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by: Winn

22

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# Chapter 6 : Continuous Random Variables and the Normal Distribution ( Part 1 ) with the Even number exercises. MGT 218

Marketplace > Marshall University > Business > MGT 218 > Chapter 6 Continuous Random Variables and the Normal Distribution Part 1 with the Even number exercises
Winn
Marshall
GPA 3.72

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Continuous Random Variables Normal Distribution value z
COURSE
Introductory Statistics
PROF.
TYPE
Class Notes
PAGES
5
WORDS
CONCEPTS
normal distribution, z scores, z value, random variables
KARMA
25 ?

## Popular in Business

This 5 page Class Notes was uploaded by Winn on Tuesday February 23, 2016. The Class Notes belongs to MGT 218 at Marshall University taught by in Spring 2016. Since its upload, it has received 22 views. For similar materials see Introductory Statistics in Business at Marshall University.

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Date Created: 02/23/16
Chapter 6 : Continuous Random Variables and the Normal Distribution ( Part 1 ) with the Even number exercises. The probability that a continuous random variable x assumes a single value is always zero. or a normal curve is a bell-shaped ( symmetric ) curve. A normal probability distribution, when plotted, gives a bell-shaped curve such that : a) The total area under the curve is 1.0 b) The curve is symmetric about the mean c) The two tails of the curve extend indefinitely. Standard normal distribution : is a special case of the normal distribution. With the normal distribution with mean = 0 and standard normal distribution = 1 z Values or z Scores : the units marked on the horizontal axis of the standard normal curve are denoted by z and are called the z values or z scores. A specific value of z gives the distance between the mean and the point represented by z in terms of the standard deviation. Even number exercises: 6.16 ) Find the area under the standard normal curve a) between z = 0 and z = 1.95 : => The area is .9744 - .5 = .4744 b) between z = .84 and z = 1.95 => the area is .9744 - .7995 = .1749 c) between z = 0 and z = -2.58 => the area is 0.5 - .0049 = .4951 d) between z = -.57 and z = -2.49 => the area is .2843 – .0064 = .2779 e) between z = - 2.15 and z = 1.87 => the area is .9693 - .0158 = .9535 6.18 ) Obtain the area under the standard normal curve : a) to the right of z = 1.43 : => the area is 1 - .9236 = .0764 b) to the right of z = -.65: => the area is 1 – 0.2578 = .7422 c) to the left of z = -1.65 : => the area is .0495 d) to the left of z = .89 : => the area is .8133 6.20 ) Obtain the area under the standard normal curve : a) from z = 0 to z = 3.94 : => the area is close to .5 b) between z = 0 and z= -5.16 : => the area is close to .5 c) to the right of z = 5.42 => the area is close to 1 d) to the left of z = -3.68: => => the area is close to 1 6.22) Determine the following probabilities for the standard normal distribution : a) P (-2.46 <= z <= 1.88 ) = .9699 - .0069 = .963 b) P ( 0 <= z <= 1.96) = .9750 - .5 = .475 c) P (-2.58 <= z <= 0) =.5 - .0049 = .4951 d) P ( z >= .73) = 1 - .7673 = 0.2327 6.24 ) Determine the following probabilities for the standard normal distribution : a) P ( z < -1.31 ) = .0951 approximately b) P ( 1.23 <= z <= 2.89 ) = .9981 – 0.8907 = .1074 c) P ( -2.24 <= z <= -1.19) = .1170 - .0125 = .1045 d) P (z < 2.02 ) = . 9783 approximately 6.26 ) Determine the following probabilities for the standard normal distribution : a) P ( z > -1.86 ) = 1- .0314 = .686 b) P ( -.68 <= z <= 1.94 ) = .9738 - .2483 = .7255 c) P ( 0 <= z <= 3.85 ) = 0.5 – 0.5 = 0 approximately d) P (-4.34 <= z <= 0) = 0.5 – 0.5 = 0 approximately. e) P ( z >4.82 ) is close to 0 f) P ( z < -6.12 ) is close to 0

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