Probability and standard deviation questions need some help

Question

A recent study by the American Highway Patrolman’s Association revealed that 65% of American drivers use their seatbelts. A sample of 12 drivers on major highways was randomly selected.

a. Find the probability that seven of the drivers are wearing seatbelts.

b. How many of the drivers would be expected to be wearing their seatbelt?

c. Calculate the standard deviation for this distribution

Answer 1

Background:

In Mathematica:

N@Binomial[12, 7] (.65^7 (1 - .65)^(12-7) 

a) ${12 \choose 7} .65^7 (1-.65)^{12-7} = .2039$

Mean[BinomialDistribution[12, .65]]

b) $ .65 \times 12 = 7.8$

Variance[BinomialDistribution[12, .65]]

c) $Var = n p (1-p) = 12 \times 0.65 (1 - 0.65) = 2.73$

Answer 2

If you like using RStudio (R statistical computing environment), then use the following command lines:

$a$. $\hspace{2mm}$ ${\tt dbinom(7,12,0.65)}$ $\hspace{2mm}$ since the binomial distribution is the probability mass function (pmf) $$ b(x) = b(x; 12,0.65) = \begin{cases} \binom{12}{x} (0.65)^x(0.35)^{12-x} &\mbox{ if } x=0,1,2,\ldots, 12, \\ \qquad \qquad 0 & \mbox{ otherwise}, \\ \end{cases} $$

$b$. $\hspace{2mm}$ ${\tt 12*.65}$ $\hspace{2mm}$ since the expected value for a binomial distribution is $E[X]=np$, where $n$ is the number of trials and $p$ is the probability of success (the percentage of American drivers using their seatbelts),

$c$. $\hspace{2mm}$ ${\tt sqrt(12*(.65)*(.35))}$ $\hspace{2mm}$ since the variance for the binomial distribution is $V(X)=np(1-p)$. So the standard deviation is the square root of the variance, i.e., $\sigma=\sigma_X = \sqrt{V(X)}$.

$\color{green}{\textbf{Plotting}.}$ If you want to plot this binomial distribution (on a plane), then use the following command lines on RStudio: $$ \begin{align*} &{\tt x <- seq(0, 12, by = 1) } \\ &{\tt y <- dbinom(x, 12, 0.65) } \\ &{\tt png(file = "dbinom.png") } \\ &{\tt plot(x,y) } \\ &{\tt dev.off() } \\ \end{align*} $$ and you'll see the following png file saved to your computer.