Question

Consider the following independent discrete random variables.

x = number of tornadoes detected in any given month for state X

x	0	1	2	3	4	5
p(x)	0.55	0.17	0.14	0.09	0.03	0.02

y = number of tornadoes detected in any given month for state Y

y	0	1	2	3	4	5
p(y)	0.40	0.30	0.15	0.11	0.01	0.03

?_x = (0)(0.55) + (1)(0.17) + (2)(0.14) + (3)(0.09) + (4)(0.03) + (5)(0.02) = 0.94

?_y = (0)(0.40) + (1)(0.30) + (2)(0.15) + (3)(0.11) + (4)(0.01) + (5)(0.03) = 1.12

(a) What is the mean of the sum of these independent discrete random variables for the combined total of tornadoes detected in any given month for these two states?

?_{x + y} = ______tornadoes

(b) What is the variance of the difference in the number of tornadoes detected in any given month for these two states?

?x ? y2     = _________ tornadoes

(c) What is the standard deviation of the difference in the number of tornadoes detected in any given month for these two states? (Round your answer to four decimal places.)

?_{x ? y} = ________tornadoes

Answer 1

X and Y are independent discrete random variable.

x = number of tornadoes detected in any given month for state X

y = number of tornadoes detected in any given month for state Y

as they are independent the mean of the total tornadoes will be the sum of ?x and ?y

a) ?(x+y) = 0.94+1.12 = 2.06

?(x-y) = ?x + ?y - cov(x,y)

but they are independent so cov(x,y) = 0

now ?x =  (0)^2(0.55) + (1)^2(0.17) + (2)^2(0.14) + (3)^2(0.09) + (4)^2(0.03) + (5)^2(0.02) =2.52

and  ?y = (0)^2(0.40) + (1)^2(0.30) + (2)^2(0.15) + (3)^2(0.11) + (4)^2(0.01) + (5)^2(0.03) = 2.8

b) ?(x-y) = 5.32

c) and standard deviation =

= = 2.30