(a) Explanation Calculation: From the given information the probability distribution of r is, P ( r ) = e − λ λ r r ! Where, λ = 1.72 is the average number of accidents per day, r is the number of accidents per day. For r = 0 , The probability that there are no accidents is, P ( 0 ) = e − 1.72 ( 1.72 ) 0 0 ! = 0.179 For r = 1 , The probability that there is one accident is, P ( 1 ) = e − 1.72 ( 1.72 ) 1 1 ! = 0.308 For r = 2 , The probability that there are two accidents is, P ( 1 ) = e − 1.72 ( 1.72 ) 2 2 ! = 0.530 2 = 0.265 For r = 3 , The probability that there are three accidents is, P ( 3 ) = e − 1 (b) To determine Find the expected number of accidents E = 90 P ( r ) for r = 0 , 1 , 2 , 3 , 4 or more . (c) To determine Find the sample statistic χ 2 = ∑ ( O − E ) 2 E . Find the degrees of freedom. (d) To determine Check whether the Poisson distribution fits the sample data or not.

Understandable Statistics: Concepts and Methods

12th Edition

ISBN: 9781337119917

Author: Charles Henry Brase, Corrinne Pellillo Brase

Publisher: Cengage Learning

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Chapter 10.2, Problem 17P

(a)

To determine

Find the probability that the number of accidents per day.

(b)

To determine

Find the expected number of accidents E=90P(r) for r=0,1,2,3,4 or more.

(c)

To determine

Find the sample statistic χ2=∑(O−E)2E.

Find the degrees of freedom.

(d)

To determine

Check whether the Poisson distribution fits the sample data or not.

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Chapter 10 Solutions

Understandable Statistics: Concepts and Methods

Ch. 10.1 - Statistical Literacy In general, are chi-square...Ch. 10.1 - Statistical Literacy For chi-square distributions,...Ch. 10.1 - Prob. 3P Ch. 10.1 - Prob. 4P Ch. 10.1 - Prob. 5P Ch. 10.1 - Prob. 6P Ch. 10.1 - Prob. 7P Ch. 10.1 - Interpretation: Test of Independence Consider...Ch. 10.1 - Prob. 9P Ch. 10.1 - For Problems 919, please provide the following...

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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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Find the probability that the number of accidents per day.

Solution Summary: The author calculates the probability distribution of r, where lambda is the average number of accidents per day.

Videos

Find the probability that the number of accidents per day.

Find the expected number of accidents E=90P(r) for r=0,1,2,3,4 or more.

Find the sample statistic χ2=∑(O−E)2E. Find the degrees of freedom.

Check whether the Poisson distribution fits the sample data or not.

Chapter 10 Solutions

Find the sample statistic χ2=∑(O−E)2E.

Find the degrees of freedom.