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:
[tex]P(X=x)={n\choose x}p^{x}(1-p)^{n-x}[/tex]
(a)
For X = 0 the probability is:
[tex]P(X=0)={12\choose 0}(0.35)^{0}(1-0.35)^{12-0}=0.006[/tex]
For X = 1 the probability is:
[tex]P(X=1)={12\choose 1}(0.35)^{1}(1-0.35)^{12-1}=0.037[/tex]
For X = 2 the probability is:
[tex]P(X=2)={12\choose 2}(0.35)^{2}(1-0.35)^{12-2}=0.109[/tex]
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:
[tex]E(X)=np[/tex]
The expected number of vehicles out of 12 that will tailgate is:
[tex]E(X)=np=12\times0.35=4.2[/tex]
Thus, the expected number of vehicles out of 12 that will tailgate is 4.2.
(c)
The standard deviation of a Binomial distribution is:
[tex]SD(X)=np(1-p)[/tex]
The standard deviation of vehicles out of 12 that will tailgate is:
[tex]SD(X)=np(1-p)=12\times0.35\times(1-0.35)=2.73\\[/tex]
Thus, the standard deviation of vehicles out of 12 that will tailgate is 2.73.
