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. The number of statistical tables
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Jan 2001 2) The random variable X is normally distributed with mean 177.0 and standard deviation 6.4. (a) Find P(166 < X < 185). (4) It is suggested that X might be a suitable random variable to model the height‚ in cm‚ of adult males. (b) Give two reasons why this is a sensible suggestion. (2) (c) Explain briefly why mathematical models can help to improve our understanding of real-world problems.
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HOMEWORK 2 FROM CHAPTER 6 and 7‚ NORMAL DISTRIBUTION AND SAMPLING Instructor: Asiye Aydilek PART 1- Multiple Choice Questions ____ 1. For the standard normal probability distribution‚ the area to the left of the mean is a. –0.5 c. any value between 0 to 1 b. 0.5 d. 1 Answer: B The total area under the curve is 1. The area on the left is the half of 1 which is 0.5. ____ 2. Which of the following is not a characteristic of the normal probability distribution? a. The mean and median
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1. A formal statement that there is an absence of relationship between variables when tested by a researcher is called: (Points : 1) | Null hypothesis Type I error Type II error Negative interval | 2. Bivariate statistics refers to the statistical analysis of the relationship between two variables. (Points : 1) | True False | 3. Positive relationships between two variables indicate that‚ as the score of one increases‚ the score of
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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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grades from a college statistics class for this question. 85 72 64 65 98 78 75 76 82 80 61 92 72 58 65 74 92 85 74 76 77 77 62 68 68 54 62 76 73 85 88 91 99 82 80 74 76 77 70 60 A. Construct two different graphs of these data B. Calculate 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]
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Chapter 13: Chi-Square Applications SHORT ANSWER 1. When samples of size n are drawn from a normal population‚ the chi-square distribution is the sampling distribution of = ____________________‚ where s2 and are the sample and population variances‚ respectively. ANS: PTS: 1 OBJ: Section 13.2 2. Find the chi-square value for each of the right-tail areas below‚ given that the degrees of freedom are 7: A) 0.95 ____________________ B) 0.01 ____________________ C) 0.025 ____________________
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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‚ the minimum acceptable thickness is 90 microns. (a) What proportion of films will be too thin? (b) To what value should the mean be set so that only 1% of the films will be too thin? (c) If the mean remains at 110‚ what must the standard deviation be so that only 1% of the films will be too thin? Question 2: [20 points] If a resistor with
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The Normal and Lognormal Distributions John Norstad j-norstad@northwestern.edu http://www.norstad.org February 2‚ 1999 Updated: November 3‚ 2011 Abstract The basic properties of the normal and lognormal distributions‚ with full proofs. We assume familiarity with elementary probability theory and with college-level calculus. 1 1 DEFINITIONS AND SUMMARY OF THE PROPOSITIONS 1 Definitions and Summary of the Propositions ∞ √ Proposition 1: −∞ 2 2 1 e−(x−µ) /2σ
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.5. The probability that John watches the show‚ given that Jane does‚ is .7. a) Find the probability that both John and Jane watch the show. Answer: A : John watches a certain television show B: Jane watches the show P(A) = 0.4 P(B) = 0.5 P( A/B) = 0.7 P(A/B) = 0.7 = P(A and B) b) Find the probability that Jane watches the show‚ given that John does. Answer: P(B/A) = P(B/A)=
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