Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Fingerprint expertise.A study published in PsychologicalScience(August 2011) tested the accuracy of experts andnovices in identifying fingerprints. Participants were presentedpairs of fingerprints and asked to judge whetherthe prints in each pair matched. The pairs were presentedunder three different conditions: prints from the same individual (match condition), non-matching but similar prints (similar distracter condition), and nonmatching and very dissimilar prints (non-similar distracter condition). The percentages of correct decisions made by the two groups under each of the three conditions are listed in the table.

Conditions

Fingerprints expert

Novices

Match similar

92.12%

74.55%

Distracter

99.32%

44.82%

Non-similar distracter

100%

77.03%

a.Given a pair of matched prints, what is the probability that an expert failed to identify the match?

b. Given a pair of matched prints, what is the probabilitythat a novice failed to identify the match?

c. Assume the study included 10 participants, 5 experts and 5 novices. Suppose that a pair of matched prints was presented to a randomly selected study participant and the participant failed to identify the match. Is the participant more likely to be an expert or a novice?

Short Answer

Expert verified

(a) The probability is 0.079.

(b) The probability is 0.2545.

(c) The participant more likely to be novice.

Step by step solution

01

Given information

The data is provided in the table,

Let

P-The professional to make the best decision

B-The beginner to make the right decision

C-The condition that is connected or matched

S-The same distracter condition

N-The non-similar distracter.

02

The probability that an expert failed to identify the match.

PPcC=1-PPC=1-0.9212=0.079

So, the probability is 0.079.

03

The probability that a novice failed to identify the match

PScC=1-PSC=1-0.7455=0.2545

Hence, the probability is 0.2545.

04

Is the participant more likely to be an expert or a novice.

From the results of part (a) and (b), its clear that the participant more likely to be a novice.

Therefore, the contestant is more likely to be inexperienced.

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

To compare the means of two populations, independent random samples of 400 observations are selected from each population, with the following results:

Sample 1

Sample 2

x¯1=5,275σ1=150

x¯2=5,240σ2=200

a. Use a 95%confidence interval to estimate the difference between the population means (μ1μ2). Interpret the confidence interval.

b. Test the null hypothesis H0:(μ1μ2)=0versus the alternative hypothesis Ha:(μ1μ2)0 . Give the significance level of the test and interpret the result.

c. Suppose the test in part b was conducted with the alternative hypothesis Ha:(μ1μ2)0 . How would your answer to part b change?

d. Test the null hypothesis H0:(μ1μ2)=25 versus Ha:(μ1μ2)25. Give the significance level and interpret the result. Compare your answer with the test conducted in part b.

e. What assumptions are necessary to ensure the validity of the inferential procedures applied in parts a–d?

4.135 Suppose xhas an exponential distribution with θ=1. Find

the following probabilities:

a.P(x>1)b.P(x3)cP(x>1.5)d.P(x5)

Solar energy generation along highways. Refer to the International Journal of Energy and Environmental Engineering (December 2013) study of solar energy generation along highways, Exercise 8.39 (p. 481). Recall that the researchers compared the mean monthly amount of solar energy generated by east-west– and north-south– oriented solar panels using a matched-pairs experiment. However, a small sample of only five months was used for the analysis. How many more months would need to be selected to estimate the difference in means to within 25 kilowatt-hours with a 90% confidence interval? Use the information provided in the SOLAR file to find an estimate of the standard error required to carry out the calculation

Find a value of the standard normal random variable z, call it z0, such that

a.P(zz0)=0.2090b.P(zz0)=0.7090c.P(-z0z<z0)=0.8472d.P(-z0z<z0)=0.1664e.P(z0zz0)=0.4798f.P(-1<z<z0)

Question: The purpose of this exercise is to compare the variability of with the variability of .

a. Suppose the first sample is selected from a population with mean and variance . Within what range should the sample mean vary about of the time in repeated samples of measurements from this distribution? That is, construct an interval extending standard deviations of on each side of .

b. Suppose the second sample is selected independently of the first from a second population with mean and variance . Within what range should the sample mean vary about the time in repeated samples of measurements from this distribution? That is, construct an interval extending standard deviations on each side .

c. Now consider the difference between the two sample means . What are the mean and standard deviation of the sampling distribution ?

d. Within what range should the difference in sample means vary about the time in repeated independent samples of measurements each from the two populations?

e. What, in general, can be said about the variability of the difference between independent sample means relative to the variability of the individual sample means?

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