83. Random digit dialing Refer to Exercise 81. Let $Y=$ the number of calls that don’t reach a live person.

(a) Find the mean of $Y$. How is it related to the mean of $X$? Explain why this makes sense.
(b) Find the standard deviation of $Y.$How is it related to the standard deviation of $X$? Explain why this makes sense

Let $Y=$the number of calls that don’t reach a live person.

Given: $n=15$, and $p=0.20$

Since, $Y$is equal to $X$, and the only difference is that the terms "success" and "failure" have been swapped.

The mean for the sample size $n$and the probability$p$ is:
$$\begin{gathered} p_Y=1-p_X \\ =1-0.20 \\ =0.80 \end{gathered}$$
Here, $n=15$
$$\begin{gathered} {\mu }_Y=n\times p \\ =15\left(0.80\right) \\ =12 \end{gathered}$$

Since, the mean of $Y$is the sample size $n$dropped by the mean of $X$.

The standard deviation of$Y$is related to the standard deviation of $X$.

Given: $n=15$, and $p=0.20$

Since, $Y$is equal to $X$, and the only distinction is that the terms "success" and "failure" have been swapped.

The standard deviation for the sample size $n$, and the probability $p$is,

$$\begin{gathered} p_Y=1-p_X \\ =1-0.20 \\ =0.80 \end{gathered}$$
Here,$n=15$
$$\begin{gathered} {\sigma }_Y=\sqrt{(n\times p)(1-p)} \\ =\sqrt{15(0.80)(1-0.80)} \\ =1.5492 \end{gathered}$$

Since, the standard deviation for $X$and $Y$are similar.