6th Edition

ISBN: 9780134685779

Author: Mario F. Triola

Publisher: PEARSON

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A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.

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Textbook Question

Chapter 5.2, Problem 43BB

Hypergeometric Distribution If we sample from a small finite population without replacement, the binomial distribution should not be used because the events are not independent. If sampling is done without replacement and the outcomes belong to one of two types, we can use the hypergeometric distribution. If a population has A objects of one type (such as lottery numbers you selected), while the remaining B objects are of the other type (such as lottery numbers you didn’t select), and if n objects are sampled without replacement (such as six drawn lottery numbers), then the probability of getting x objects of type A and n − x objects of type B is

P ( x ) = A ! ( A − x ) ! x ! • B ! ( B − n + x ) ! ( n − x ) ! ÷ ( A + B ) ! ( A + B − n ) ! n !

In New Jersey’s Pick 6 lottery game, a bettor selects six numbers from 1 to 49 (without repetition), and a winning six-number combination is later randomly selected. Find the probabilities of getting exactly two winning numbers with one ticket. (Hint: Use A = 6. B = 43, n = 6, and x = 2.)

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Chapter 5.2, Problem 42BB

Chapter 5.3, Problem 1BSC

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Chapter 5 Solutions

Essentials of Statistics (6th Edition)

Ch. 5.1 - Random Variable The accompanying table lists...Ch. 5.1 - Discrete or Continuous? Is the random variable...Ch. 5.1 - Probability Distribution For the accompanying...Ch. 5.1 - Significant For 100 births, P(exactly 56 girls) =...Ch. 5.1 - Identifying Discrete and Continuous Random...Ch. 5.1 - Identifying Discrete and Continuous Random...Ch. 5.1 - Identifying Probability Distributions. In...Ch. 5.1 - Identifying Probability Distributions. In...Ch. 5.1 - Identifying Probability Distributions. In...Ch. 5.1 - Identifying Probability Distributions. In...

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Solution Summary: The author explains that the probability of getting exactly two winning numbers with one ticket is 0.1324.

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