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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![P(X)={ ^nC_xp^xq^{n-x}} \\ \\ \Rightarrow P(15)={ ^{15}C_{15}(0.85)^{15}(0.15)^0} \\ \\ =1\times0.0874\times1=0.0874](https://tex.z-dn.net/?f=P%28X%29%3D%7B%20%5EnC_xp%5Exq%5E%7Bn-x%7D%7D%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20P%2815%29%3D%7B%20%5E%7B15%7DC_%7B15%7D%280.85%29%5E%7B15%7D%280.15%29%5E0%7D%20%5C%5C%20%20%5C%5C%20%3D1%5Ctimes0.0874%5Ctimes1%3D0.0874)
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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:
![P(X)={ ^nC_xp^xq^{n-x}} \\ \\ \Rightarrow P(X\geq1)=1-P(0) \\ \\ =1-{ ^{15}C_0(0.15)^0(0.85)^{15}} \\ \\ =1-1\times1\times0.0874=1-0.0874 \\ \\ =0.9126](https://tex.z-dn.net/?f=P%28X%29%3D%7B%20%5EnC_xp%5Exq%5E%7Bn-x%7D%7D%20%5C%5C%20%5C%5C%20%5CRightarrow%20P%28X%5Cgeq1%29%3D1-P%280%29%20%5C%5C%20%20%5C%5C%20%3D1-%7B%20%5E%7B15%7DC_0%280.15%29%5E0%280.85%29%5E%7B15%7D%7D%20%5C%5C%20%5C%5C%20%3D1-1%5Ctimes1%5Ctimes0.0874%3D1-0.0874%20%5C%5C%20%20%5C%5C%20%3D0.9126)
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:
![\mu=npq \\ \\ =15\times0.15\times0.85 \\ \\ =1.9125](https://tex.z-dn.net/?f=%5Cmu%3Dnpq%20%5C%5C%20%20%5C%5C%20%3D15%5Ctimes0.15%5Ctimes0.85%20%5C%5C%20%20%5C%5C%20%3D1.9125)
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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![P(1)={ ^{15}C_1(0.15)^1(0.85)^{14}} \\ \\ =15\times0.15\times0.1028=0.2312](https://tex.z-dn.net/?f=P%281%29%3D%7B%20%5E%7B15%7DC_1%280.15%29%5E1%280.85%29%5E%7B14%7D%7D%20%5C%5C%20%20%5C%5C%20%3D15%5Ctimes0.15%5Ctimes0.1028%3D0.2312)
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