91Ó°ÊÓ

Let $Y=$the number of people who the lie detector says are telling the truth. And to find $P(Yâ‰\yen 10)$ is this related to$P(X≤2)$ .

Given: $n=12$, and $p=0.20$
The Binomial Probability:
$\mathrm{P}(\mathrm{X}=\mathrm{k})=\left(\begin{array}{l}n\\ k\end{array}\right)×p^k×(1-p{)}^{n-k}$

If a lie detector identifies $10$or more persons as speaking the truth, then the number of people deceiving is $2$or less than $2$. There are a total of $12$persons in the group.

localid="1650029337564" $$\begin{gathered} \mathrm{P}(\mathrm{Y}â‰\yen 10)=\mathrm{P}(\mathrm{X}≤2) \\ =\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2) \\ ≈0.5583 \\ ≈55.83\mathrm{\%} \end{gathered}$$

Calculate $\mathrm{μ}Y$ and $\mathrm{σ}Y$ and compare with $\mathrm{μ}X$ and $\mathrm{σ}X$.

Given: $n=12,$and $p=0.20$

Let, which is equivalent to, with the exception that the values of success and failure are swapped.

$$\begin{gathered} p_Y=1-p_X \\ =1-0.20 \\ =0.80 \end{gathered}$$

Here$n=12$

The mean is:

$$\begin{gathered} {\mathrm{μ}}_Y=n×p \\ =12(0.80) \\ =9.6 \end{gathered}$$

The standard deviation is:
$$\begin{gathered} {\mathrm{σ}}_Y=\sqrt{(n×p)(1-p)} \\ =\sqrt{12(0.80)(1-0.80)} \\ =1.3856 \end{gathered}$$.