91Ó°ÊÓ

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:

$Mean\left(\mathrm{μ}\right)=n×p$

Standard deviation $\left(\mathrm{σ}\right)=\sqrt{n×p×(1-p)}$

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:

$$\begin{gathered} P(Y≤3)=P(Y=0)+P(Y=1)+P(Y=2) \\ =\underset{r=0}{\overset{12}{∑}}{C}_r×(0.2{)}^r×(1-0.2{)}^{12-r} \\ =0.5583 \end{gathered}$$

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

$\begin{array}{rcl}{\mathrm{μ}}_Y& =& n×p\\ & =& 12(0.2)\\ & =& 2.4\end{array}$

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:

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