Part A:
Given that lie <span>detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.
The case that the lie detector correctly determined that a selected person is saying the truth has a probability of 0.85
Thus p = 0.85
Thus, the probability that </span>the lie detector will conclude that all 15 are telling the truth if <span>all 15 applicants tell the truth is given by:
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
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</span>Part B:
Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.
The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.25
Thus p = 0.15
Thus, the probability that the lie detector will conclude that at least 1 is lying if all 15 applicants tell the truth is given by:

Part C:
Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.
The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.15
Thus p = 0.15
The mean is given by:

Part D:
Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.
The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.15
Thus p = 0.15
The <span>probability that the number of truthful applicants classified as liars is greater than the mean is given by:
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![P(X\ \textgreater \ \mu)=P(X\ \textgreater \ 1.9125) \\ \\ 1-[P(0)+P(1)]](https://tex.z-dn.net/?f=P%28X%5C%20%5Ctextgreater%20%5C%20%5Cmu%29%3DP%28X%5C%20%5Ctextgreater%20%5C%201.9125%29%20%5C%5C%20%20%5C%5C%201-%5BP%280%29%2BP%281%29%5D)
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
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