/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q30E Video game players and divided a... [FREE SOLUTION] | 91影视

91影视

Video game players and divided attention tasks. Human Factors (May 2014) published the results of a study designed to determine whether video game players are better than non鈥搗ideo game players at crossing the street when presented with distractions. Participants (college students) entered a street-crossing simulator. The simulator was designed to have cars traveling at various high rates of speed in both directions. During the crossing, the students also performed a memory task as a distraction. The researchers found that students who are video game players took an average of 5.1 seconds to cross the street, with a standard deviation of .8 second. Assume that the time, x, to cross the street for the population of video game players has , Now consider a sample of 30 students and let x represent the sample mean time (in seconds) to cross the street in the simulator.

a. Find Px>5.5

b. The 30 students in the sample are all non鈥搗ideo game players. What inference can you make about and/or for the population of non鈥搗ideo game players? Explain.

Short Answer

Expert verified

a. Probability of x greater than 5.5 is 0.0031.

b. The population mean for the non-video game players is greater than 5.1 and the population standard deviation for the non-video game players is greater than 0.8.

Step by step solution

01

Given information

A sample of 30 students is selected with mean 5.1 and the standard deviation 0.8.

02

Calculating the probability

a.

Let X be the time to cross the street in the simulator.

From the given problem =5.1,=0.8 and sample size is n=30.

According to Central limit theorem, if the sample size is large, then the sampling distribution of the sample meanx becomes approximately normal.

Then

Px>5.5=Px-/n>5.5-5.10.8/30=PZ>0.40.1461=PZ>2.74

=1-PZ2.74=1-0.5+P0<Z<2.74=1-0.5-0.4969=0.0031

Thus, Px>5.5=0.0031

03

Interpretation

b.

From part a., the probability that sample mean is greater than 5.5 is 0.0031. Moreover, the probability for the video game players is very low than the non-video game players. Thus, the population mean for the non-video game players is greater than 5.1 and the population standard deviation for the non-video game players is greater than 0.8.

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 91影视!

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

Length of job tenure. Researchers at the Terry College ofBusiness at the University of Georgia sampled 344 business students and asked them this question: 鈥淥ver the course of your lifetime, what is the maximum number of years you expect to work for any one employer?鈥 The sample resulted in x= 19.1 years. Assume that the sample of students was randomly selected from the 6,000 undergraduate students atthe Terry College and that = 6 years.

  1. Describe the sampling distribution of X.
  2. If the mean for the 6,000 undergraduate students is= 18.5 years, findPx>19.1.
  3. If the mean for the 6,000 undergraduate students is= 19.5 years, findPx>19.1.
  4. If,P(x>19.1)=0.5 what is?
  5. If,Px>19.1=0.2 isgreater than or less than 19.1years? Explain.

Soft-drink bottles. A soft-drink bottler purchases glass bottles from a vendor. The bottles are required to have an internal pressure of at least 150 pounds per square inch (psi). A prospective bottle vendor claims that its production process yields bottles with a mean internal pressure of 157 psi and a standard deviation of 3 psi. The bottler strikes an agreement with the vendor that permits the bottler to sample from the vendor鈥檚 production process to verify the vendor鈥檚 claim. The bottler randomly selects 40 bottles from the last 10,000 produced, measures the internal pressure of each, and finds the mean pressure for the sample to be 1.3 psi below the process mean cited by the vendor.

a. Assuming the vendor鈥檚 claim to be true, what is the probability of obtaining a sample mean this far or farther below the process mean? What does your answer suggest about the validity of the vendor鈥檚 claim?

b. If the process standard deviation were 3 psi as claimed by the vendor, but the mean were 156 psi, would the observed sample result be more or less likely than in part a? What if the mean were 158 psi?

c. If the process mean were 157 psi as claimed, but the process standard deviation were 2 psi, would the sample result be more or less likely than in part a? What if instead the standard deviation were 6 psi?

Switching banks after a merger. Banks that merge with others to form 鈥渕ega-banks鈥 sometimes leave customers dissatisfied with the impersonal service. A poll by the Gallup Organization found 20% of retail customers switched banks after their banks merged with another. One year after the acquisition of First Fidelity by First Union, a random sample of 250 retail customers who had banked with First Fidelity were questioned. Letp^ be the proportion of those customers who switched their business from First Union to a different bank.

  1. Find the mean and the standard deviation of role="math" localid="1658320788143" p^.
  2. Calculate the interval Ep^2p^.
  3. If samples of size 250 were drawn repeatedly a large number of times and determined for each sample, what proportion of the values would fall within the interval you calculated in part c?

Do social robots walk or roll? Refer to the International Conference on Social Robotics (Vol. 6414, 2010) study of the trend in the design of social robots, Exercise 2.5 (p. 72). The researchers obtained a random sample of 106 social robots through a Web search and determined the number that was designed with legs but no wheels. Let p^represent the sample proportion of social robots designed with legs but no wheels. Assume that in the population of all social robots, 40% are designed with legs but no wheels.

a. Give the mean and standard deviation of the sampling distribution of p^.

b. Describe the shape of the sampling distribution of p^.

c. Find P(p^>.59).

d. Recall that the researchers found that 63 of the 106 robots were built with legs only. Does this result cast doubt on the assumption that 40% of all social robots are designed with legs but no wheels? Explain.

The probability distribution shown here describes a population of measurements that can assume values of 0, 2, 4, and 6, each of which occurs with the same relative frequency:

  1. List all the different samples of n = 2 measurements that can be selected from this population. For example, (0, 6) is one possible pair of measurements; (2, 2) is another possible pair.
  2. Calculate the mean of each different sample listed in part a.
  3. If a sample of n = 2 measurements is randomly selected from the population, what is the probability that a specific sample will be selected.
  4. Assume that a random sample of n = 2 measurements is selected from the population. List the different values of x found in part b and find the probability of each. Then give the sampling distribution of the sample mean x in tabular form.
  5. Construct a probability histogram for the sampling distribution ofx.
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.