Question

1. Let the random variable x represent the number of cars owned by a family. Assume that x can take on five values: 0, 1, 2, 3, 4. A partial probability distribution is shown below:

        

x	0	1	2	3	4
p(x)	0.2	0.1	0.3	?	0.1

         i.       The probability that a family owns three cars equals _______

         ii.       The probability that a family owns between 1 and 3 cars, inclusive, equals _______

         iii.      The expected number of cars equals _______

Answer 1

i) Since  $\sum_{x=0}^{4}p(x) = 1$

$\Rightarrow 0.2 + 0.1 + 0.3 + p(3) +0.1 = 1$

$\Rightarrow 0.7 + p(3) = 1$

$\Rightarrow p(3) = 1 - 0.7 = 0.3$

ii) The probability that a family owns between 1 and 3 cars, inclusive, equals

$=p(1) +p(2) +p(3)$

$= 0.1 + 0.3 + 0.3$

$= 0.7$

iii) The expected number of cars equals

$E(X)= \sum_{x=0}^{4}x*p(x) = (0*0.2) +(1*0.1)+(2*0.3) +(3*0.3) +(4*0.1) = 2$