91Ó°ÊÓ

The given statement from Exercise 5.166 is:

The National Institute of Mental Health reports that there is a 20% chance of an adult American suffering from a psychiatric disorder.

Four randomly selected adult Americans are examined for psychiatric disorders.

First, identify the success:

Adults in the United States suffer from psychiatric disorders.

Now determine the probability of success (p):

The chances of an adult having a psychological condition are 20%.

Therefore,$p=0.2$.

Determine the number of trials (n).

Four adult Americans are chosen at random and tested for psychiatric illnesses.

Therefore,$n=4$.

The formula is:

$P\left(Y=y\right)=\left(\begin{array}{c}n\\ y\end{array}\right)P^y{\left(1-p\right)}^{n-y}$

Insert the values in the formula,

$P\left(Y=y\right)=\left(\begin{array}{c}4\\ y\end{array}\right){\left(0.2\right)}^y{\left(1-0.2\right)}^{4-y}$

We can see that Yis a binomial random variable with parameters and the binomial distribution $n=4\mathrm{and}p=0.2$.

We must now calculate the probabilities of all Yvalues. That is, Yis a random variable with values ranging from 0 to 4.

$$\begin{gathered} P\left(Y=0\right)=\left(\begin{array}{c}4\\ 0\end{array}\right){\left(0.2\right)}^0{\left(0.8\right)}^4 \\ =1×{\left(0.8\right)}^4 \\ =0.4096 \\ P\left(Y=1\right)=\left(\begin{array}{c}4\\ 1\end{array}\right){\left(0.2\right)}^1{\left(0.8\right)}^3 \\ =4×0.2×{\left(0.8\right)}^3 \\ =0.4096 \\ P\left(Y=2\right)=\left(\begin{array}{c}4\\ 2\end{array}\right){\left(0.2\right)}^2{\left(0.8\right)}^2 \\ =6×{\left(0.2\right)}^2{\left(0.8\right)}^2 \\ =0.1536 \\ P\left(Y=3\right)=\left(\begin{array}{c}4\\ 3\end{array}\right){\left(0.2\right)}^3{\left(0.8\right)}^1 \\ =4×{\left(0.2\right)}^3{\left(0.8\right)}^1 \\ =0.0256 \\ P\left(Y=4\right)=\left(\begin{array}{c}4\\ 4\end{array}\right){\left(0.2\right)}^4{\left(0.8\right)}^0 \\ =1×{\left(0.2\right)}^4 \\ =0.0016 \end{gathered}$$