Question

We toss a fair 6 sided die and flip a fair coin twice. Define the random variable Y1 to be the number of sports atop the die. Define Y2 to be the total number of heads obtained on the two flips of the coin.

a) Find the mean value and standard deviation of Y1.

b) Find the mean value and standard deviation of Y2.

c) Find the mean value and standard deviation of Y1+Y2.

d) What are the mean value and standard deviation of 5+Y1?

e) What are the mean value and standard deviation of 5(Y1)?

Answer 1

a)

here as probabiolity of an outcome on a die p=1/6

hence

Y1	F(Y1)	Y1P(Y1)	Y12P(Y1)
1	1/6	0.167	0.167
2	1/6	0.333	0.667
3	1/6	0.500	1.500
4	1/6	0.667	2.667
5	1/6	0.833	4.167
6	1/6	1.000	6.000
	total	3.500	15.167
	E(Y1) =μ=	ΣY1P(Y1) =	3.5000
	E(Y1^2) =	ΣY12P(Y1) =	15.1667
	Var(Y1)=σ2 =	E(Y12)-(E(Y1))2=	2.9167
	std deviation=	σ= √σ2 =	1.7078

mean of Y1 =E(Y1)=3.5

Std deviaiton of Y1 SD(Y1)=1.7078

b)

for P(Y2=0)=P(0 heads)=(1/2)*(1/2)=1/4

P(Y2=1)=P(1st toss head and secnd tail)+P(1st tail and 2nd head)=(1/2)*(1/2)+(1/2)*(1/2)=1/2

P(Y2=2)=P(both heads)=(1/2)*(1/2)=1/4

Y2	F(Y2)	Y2P(Y2)	Y22P(Y2)
0	1/4	0.000	0.000
1	1/2	0.500	0.500
2	1/4	0.500	1.000
	total	1.000	1.500
	E(Y2) =μ=	ΣY2P(Y2) =	1.0000
	E(Y2^2) =	ΣY22P(Y2) =	1.5000
	Var(Y2)=σ2 =	E(Y22)-(E(Y2))2=	0.5000
	std deviation=	σ= √σ2 =	0.7071

Here mean of Y2 =E(Y2)=1

std deviation of Y2 =0.7071

c)

mean of Y1+Y2 =E(Y1)+E(Y2)=3.5+1 =4.5

std deviaiton of Y1+Y2 =sqrt(Var(Y1)+Var(Y2))=sqrt(2.9167+0.5)=1.8484

d)

mean of 5+Y1 =5+E(Y1)=5+3.5=8.5

std deviaiton of 5+Y1 =SD(Y1)=1.7078

e)

mean of 5Y1 =5*E(Y1)=5*3.5=17.5

std deviation of 5Y1 =5*SD(Y1)=5*1.7078=8.539