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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