MBA 570 Homework Questions Chapter 1 and 2 Chapter 1 (1-20) Mysti Farris (See problem 1-19) is considering raising the selling price of each cue to $50 instead of $40. If this is done while the costs remain the same‚ what would the new breakeven point be? What would the total revenue be at this breakeven point? (Given in problem 1-19: fc of 2400 and vc of 25) (1-21) Mysi Farris (see problem 1-19) believes that there is a high probability that 120 pool cues can be sold if the selling price
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http://www.nbcc.org/NCE/Sample n attempting to understand the life perspectives and characteristics of their clients‚ some counselors use Kohlberg’s theory of moral development as a theoretical framework. These counselors know that Kohlberg’s theory includes three progressive levels culminating in: principled thought‚ wherein the individual adopts a self-accepted set of standards of behavior personhood‚ wherein the individual is free from moral dilemmas self-actualization‚ wherein the individual
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McGill University Advanced Business Statistics MGSC-372 Review Normal Distribution The Normal Distribution aka The Gaussian Distribution The Normal Distribution y 1 f ( x) e 2 1 x 2 2 x Areas under the Normal Distribution curve -3 -2 - 68% 95% 99.7% + +2 +3 X = N( ‚ 2 ) Determining Normal Probabilities Since each pair of values for and represents a different distribution‚ there are an infinite number of possible normal distributions
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NORMAL DISTRIBUTION 1. Find the distribution: a. b. c. d. e. f. following probabilities‚ the random variable Z has standard normal P (0< Z < 1.43) P (0.11 < Z < 1.98) P (-0.39 < Z < 1.22) P (Z < 0.92) P (Z > -1.78) P (Z < -2.08) 2. Determine the areas under the standard normal curve between –z and +z: ♦ z = 0.5 ♦ z = 2.0 Find the two values of z in standard normal distribution so that: P(-z < Z < +z) = 0.84 3. At a university‚ the average height of 500 students of a course is 1.70 m; the standard
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these cases since they are on the area below the curve. The portfolios that are closest to the efficient frontier are more diversified since the portfolios on the efficient frontier are the most diversified. As you can see in exhibit 7 the minimum variances that you obtained by making up a portfolio of 7 securities are all below the individual standard deviations for the same level of expected return while they are
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Lab 1: Calibration of Pipettes and Burettes Names TA Name‚ Thursday Lab: 1:40 P.M. - 4:10 P.M. Aim: This experiment sought to gather data regarding water volume gathered in glassware‚ then to statistically analyze the data‚ focusing on average and standard deviation‚ along with relative error gleaned from linear regression. Introduction: This experiment was divided into two parts. For part one of the experiment‚ 25 mL were transferred from a glass pipette to a tarred beaker and weighed
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minutes. (b) Mr. Jones spends more time traveling than in the library. 6. It is estimated that of the population of England watched last year’s Cup Final on television. If random samples of 100 people are interviewed‚ (a) calculate the mean and variance of the number of people from these samples who watched the Cup Final on television. (b) calculate the probability that more than 30 people watched the Cup Final on television. 7. Four hundred pupils sit a test which consists of 80 true-false
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1. This problem is in reference to students who may or may not take advantage of the opportunities provided in QMB such as homework. Some of the students pass the course‚ and some of them do not pass. Research indicates that 40% of the students do the assigned homework. Of the students who do homework‚ there is an 80% chance they will pass the course. The probability of not passing if the student does not do the home work is 90%. What is the probability of a student not doing homework or
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tossed 100 times. Estimate the probability that the number of heads lies between 40 and 60 using central limit theorem(the word between in mathematics means inclusive of the endpoints). Solution: The expected number of heads is 100 1 = 50‚ and the variance for the number of heads is 2 11 100 2 2 = 25. Thus‚ since n = 100 is reasonably large‚ we have Sn − 50 ∼ N (0‚ 1) 5 40 − 50 60 − 50 ∗ P (40 ≤ Sn ≤ 60) = P ≤ Sn ≤ 5 5 ∗ = P (−2 ≤ Sn ≤ 2) ∗ Sn = = 2[φ(2) − 0.5] = 2 ∗ (.4772) = .9544 Problem 2 An
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Business Statistics I: QM 1 Lecture N otes by Stefan W aner (5th printing: 2003) Department of Mathematics‚ Hofstra University BUSINESS STATISTCS I: QM 001 (5th printing: 2003) LECTURE NOTES BY STEFAN WANER TABLE OF CONTENTS 0. Introduction................................................................................................... 2 1. Describing Data Graphically ...................................................................... 3 2. Measures of Central Tendency
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