Explanation Given info: The data represents the boxplot of all of the domestic load factors and the boxplot shows single outlier that corresponds to September of 2001. Justification: Exercise 6 represents the boxplots with the same scale for the domestic load factors from 2000 to 2015 years. From the boxplots in Exercise 6, it is seen that there is no outlier for the domestic load factors in the year 2001. And the only outlier is in the year 2015. In the present Exercise, it is said that the outlier belongs to the domestic load factors in the month of September of 2001. From Exercise 6, it is seen that the values of load factors in the year 2001 are lower when compared to the load factors of the remaining years...

Student's Solutions Manual For Intro Stats

5th Edition

ISBN: 9780134265353

Author: William B Craine

Publisher: PEARSON

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Contingency Table

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Binomial Distribution

Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.

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Chapter 4, Problem 8E

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Explain the reason for the difference in the location of an outlier in Exercise-6 and the present boxplot.

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Chapter 4, Problem 7E

Chapter 4, Problem 9E

Chapter 4 Solutions

Student's Solutions Manual For Intro Stats

Ch. 4.1 - The Bureau of Transportation Statistics of the...Ch. 4.1 - The Bureau of Transportation Statistics of the...Ch. 4.1 - Prob. 3JC Ch. 4 - Load factors 2016 The Research and Innovative...Ch. 4 - Load factors, 2016 by season We can also compare...Ch. 4 - Prob. 3E Ch. 4 - Prisons 2014 In a way, boxplots are the opposite...Ch. 4 - Load factors 2016 by month Here is a display of...Ch. 4 - Prob. 6E Ch. 4 - Prob. 7E

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Population Genetics In the study of population genetics, an important measure of inbreeding is the proportion of homozygous genotypesthat is, instances in which the two alleles carried at a particular site on an individuals chromosomes are both the same. For population in which blood-related individual mate, them is a higher than expected frequency of homozygous individuals. Examples of such populations include endangered or rare species, selectively bred breeds, and isolated populations. in general. the frequency of homozygous children from mating of blood-related parents is greater than that for children from unrelated parents Measured over a large number of generations, the proportion of heterozygous genotypesthat is, nonhomozygous genotypeschanges by a constant factor 1 from generation to generation. The factor 1 is a number between 0 and 1. If 1=0.75, for example then the proportion of heterozygous individuals in the population decreases by 25 in each generation In this case, after 10 generations, the proportion of heterozygous individuals in the population decreases by 94.37, since 0.7510=0.0563, or 5.63. In other words, 94.37 of the population is homozygous. For specific types of matings, the proportion of heterozygous genotypes can be related to that of previous generations and is found from an equation. For mating between siblings 1 can be determined as the largest value of for which 2=12+14. This equation comes from carefully accounting for the genotypes for the present generation the 2 term in terms of those previous two generations represented by for the parents generation and by the constant term of the grandparents generation. a Find both solutions to the quadratic equation above and identify which is 1 use a horizontal span of 1 to 1 in this exercise and the following exercise. b After 5 generations, what proportion of the population will be homozygous? c After 20 generations, what proportion of the population will be homozygous?

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Explain the reason for the difference in the location of an outlier in Exercise-6 and the present boxplot.

Solution Summary: The author explains the difference in the location of an outlier in Exercise-6 and the present boxplot.

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Explain the reason for the difference in the location of an outlier in Exercise-6 and the present boxplot.

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Chapter 4 Solutions