4) Dan Ariely and colleagues have conducted some extremely interesting studies on cheating (see, for example Mazar, Amir, & Ariely, 2008). To provide an opportunity for cheating, participants are given a packet of problems and told they will be paid $0.50 for each correct answer. The experiment is setup, however, so that the participant doesn’t have to turn in their work, they simply have to state how many problems they solved to get paid. This provides a prime opportunity to cheat to earn extra money. In previous tests where the problems are properly scored, so no cheating, Ariely and colleagues found that scores are normally distributed:  = 3.3,  = 1.0. Here is a list of scores for participants who knew that their work would not be checked. Which of these would you suspect of cheating? Why?

I've attached the complete question.

Only participant 1 is not cheating while the rest are cheating.

Because only participant 1 has a z-score that falls within the 95% confidence interval.

Step-by-step explanation:

We are given;

Mean; μ = 3.3

Standard deviation; s = 1

Participant 1: X = 4

Participant 2: X = 6

Participant 3: X = 7

Participant 4: X = 0

Z - score for participant 1:

z = (x - μ)/s

z = (4 - 3.3)/1

z = 0.7

Z-score for participant 2;

z = (6 - 3.3)/1

z = 2.7

Z-score for participant 3;

z = (7 - 3.3)/1

z = 3.7

Z-score for participant 4;

z = (0 - 3.3)/1

z = -3.3

Now from tables, the z-score value for confidence interval of 95% is between -1.96 and 1.96

Now, from all the participants z-score only participant 1 has a z-score that falls within the 95% confidence interval.

Thus, only participant 1 is not cheating while the rest are cheating.

75%

step-by-step explanation:

answer: -0.5 would be the most likley outcome

step-by-step explanation: