Social network densities. Social networking sites for business professionals are used to promote one’s business. Each social network involves interactions (connections) between members of the network. Researchers define network density as the ratio of actual network connections to the number of possible one-to-one connections.

For example, a network with 10 members has a(102)=45 total possible connections. If that network has only 5 connections, the network density is 545=0.111. Sociologists at the University of Michigan assumed that the density x of a social network would follow a uniform distribution between 0 and 1 (Social Networks, 2010).

a. On average, what is the density of a randomly selected social network?

b. What is the probability that the randomly selected network has a density higher than .7?

c. Consider a social network with only 2 members. Explain why the uniform model would not be a good approximation for the distribution of network density.

Short Answer

Expert verified

a. On average, the density of a randomly selected social network is 0.5.

b. The probability that the randomly selected network has a density higher than .7 is .3.

c. There is only one possible connection for a social network with only 2 members. The network density for no connection is 01=0and for one connection is11=1 . In this case, it takes only two possible densities, 0 and 1; since it does not take the values between 0 and 1, the uniform model would not be a good approximation for the distribution of network density.

Step by step solution

01

Given information

The density x of a social network follows a uniform distribution between 0 and 1.

02

Define the probability density function

The probability density function of a random variable x is:

fx=1;0x1

03

Computing the mean value

a.

The mean of a random variable x is obtained as:

μ=12=0.5.

Since the mean of a uniform distribution is: μ=c+d2 .

Therefore, on average, the density of a randomly selected social network is 0.5.

04

Computing the required probability

b.

The probability that the randomly selected network has a density higher than .7 is obtained as:

Px>.7=P.7<x<1=1-.71-0=.31=.3

For a uniform distribution: P(a<x<b)=b-ad-c,ca<bd.

Hence, the probability that the randomly selected network has a density higher than .7 is .3.

05

Stating the reason why a uniform model would not be a good approximation

c.

There is only one possible connection for a social network with only 2 members.

The network density for no connection is 01=0and for one connection is11=1 .

In this case, it takes only two possible densities, 0 and 1; since it does not take the values between 0 and 1, the uniform model would not be a good approximation for the distribution of network density.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Ages of “dot-com” employees. The age (in years) distribution for the employees of a highly successful “dot-com” company headquartered in Atlanta is shown in the next table. An employee is to be randomly selected from this population.

  1. Can the relative frequency distribution in the table be interpreted as a probability distribution? Explain.
  2. Graph the probability distribution.
  3. What is the probability that the randomly selected employee is over 30 years of age? Over 40 years of age? Under 30 years of age?
  4. What is the probability that the randomly selected employee will be 25 or 26 years old?

If x is a binomial random variable, compute for each of the following cases:

  1. n = 4, x = 2, p = .2
  2. n = 3, x = 0, p = .7
  3. n = 5, x = 3, p = .1
  4. n = 3, x = 1, p = .9
  5. n = 3, x = 1, p = .3
  6. n = 4, x = 2, p = .6

4.138 The random variable xcan be adequately approximated by an exponential probability distribution withθ=2 . Find the probability that xassumes a value

a. More than 3 standard deviations fromμ

b. Less than 2 standard deviations fromμ

c. Within half a standard deviation ofμ

The binomial probability distribution is a family of probability distributions with every single distribution depending on the values of n and p. Assume that x is a binomial random variable with n = 4.

  1. Determine a value of p such that the probability distribution of x is symmetric.
  2. Determine a value of p such that the probability distribution of x is skewed to the right.
  3. Determine a value of p such that the probability distribution of x is skewed to the left.
  4. Graph each of the binomial distributions you obtained in parts a, b, and c. Locate the mean for each distribution on its graph.\
  5. In general, for what values of p will a binomial distribution be symmetric? Skewed to the right? Skewed to the left?

Hotels’ use of ecolabels. Refer to the Journal of Vacation Marketing (January 2016) study of travelers’ familiarity with ecolabels used by hotels, Exercise 2.64 (p. 104). Recall that adult travelers were shown a list of 6 different ecolabels, and asked, “Suppose the response is measured on a continuous scale from 10 (not familiar at all) to 50 (very familiar).” The mean and standard deviation for the Energy Star ecolabel are 44 and 1.5, respectively. Assume the distribution of the responses is approximately normally distributed.

a. Find the probability that a response to Energy Star exceeds 43.

b. Find the probability that a response to Energy Star falls between 42 and 45.

c. If you observe a response of 35 to an ecolabel, do you think it is likely that the ecolabel was Energy Star? Explain.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free