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# Estadistica Matematica Con Aplicaciones Solucionario Freundgolkes

## Estadistica Matematica con Aplicaciones Solucionario Freundgolkes

Estadistica Matematica con Aplicaciones Solucionario Freundgolkes is a book that provides solutions and answers to the exercises and problems of the textbook Estadistica Matematica con Aplicaciones, 7th Edition, by Dennis Wackerly, Richard L. Scheaffer, and William Mendenhall. This book is a useful resource for students and instructors of mathematical statistics, as it covers topics such as probability, random variables, distributions, estimation, hypothesis testing, regression, and analysis of variance.

## estadistica matematica con aplicaciones solucionario freundgolkes

In this article, we will review some of the main features and benefits of Estadistica Matematica con Aplicaciones Solucionario Freundgolkes, and provide some examples of how it can help you learn and master mathematical statistics with applications.

## Features and Benefits of Estadistica Matematica con Aplicaciones Solucionario Freundgolkes

Estadistica Matematica con Aplicaciones Solucionario Freundgolkes has the following features and benefits:

• It contains step-by-step solutions and answers to all the exercises and problems of the textbook Estadistica Matematica con Aplicaciones, 7th Edition. You can check your work and learn from your mistakes by comparing your solutions with the ones provided in the book.

• It provides detailed explanations and reasoning for each solution and answer. You can understand the concepts and methods behind each problem and exercise, and improve your skills and confidence in solving mathematical statistics problems.

• It covers a wide range of topics and applications of mathematical statistics, such as descriptive statistics, probability theory, discrete and continuous random variables, sampling distributions, point and interval estimation, hypothesis testing, linear regression, analysis of variance, nonparametric methods, and more. You can learn how to apply mathematical statistics to real-world situations and data sets.

• It is organized by chapters and sections that correspond to the textbook Estadistica Matematica con Aplicaciones, 7th Edition. You can easily find the solutions and answers that you need for your homework assignments or exams.

• It is available online for free. You can access it anytime and anywhere with an internet connection.

## Examples of Estadistica Matematica con Aplicaciones Solucionario Freundgolkes

To give you an idea of how Estadistica Matematica con Aplicaciones Solucionario Freundgolkes can help you learn and master mathematical statistics with applications, here are some examples of solutions and answers from the book:

### Example 1: Chapter 1, Section 1.3, Exercise 1a

The exercise asks you to calculate the mean of the following data set: 2.5 3.0 2.8 3.4 2.9 3.1 3.0 2.7

The solution is:

The mean of a data set is the sum of the values divided by the number of values. In this case, we have:

mean = (2.5 + 3.0 + 2.8 + 3.4 + 2.9 + 3.1 + 3.0 + 2.7) / 8

mean = 23.4 / 8

mean = 2.925

### Example 3: Chapter 3, Section 3.5, Exercise 6a

The exercise asks you to find the probability that a randomly selected adult has a systolic blood pressure between 120 and 140 mm Hg, given that the mean and standard deviation of systolic blood pressure for adults are 133 and 24 mm Hg, respectively.

The solution is:

We can use the standard normal distribution to approximate the probability, since the sample size is large enough. First, we need to standardize the values of 120 and 140 by subtracting the mean and dividing by the standard deviation:

z1 = (120 - 133) / 24 = -0.54

z2 = (140 - 133) / 24 = 0.29

Then, we can use a table or a calculator to find the area under the standard normal curve between z1 and z2:

P(-0.54

P(-0.54

P(-0.54

Therefore, the probability that a randomly selected adult has a systolic blood pressure between 120 and 140 mm Hg is about 0.3195.

### Example 4: Chapter 4, Section 4.7, Exercise 8a

The exercise asks you to find the expected value and variance of the number of heads obtained when three fair coins are tossed.

The solution is:

We can use the binomial distribution to model the number of heads obtained when three fair coins are tossed. The parameters of the binomial distribution are n = 3 (the number of trials) and p = 0.5 (the probability of success, which is getting a head). The expected value and variance of a binomial random variable are given by:

E(X) = np

V(X) = np(1 - p)

Substituting n = 3 and p = 0.5, we get:

E(X) = (3)(0.5) = 1.5

V(X) = (3)(0.5)(1 - 0.5) = 0.75

Therefore, the expected value of the number of heads obtained when three fair coins are tossed is 1.5, and the variance is 0.75.

### Example 5: Chapter 5, Section 5.4, Exercise 4a

The exercise asks you to find the probability that a randomly selected person has a cholesterol level between 200 and 240 mg/dL, given that the mean and standard deviation of cholesterol levels for adults are 215 and 25 mg/dL, respectively.

The solution is:

We can use the normal distribution to model the cholesterol levels for adults, since the sample size is large enough and the data is approximately symmetric. First, we need to standardize the values of 200 and 240 by subtracting the mean and dividing by the standard deviation:

z1 = (200 - 215) / 25 = -0.6

z2 = (240 - 215) / 25 = 1

Then, we can use a table or a calculator to find the area under the normal curve between z1 and z2:

P(-0.6

P(-0.6

P(-0.6

Therefore, the probability that a randomly selected person has a cholesterol level between 200 and 240 mg/dL is about 0.567.

### Example 6: Chapter 6, Section 6.5, Exercise 2a

The exercise asks you to find the mean and variance of the number of defective items in a shipment of 100 items, given that each item has a probability of being defective of 0.02.

The solution is:

We can use the Poisson distribution to approximate the number of defective items in a shipment of 100 items, since the probability of being defective is small and the number of items is large. The parameter of the Poisson distribution is λ = np, where n is the number of items and p is the probability of being defective. The mean and variance of a Poisson random variable are both equal to λ. Substituting n = 100 and p = 0.02, we get:

λ = (100)(0.02) = 2

E(X) = λ = 2

V(X) = λ = 2

Therefore, the mean and variance of the number of defective items in a shipment of 100 items are both equal to 2.

### Example 7: Chapter 7, Section 7.4, Exercise 6a

The exercise asks you to find the probability that a randomly selected person has a height between 66 and 72 inches, given that the mean and standard deviation of heights for adults are 68 and 2.5 inches, respectively.

The solution is:

We can use the normal distribution to model the heights for adults, since the sample size is large enough and the data is approximately symmetric. First, we need to standardize the values of 66 and 72 by subtracting the mean and dividing by the standard deviation:

z1 = (66 - 68) / 2.5 = -0.8

z2 = (72 - 68) / 2.5 = 1.6

Then, we can use a table or a calculator to find the area under the normal curve between z1 and z2:

P(-0.8

P(-0.8

P(-0.8

Therefore, the probability that a randomly selected person has a height between 66 and 72 inches is about 0.7333.

### Example 8: Chapter 8, Section 8.5, Exercise 4a

The exercise asks you to find the probability that a randomly selected sample of size n = 25 has a mean cholesterol level between 200 and 240 mg/dL, given that the mean and standard deviation of cholesterol levels for adults are 215 and 25 mg/dL, respectively.

The solution is:

We can use the central limit theorem to approximate the probability, since the sample size is large enough. The central limit theorem states that the sampling distribution of the sample mean is approximately normal with mean μ and standard deviation σ/n, where μ and σ are the population mean and standard deviation, respectively. In this case, we have:

μ = 215

σ/n = 25/25 = 5

First, we need to standardize the values of 200 and 240 by subtracting μ and dividing by σ/n:

z1 = (200 - 215) / 5 = -3

z2 = (240 - 215) / 5 = 5

Then, we can use a table or a calculator to find the area under the normal curve between z1 and z2:

P(-3

P(-3

P(-3

Therefore, the probability that a randomly selected sample of size n = 25 has a mean cholesterol level between 200 and 240 mg/dL is about 0.9984.

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### Example 10: Chapter 10, Section 10.4, Exercise 2a

The exercise asks you to find the probability that a randomly selected sample of size n = 100 has a proportion of defective items between 0.01 and 0.03, given that the true proportion of defective items is 0.02.

The solution is:

We can use the central limit theorem to approximate the probability, since the sample size is large enough and the true proportion is not too close to 0 or 1. The central limit theorem states that the sampling distribution of the sample proportion is approximately normal with mean p and standard deviation (p(1 - p)/n), where p is the true proportion and n is the sample size. In this case, we have:

p = 0.02

(p(1 - p)/n) = ((0.02)(0.98)/100) = 0.014

First, we need to standardize the values of 0.01 and 0.03 by subtracting p and dividing by (p(1 - p)/n):

z1 = (0.01 - 0.02) / 0.014 = -0.71

z2 = (0.03 - 0.02) / 0.014 = 0.71

Then, we can use a table or a calculator to find the area under the normal curve between z1 and z2:

P(-0.71

P(-0.71

P(-0.71

Therefore, the probability that a randomly selected sample of size n = 100 has a proportion of defective items between 0.01 and 0.03 is about 0.5222.

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