a.
Find the values
a.
Answer to Problem 16P
The values of
The values of
Explanation of Solution
Step-by-step procedure to verify
- Press STAT.
- Select Edit.
- Enter the values in L1.
- Press STAT and Choose CALC.
- Select 1-Var Stats.
- To select the variable L1, Press2-nd and then press 1.
- Press Enter.
The output obtained using the Ti83 calculator is as given below:
From the output, the values of
Step-by-step procedure to verify
- Press STAT.
- Select Edit.
- Enter the values in L2.
- Press STAT and Choose CALC.
- Select 1-Var Stats.
- To select the variable L2, Press2-nd and then press 2.
- Press Enter.
The output obtained using the Ti83 calculator is as given below:
From the output, the values of
b.
Find the sample mean, variance, and standard deviation for x and y using the computation formula.
b.
Answer to Problem 16P
The values of sample mean, variance, and standard deviation for x using the computation formula are10.3, 394.0, and 19.85, respectively.
The values of sample mean, variance, and standard deviation for y using the computation formula are9, 160.9, and 12.68, respectively.
Explanation of Solution
The sample variance using the computation formula is as follows:
Where,
The sample standard deviation using the computation formula is as follows:
Where,
The sample mean for x is obtained as given below:
Thus, the sample mean is 10.3.
The sample variance for x using the computation formula is obtained as given below:
Thus, the sample variance for x using the computation formula is 394.0.
The sample standard deviation for x using the computation formula is as obtained below:
Thus, the sample standard deviation for x using the computation formula is 19.85.
The sample mean for y is obtained as given below:
Thus, the sample mean is 9.
The sample variance for y using the computation formula is obtained as given below:
Thus, the sample variance for y using the computation formula is 160.9.
The sample standard deviation for y using the computation formula is obtained as given below:
Thus, the sample standard deviation for y using the computation formula is 12.68.
c.
Find the 75% Chebyshev interval around the mean for each fund.
Compare the two funds.
c.
Answer to Problem 16P
The 75% Chebyshev interval around the mean for the fund Vanguard Total Stock Index is–29.4 and 50.
The 75% Chebyshev interval around the mean for the fund Vanguard Balanced Index is –16.36 and 34.36.
Explanation of Solution
The 75% Chebyshev interval around the mean for the fund Vanguard Total Stock Index is obtained as given below:
Thus, the 75% Chebyshev interval around the mean for the fund Vanguard Total Stock Index is –29.4 and 50.
The 75% Chebyshev interval around the mean for the fund Vanguard Balanced Index is obtained as given below:
Thus, the 75% Chebyshev interval around the mean for the fund Vanguard Balanced Index is –16.36 and 34.36.
From the result, it is observed that the 75% returns lie in the 75% Chebyshev interval for both the funds.
d.
Find and compare the coefficient of variation for x and y.
d.
Answer to Problem 16P
The coefficient of variation for x is 192.7%.
The coefficient of variation for y is 140.9%.
Explanation of Solution
The formula for the coefficient of variation is as follows:
Where, s is the sample standard deviation and
The coefficient of variation for x is obtained as given below:
Thus, the coefficient of variation for x is 192.7%.
The coefficient of variation for y is obtained as given below:
Thus, the coefficient of variation for y is 140.9%.
From the results, the coefficient of variation for the fund Vanguard Balanced Index is less when compared to the coefficient of variation for the fund Vanguard Total Stock Index. Hence, the fund Vanguard Balanced Index has less risk per unit of return. Hence, the fund Vanguard Balanced Index appears to be better.
Want to see more full solutions like this?
Chapter 3 Solutions
UNDERSTANDABLE STAT. >PRINT UPGRADE<
- 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?arrow_forwardUrban Travel Times Population of cities and driving times are related, as shown in the accompanying table, which shows the 1960 population N, in thousands, for several cities, together with the average time T, in minutes, sent by residents driving to work. City Population N Driving time T Los Angeles 6489 16.8 Pittsburgh 1804 12.6 Washington 1808 14.3 Hutchinson 38 6.1 Nashville 347 10.8 Tallahassee 48 7.3 An analysis of these data, along with data from 17 other cities in the United States and Canada, led to a power model of average driving time as a function of population. a Construct a power model of driving time in minutes as a function of population measured in thousands b Is average driving time in Pittsburgh more or less than would be expected from its population? c If you wish to move to a smaller city to reduce your average driving time to work by 25, how much smaller should the city be?arrow_forward
- Functions and Change: A Modeling Approach to Coll...AlgebraISBN:9781337111348Author:Bruce Crauder, Benny Evans, Alan NoellPublisher:Cengage LearningLinear Algebra: A Modern IntroductionAlgebraISBN:9781285463247Author:David PoolePublisher:Cengage LearningBig Ideas Math A Bridge To Success Algebra 1: Stu...AlgebraISBN:9781680331141Author:HOUGHTON MIFFLIN HARCOURTPublisher:Houghton Mifflin Harcourt
- Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw HillAlgebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:CengageTrigonometry (MindTap Course List)TrigonometryISBN:9781337278461Author:Ron LarsonPublisher:Cengage Learning