Question

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 $\sigma {\mathrm{Y}}^{{\sigma }_Y}$.

Short answer

(a) The resultant percentage of chance is 0.5583

(b) According to the estimated mean, the lie detector will identify deception in an average of 12 people.

(c) The total number of people detected lying by a lie detector will vary by 1.386 people when compared to a mean of 2.4.

Step-by-step solution

Part (a) Step 1: Given Information

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

Part (a) Step 2: Simplification

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 $0.5583$

Part (b) Step 1: Given information

Given:

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

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

Part (b) step 2: Simplification

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.

Part (c) Step 1: Given information

Given:

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

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

Part(c) Step 2: Simplification

Y's standard deviation can be computed as follows:

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