84. Lie detectors Refer to Exercise 82. Let $Y=$ the number of people who the lie detector says are telling the truth.
(a) Find $P(Y\ge 10)$. How is this related to$P(X\le 2)$? Explain.
(b) Calculate $\mu Y$and$\sigma Y$. How do they compare with $\mu X$and $\sigma X$? Explain why this makes sense.

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

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

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" $$\begin{gathered} \mathrm{P}(\mathrm{Y}\ge 10)=\mathrm{P}(\mathrm{X}\le 2) \\ =\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2) \\ \approx 0.5583 \\ \approx 55.83\mathrm{\%} \end{gathered}$$

Calculate $\mu Y$ and $\sigma Y$ and compare with $\mu X$ and $\sigma X$.

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:

$$\begin{gathered} {\mu }_Y=n\times p \\ =12(0.80) \\ =9.6 \end{gathered}$$

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