBUTIONS I. Concept of probability (3%) 1. Explain why the distribution B(n‚p) can be approximated by Poisson distribution with parameter if n tends to infinity‚ p 0‚ and = np can be considered constant. 2. Show that – and + are the turning points in the graph of the p.d.f. of normal distribution with mean and standard deviation . 3. What is the relationship between exponential distribution and Poisson distribution? II. Computation
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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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Discrete and Continuous Probability All probability distributions can be categorized as discrete probability distributions or as continuous probability distributions (stattrek.com). A random variable is represented by “x” and it is the result of the discrete or continuous probability. A discrete probability is a random variable that can either be a finite or infinite of countable numbers. For example‚ the number of people who are online at the same time taking a statistics class at CTU on
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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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