91Ó°ÊÓ

The given statement is:

A drunk or alcohol-impaired driver or nonoccupant has a 0.40 chance of causing a traffic fatality.

$p=0.40,n=8$

The binomial probability formula is:

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

The likelihood that the number of intoxicated or alcohol-impaired drivers or nonoccupants involved is exactly three:

$$\begin{gathered} P\left(Y=3\right)=\left(\begin{array}{c}8\\ 3\end{array}\right){\left(0.40\right)}^3{\left(0.60\right)}^{8-3} \\ =\frac{8!}{3!×\left(8-3\right)!}{\left(0.40\right)}^3{\left(0.60\right)}^5 \\ =0.2787 \end{gathered}$$

The likelihood that the number of intoxicated or alcohol-impaired drivers or nonoccupants involved is at least three:

$$\begin{gathered} P\left(Yâ‰\yen 3\right)=1-P\left(Yâ‰\yen 2\right) \\ =1-\left[P\left(Y=0\right)+P\left(Y=1\right)+P\left(Y=2\right)\right] \\ P\left(Y=0\right)=\left(\begin{array}{c}8\\ 0\end{array}\right){\left(0.40\right)}^0{\left(0.60\right)}^{8-0} \\ =\frac{8!}{0!×\left(8-0\right)!}{\left(0.40\right)}^0{\left(0.60\right)}^8 \\ =0.0168 \\ P\left(Y=1\right)=\left(\begin{array}{c}8\\ 1\end{array}\right){\left(0.40\right)}^1{\left(0.60\right)}^{8-1} \\ =\frac{8!}{1!×\left(8-1\right)!}{\left(0.40\right)}^1{\left(0.60\right)}^7 \\ =0.0896 \\ P\left(Y=2\right)=\left(\begin{array}{c}8\\ 2\end{array}\right){\left(0.40\right)}^2{\left(0.60\right)}^{8-2} \\ =\frac{8!}{2!×\left(8-2\right)!}{\left(0.40\right)}^2{\left(0.60\right)}^6 \\ =0.2090 \\ P\left(Yâ‰\yen 3\right)=1-\left[P\left(Y=0\right)+P\left(Y=1\right)+P\left(Y=2\right)\right] \\ =1-\left[0.0168+0.0896+0.2090\right] \\ =1-0.3154 \\ =0.6846 \end{gathered}$$

The likelihood that the number of intoxicated or alcohol-impaired drivers or nonoccupants involved is at most three:

$$\begin{gathered} P\left(Y≤3\right)=P\left(Y=0\right)+P\left(Y=1\right)+P\left(Y=2\right)+P\left(Y=3\right) \\ P\left(Y=0\right)=0.0168 \\ P\left(Y=1\right)=0.0896 \\ P\left(Y=2\right)=0.2090 \\ P\left(Y=3\right)=0.2787 \\ P\left(Y≤3\right)=P\left(Y=0\right)+P\left(Y=1\right)+P\left(Y=2\right)+P\left(Y=3\right) \\ =0.0168+0.0896+0.2090+0.2787 \\ =0.5941 \end{gathered}$$

$$\begin{gathered} P\left(2≤y≤4\right)=P\left(Y=2\right)+P\left(Y=3\right)+P\left(Y=4\right) \\ P\left(Y=2\right)=0.2090 \\ P\left(Y=3\right)=0.2787 \\ P\left(Y=4\right)=\left(\begin{array}{c}8\\ 4\end{array}\right){\left(0.40\right)}^4{\left(0.60\right)}^{8-4} \\ =\frac{8!}{4!×\left(8-4\right)!}{\left(0.40\right)}^4{\left(0.60\right)}^4 \\ =0.2322 \\ P\left(2≤y≤4\right)=P\left(Y=2\right)+P\left(Y=3\right)+P\left(Y=4\right) \\ =0.2090+0.2787+0.2322 \\ =0.7199 \end{gathered}$$

The mean formula is:

$$\begin{gathered} \mathrm{μ}=np \\ =8×0.40 \\ =3.2 \end{gathered}$$

The average number is 3.20, which means that for every eight road fatalities, an alcoholic or alcohol-impaired driver or non-occupant is involved 3.20 times.

A binomial random variable with parameters nand phas a standard deviation of:

$$\begin{gathered} \mathrm{σ}=\sqrt{np\left(1-p\right)} \\ =\sqrt{8×0.40×0.60} \\ =\sqrt{1.92} \\ =1.3856 \end{gathered}$$