Answer:
(a) The histogram is shown below.
(b) E (X) = 4.2
(c) SD (X) = 2.73
Step-by-step explanation:
Let <em>X</em> = <em>r</em><em> </em>= 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) = <em>p</em> = 0.35.
The sample selected is of size <em>n</em> = 12.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 12 and <em>p</em> = 0.35.
The probability function of a binomial random variable is:
![P(X=x)={n\choose x}p^{x}(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%3Dx%29%3D%7Bn%5Cchoose%20x%7Dp%5E%7Bx%7D%281-p%29%5E%7Bn-x%7D)
(a)
For <em>X</em> = 0 the probability is:
![P(X=0)={12\choose 0}(0.35)^{0}(1-0.35)^{12-0}=0.006](https://tex.z-dn.net/?f=P%28X%3D0%29%3D%7B12%5Cchoose%200%7D%280.35%29%5E%7B0%7D%281-0.35%29%5E%7B12-0%7D%3D0.006)
For <em>X</em> = 1 the probability is:
![P(X=1)={12\choose 1}(0.35)^{1}(1-0.35)^{12-1}=0.037](https://tex.z-dn.net/?f=P%28X%3D1%29%3D%7B12%5Cchoose%201%7D%280.35%29%5E%7B1%7D%281-0.35%29%5E%7B12-1%7D%3D0.037)
For <em>X</em> = 2 the probability is:
![P(X=2)={12\choose 2}(0.35)^{2}(1-0.35)^{12-2}=0.109](https://tex.z-dn.net/?f=P%28X%3D2%29%3D%7B12%5Cchoose%202%7D%280.35%29%5E%7B2%7D%281-0.35%29%5E%7B12-2%7D%3D0.109)
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](https://tex.z-dn.net/?f=E%28X%29%3Dnp)
The expected number of vehicles out of 12 that will tailgate is:
![E(X)=np=12\times0.35=4.2](https://tex.z-dn.net/?f=E%28X%29%3Dnp%3D12%5Ctimes0.35%3D4.2)
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)](https://tex.z-dn.net/?f=SD%28X%29%3Dnp%281-p%29)
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\\](https://tex.z-dn.net/?f=SD%28X%29%3Dnp%281-p%29%3D12%5Ctimes0.35%5Ctimes%281-0.35%29%3D2.73%5C%5C)
Thus, the standard deviation of vehicles out of 12 that will tailgate is 2.73.