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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Normal Distribution Normal distribution is a statistics‚ which have been widely applied of all mathematical concepts‚ among large number of statisticians. Abraham de Moivre‚ an 18th century statistician and consultant to gamblers‚ noticed that as the number of events (N) increased‚ the distribution approached‚ forming a very smooth curve. He insisted that a new discovery of a mathematical expression for this curve could lead to an easier way to find solutions to
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Pages C H A P T E R 6 The Normal Distribution Objectives Outline After completing this chapter‚ you should be able to 1 2 3 Identify distributions as symmetric or skewed. 4 Find probabilities for a normally distributed variable by transforming it into a standard normal variable. Introduction 6–1 Normal Distributions Identify the properties of a normal distribution. Find the area under the standard normal distribution‚ given various z values. 5 Find
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standard deviation = square root of variance = sqrt(846) = 29.086 4. If we have the following data 34‚ 38‚ 22‚ 21‚ 29‚ 37‚ 40‚ 41‚ 22‚ 20‚ 49‚ 47‚ 20‚ 31‚ 34‚ 66 Draw a stem and leaf. Discuss the shape of the distribution. Solution: 2 3 4 5 6 | | | | | 219200 48714 0197 6 This distribution is right skewed (positively skewed) because the “tail” extends to the right. 5. What type of relationship is shown by this scatter plot? 45 40 35 30 25 20 15 10 5 0 0 5 10 15 20 Solution: Weak positive linear
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decimal places) 2. Find the value of z if the area under a Standard Normal curve a) to the right of z is 0.3632; b) to the left of z is 0.1131; c) between 0 and z‚ with z > 0‚ is 0.4838; d) between -z and z‚ with z > 0‚ is 0.9500. Ans : a) z = + 0.35 ( find 0.5- 0.3632 = 0.1368 in the normal table) b) z = -1.21 ( find 0.5 – 0.1131 = 0.3869 in the normal table) c ) the area between 0 to z is 0.4838‚ z = 2.14 d) the area to the
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random will be between 19 and 31 is about 0.95. This area (probability) is shown fir the X values and for the z values. σ = 3 0.95 σ = 1 0.95 X 19 25 31 -2 0 +2 Normal curve showing Standard normal curve showing area between 19 and 31 area between -2 and +2 Entry to a certain University is determined by a national test. The scores on this test are normally distributed with a mean of 500 and a standard deviation of 100
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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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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 are
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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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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
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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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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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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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Normal Distribution:- A continuous random variable X is a normal distribution with the parameters mean and variance then the probability function can be written as f(x) = - < x < ‚ - < μ < ‚ σ > 0. When σ2 = 1‚ μ = 0 is called as standard normal. Normal distribution problems and solutions – Formulas: X < μ = 0.5 – Z X > μ = 0.5 + Z X = μ = 0.5 where‚ μ = mean σ = standard deviation X = normal random variable Normal Distribution Problems and Solutions – Example
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skewed-right distribution with a mean of 10 minutes and a standard deviation of 8 minutes. Suppose 100 flights have been randomly sampled. Describe the sampling distribution of the mean waiting time between when the airplane taxis away from the terminal until the flight takes off for these 100 flights. a) Distribution is skewed-right with mean = 10 minutes and standard error = 0.8 minutes. b) Distribution is skewed-right with mean = 10 minutes and standard error = 8 minutes. c) Distribution is approximately
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points) 3. Suppose that a random sample of size 64 is to be selected from a population having [pic] and standard deviation 5. (a) What are the mean and standard deviation of the [pic] sampling distribution? Can we say that the shape of the distribution is approximately normal? Why or why not? (10 points) (b) What is the probability that [pic] will be within 0.5 of the population mean? (5 points) (c) What is the probability that [pic] will differ from the population
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•H.P.Gautam The purpose of this article is not to explain any more the usefulness of normal distribution in decision-making process no matter whether in social sciences or in natural sciences. Nor is the purpose of making any discussions on the theory of how it can be derived. The only objective of writing this article is to acquaint the enthusiastic readers (specially students) with the simple procedure ( iterative procedure) for finding the numerical value of a normally distributed variable. The
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at Weston Materials‚ Inc.‚ a national manufacturer of unattached garages‚ reports that it takes two construction workers a mean of 32 hours and a standard deviation of 2 hours to erect the Red Barn model. Assume the assembly times follow the normal distribution. a. Determine the z values for 29 and 34 hours. What percent of the garages take between 32 hours and 34 hours to erect? z(29) = (29-32)/2 = -3/2 z(34) = (34-32)/2 = 1 z(32) = 0 P(32 < x < 34) = P(0< z < 1) = 0.34 b. What percent of
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completion of this course will provide students with a working knowledge of the principles of statistics‚ the ability to analyze and solve problems involving probability‚ and a working knowledge of averages and variations‚ normal probability distributions‚ sampling distributions‚ confidence intervals and testing statistical hypotheses. The emphasis of the course will be on the proper use of statistical techniques and their implementation rather than on mathematical proofs. (Prerequisite: MATH110 formerly
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Normal Distribution It is important because of Central Limit Theorem (CTL)‚ the CTL said that Sum up a lot of i.i.d random variables the shape of the distribution will looks like Normal. Normal P.D.F Now we want to find c This integral has been proved that it cannot have close form solution. However‚ someone gives an idea that looks stupid but actually very brilliant by multiply two of them. reminds the function of circle which we can replace them to polar coordinate Thus Mean
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