Question #283933

There are 4 traffic lights along the road, each prohibits further movement of the car with a probability of 0.5. Find the

distribution of the number of the traffic lights that car passed to the first stop. What is the variance of the random

variable? 



1
Expert's answer
2022-01-10T14:25:13-0500

Let X=X= the number of the traffic lights that car passed to the first stop.

Then


P(X=0)=0.5(0.5)0=0.5P(X=0)=0.5(0.5)^0=0.5

P(X=1)=0.5(0.5)1=0.25P(X=1)=0.5(0.5)^1=0.25

P(X=2)=0.5(0.5)2=0.125P(X=2)=0.5(0.5)^2=0.125

P(X=3)=0.5(0.5)3=0.0625P(X=3)=0.5(0.5)^3=0.0625

P(X=4)=0.5(0.5)4=0.03125P(X=4)=0.5(0.5)^4=0.03125

x01234p0.50.250.1250.06250.03125\def\arraystretch{1.5} \begin{array}{c:c} x & 0 & 1 & 2 & 3 & 4 \\ \hline p & 0.5 & 0.25 & 0.125 & 0.0625 & 0.03125 \\ \end{array}

E(X)=0(0.5)+1(0.25)+2(0.125)E(X)=0(0.5)+1(0.25)+2(0.125)

+3(0.0625)+4(0.03125)=0.8125+3(0.0625)+4(0.03125)=0.8125

E(X2)=02(0.5)+12(0.25)+22(0.125)E(X^2)=0^2(0.5)+1^2(0.25)+2^2(0.125)

+32(0.0625)+42(0.03125)=1.8125+3^2(0.0625)+4^2(0.03125)=1.8125

Var(X)=E(X2)(E(X))2Var(X)=E(X^2)-(E(X))^2

=1.8125(0.8125)2=1.15234375=1.8125-(0.8125)^2=1.15234375


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