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by: Winn

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# Chapter 5 : Discrete Random Variables and their Probability Distributions with Exercise MGT 218

Marketplace > Marshall University > Business > MGT 218 > Chapter 5 Discrete Random Variables and their Probability Distributions with Exercise
Winn
Marshall
GPA 3.72

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_Discrete Random Variables _ Probability distributions
COURSE
Introductory Statistics
PROF.
TYPE
Class Notes
PAGES
4
WORDS
CONCEPTS
random, variable, probability distributions
KARMA
Free

## Popular in Introductory Statistics

This 4 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 18 views. For similar materials see Introductory Statistics in Business at Marshall University.

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Date Created: 02/23/16
Chapter 5 : Discrete Random Variables and their Probability Distributions Exercise ( Even number exercises ) Theory:  Random variable is a variable whose value is determined by the outcome of random experiment. Ex : the number of fish caught on a fishing trip.  Discrete Random variable : that assumes countable values Ex : the number of heads obtained in three tosses of a coin  Continuous Random variable : a random variable that can assume any value contained in one or more intervals.  Probability Distribution of a discrete random variable lists all the possible values that the random variable can assume and their corresponding probabilities 5.24) Find the mean and standard deviation for each of the following probability distributions: \ a) x P(x) x. P(x) x*x x*x*P (x) 3 .09 .27 9 .81 4 .21 .84 16 3.36 5 .34 1.7 25 8.5 6 .23 1.38 36 8.28 7 .13 .91 49 6.37 Mean = 5.1 Standard deviation = 1.14 b) x P(x) x . P(x) x*x x*x* P(x) 0 .43 0 0 0 1 .31 .31 1 .31 2 .17 .34 4 .68 3 .09 .27 9 .81 Mean = 0.92 Standard deviation = 0.98 5.26 ) Let x be the number of magazines a person reads every week. Based on a sample survey of adults, the following probability distribution table was prepared.Find the mean and the standard deviation of x x P(x) x . P(x) x*x x*x* P(x) 0 .36 0 0 0 1 .24 .24 1 .24 2 .18 .36 4 .72 3 .1 .3 9 .9 4 .07 .28 16 1.12 5 .05 .25 25 1.25 Mean = 1.43 Standard deviation = 1.48 5.28 ) The following table gives the probability distribution of the number of camcorders sold on a given day at an electronics store : Patients per x . P hour Probability (x) x * x x *x * P(x) 0 0.2725 0 0 0 0.354 1 0.3543 3 1 0.3543 0.460 2 0.2303 6 4 0.9212 0.299 3 0.0998 4 9 0.8982 0.129 4 0.0324 6 16 0.5184 5 0.0084 0.042 25 0.21 6 0.0023 0.013 36 0.0828 8 Mean = 1.3 Standard deviation = 1.1 5.30 ) Let x be the number of potential weapons detected by a metal detector at an airport on a given day. The following table lists the probability of distribution of x. x . P X P (x) (x) x * x x *x * P(x) 0 0.14 0 0 0 1 0.28 0.28 1 0.28 2 0.22 0.44 4 0.88 3 0.18 0.54 9 1.62 4 0.12 0.48 16 1.92 5 0.06 0.3 25 1.5 Mean = 2.04 Standard deviation = 1.43 5.32 ) Refer to Exercise 5.15. Find the mean and standard deviation of the probability distribution you developed for the number of remote starting systems installed per day by Al’s Auto Securtity Shop over the past 80 days. Give a brief of the values of mean and standard deviation. x . P x P(x) (x) x * x x *x * P(x) 1 0.1 0.1 1 0.1 2 0.25 0.5 4 1 3 0.3 0.9 9 2.7 4 0.2 0.8 16 3.2 5 0.15 0.75 25 3.75 Mean = 3.05 Standard deviation = 1.21 5.34) Refer to the probability distribution you developed in Exercise 5.17 for the number of uninsured motorists in a sample of two motorists. Calculate the mean and standard deviation of x for that probability distribution. x P(x) x . P (x) x * x x *x * P(x) 0 0.703921 0 0 0 1 0.135079 0.135079 1 0.135079 2 0.025921 0.051842 4 0.103684 Mean = 0.19 Standard deviation = 0.45 5.36 )** 8 An instant lottery ticket costs \$2. Out of a total of 10,000 tickets printed for this lottery, 1000 tickets contain a prize of \$5 each, 100 tickets have a prize of \$10 each, 5 tickets have a prize of \$1000 each, and 1 ticket has a prize of \$5000. Let x be random variable that denotes the net amount a player wins by playing this lottery. Write the distribution of x. Determine the mean and standard deviation of x. How will you interpret the values of the mean and standard deviation of x. x P(x) x . P (x) x * x x *x * P(x) 5-2 1000/10000 0.3 9 0.9 10-2 100/10000 0.08 64 0.64 1000-2 5/10000 0.499 996004 498.002 5000-2 1/10000 0.4998 24980004 2498.0004 Mean = 1.38 Standard deviation = 54.73

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