Question

Psychiatric Disorders. Use Procedure 5.1 on page 236 to solve part (g) of Exercise 5.166.

Short answer

The probabilities are:

y	0	1	2	3	4
$P\left(Y=y\right)$	0.4096	0.4096	0.1536	0.0256	0.0016

Step-by-step solution

Step 1. Given information.

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.

Step 2. Obtain the probability distribution of the random variable Y, the number of adults out of four who have a psychiatric disorder.

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,$n=4$.

Step 3. Use the binomial probability formula.

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.

Step 3. Use the binomial probability formula.

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