Lie detectors A federal report finds that lie detector tests given to truthful persons have probability 0.2 of suggesting that the person is deceptive. 11 A company asks 12 job applicants about thefts from previous employers, using a lie detector to assess their truthfulness. Suppose that all 12 answer truthfully. Let Y= the number of people whom the lie detector indicates are being deceptive.

a. Find the probability that the lie detector indicates that at least 10 of the people are being honest.

b. Calculate and interpret $\mu {\mathrm{Y}}^{{\mu }_Y}$.

c. Calculate and interpret $\sigma {\mathrm{Y}}^{{\sigma }_Y}$.

The likelihood of success $\left(p\right)=0.2$

The total number of trials $\left(n\right)=12$

The following formula was used:

To calculate the mean and standard deviation, use the following formula:

$Mean\left(\mu \right)=n\times p$

Standard deviation $\left(\sigma \right)=\sqrt{n\times p\times (1-p)}$

Consider the random number Y, which reflects the number of persons who are flagged as lying by a lie detector.

The following formula can be used to compute the probability:

$$\begin{gathered} P(Y\le 3)=P(Y=0)+P(Y=1)+P(Y=2) \\ =\sum _{r=0}^{12}{C}_r\times (0.2{)}^r\times (1-0.2{)}^{12-r} \\ =0.5583 \end{gathered}$$

Thus, the resultant probability is $0.5583$

Given:

Probability of success $\left(p\right)=0.2$

Number of trials $\left(n\right)=12$

Y's mean can be calculated as follows

$\begin{array}{rcl}{\mu }_Y& =& n\times p\\ & =& 12(0.2)\\ & =& 2.4\end{array}$

Hence, the required mean is 2.4.

Given:

The Probability of success $\left(p\right)=0.2$

The Number of trials $\left(n\right)=12$

Y's standard deviation can be computed as follows:

$\begin{array}{rcl}{\sigma }_Y& =& \sqrt{n\times p\times (1-p)}\\ & =& \sqrt{12(0.2)(1-0.2)}\\ & =& 1.386\end{array}$