Question

Do you tailgate the car in front of you? About 35% of all drivers will tailgate before passing, thinking they can make the car in front of them go faster. Suppose that you are driving a considerable distance on a two-lane highway and are passed by 12 vehicles.
(a) Let r be the number of vehicles that tailgate before passing. Make a histogram showing the probability distribution of r for r = 0 through r = 12.

(b) Compute the expected number of vehicles out of 12 that will tailgate. (Round your answer to two decimal places.)
vehicles

(c) Compute the standard deviation of this distribution. (Round your answer to two decimal places.)
vehicles

Answer 1

Answer:

(a) The histogram is shown below.

(b) E (X) = 4.2

(c) SD (X) = 2.73

Step-by-step explanation:

Let X = r = a driver will tailgate the car in front of him before passing.

The probability that a driver will tailgate the car in front of him before passing is, P (X) = p = 0.35.

The sample selected is of size n = 12.

The random variable X follows a Binomial distribution with parameters n = 12 and p = 0.35.

The probability function of a binomial random variable is:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x}$

(a)

For X = 0 the probability is:

$P(X=0)={12\choose 0}(0.35)^{0}(1-0.35)^{12-0}=0.006$

For X = 1 the probability is:

$P(X=1)={12\choose 1}(0.35)^{1}(1-0.35)^{12-1}=0.037$

For X = 2 the probability is:

$P(X=2)={12\choose 2}(0.35)^{2}(1-0.35)^{12-2}=0.109$

Similarly the remaining probabilities will be computed.

The probability distribution table is shown below.

The histogram is also shown below.

(b)

The expected value of a Binomial distribution is:

$E(X)=np$

The expected number of vehicles out of 12 that will tailgate is:

$E(X)=np=12\times0.35=4.2$

Thus, the expected number of vehicles out of 12 that will tailgate is 4.2.

(c)

The standard deviation of a Binomial distribution is:

$SD(X)=np(1-p)$

The standard deviation of vehicles out of 12 that will tailgate is:

$SD(X)=np(1-p)=12\times0.35\times(1-0.35)=2.73$

Thus, the standard deviation of vehicles out of 12 that will tailgate is 2.73.