Question

E(Y) = 4, E(X) = E(Y+2), Var(X) = 5, Var(Y) = Var(2X+2)

a. What is E(2X -2Y)?
b. What is Var(2X-Y+2)?
c. What is SD(3Y-3X)?

2)

In a large community, 65% of the residents want to host an autumn community fair.
The rest of the residents answer either disagree or no opinion on this proposal.
A sample of 18 residents was randomly chosen and asked for their opinions.
a) What is th probability that the residents answer either disagree or no opinion on this proposal.
b) what is the probability that all of them want to host an autumn community fair?
c) what is the probability that at most 15 residents want to host an autumn community fair?
d) what is the probability that less than 5 residents want to host an autumn community fair?

3)

The following information is applied to both question 5 and question 6.

A drink manufactory produce a drink with 12 oz bottle. The production line will fill the bottle with the drink.

The machine has a probability of 0.03 to fail to fill in the drink in the bottle properly. Suppose this probability is independent of the others.

80 bottles of drink were produced.

Question 5

a) What is the probability that all the bottles of drink are filled in properly?

b) What is the probabilty that at most 3 bottle of drinks are not filled in properly?

c) What is the expeted value of the number of bottle of drink are not filled in properly?

d) What is the variance of the number of bottle of drink are not filled in properly?

Answer 1

solution:

1) Given that

E[Y] = 4

E[X] = E[Y+2] = E[Y]+2 = 4+2 = 6

Var[X] = 5

Var[Y] = Var[2X+2]  

= Var[2X] + var[2]

= 4 Var[X] + 0 [since, Var[kX] = k^2 Var[X] , Var[k] = 0 ]

= 4*5

Var[Y]    = 20

a) E [2X-2Y] = E[2X]+E[-2Y]

= 2E[X] - 2E[Y]

= 2 (E[X] - E[Y] )

= 2 ( 6-4)

= 4

   E [2X-2Y] = 4

b) Var[2X-Y+2] = Var[2X]+ Var[-Y]+ Var[2]

= 4* Var[X] + Var[Y]

= 4*5 + 20

= 40

     Var[2X-Y+2] = 40

c) SD( 3Y - 3X) = √ Var[3Y-3X]

= √ Var[3Y]+Var[-3X]

= √ 9Var[Y] + 9 Var[X]

= √ 9(Var[Y] + Var[X] )

= √ 9*(20+5)

= √ 9*25

= 15

    SD( 3Y - 3X) = 15

2) Given that

percent of  residents want to host an autumn community fair = 65%

percent of  residents  either disagree or no opinion on this proposal = 100%- 65% = 35%

No.of residents (n) = 18

a)

Let p = Probability of residents want to host an autumn community fair = 65% = 0.65

1-p = probability of  residents  either disagree or no opinion on this proposal = 1- 0.65 = 0.35

b) Here , No.of residents (n) is fixed and probability p remains same for every trial

   X ~ B(18,0.65)

By using binomial distribution, we have

P(X=x) = nCx * (p)^x * (1-p)^(n-x)

Probability that all 18 residents want to host an autumn community fair = P(X=18)

= 18C18 * (0.65)^18 * (0.35)^0

= 0.0004

c) probability that at most 15 residents want to host an autumn community fair = P(X<=15)

    = 1 - P(X>15)

   = 1 - [ P(X=16)+P(X=17)+P(X=18) ]

= 1 - [ 18C16 * (0.65)^16 * (0.35)^2 + 18C17 * (0.65)^17 * (0.35)^1 + 18C18 * (0.65)^18 * (0.35)^0]

= 1 - [ 0.0190+0.0042+0.00004]

= 1 - [ 0.0236]

= 0.9764   

d) probability that less than 5 residents want to host an autumn community fair = P(X<5)

= P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)

= [ 18C0 * (0.65)^0 * (0.35)^18 +18C1 * (0.65)^1 * (0.35)^17 +18C2 * (0.65)^2 * (0.35)^16

+18C3 * (0.65)^3 * (0.35)^15 +18C4 * (0.65)^4 * (0.35)^14 ]

= [ 0 + 0000002 + 0000033 + 0000325 + 000226 ]

= 0.00026

3) Given that

No.of bottles to be filled (n)= 80

Probability of fail to fill (p)= 0.03

Here, = n* p = 80*0.03 = 2.4 [ since, Poisson distribution is limiting case of Binomial Distribution ]

Let X be the no.of bottles not filled properly

By using poisson distribution,we have

P(X=x) = e^- * ^x / x!

a) Probability that all bottles are filled properly = 1 - Probability that all bottles are not filled properly

= 1 - [   e^-2.4 * 2.4^80 / 80! -(negligible)]

= 1 - 0

= 1

b) probabilty that at most 3 bottle of drinks are not filled in properly =  P(X<=3)

= P(X=0)+P(X=1)+P(X=2)+P(X=3)

= e^-2.4 * 2.4^0 / 0! + e^-2.4 * 2.4^1 / 1! + e^-2.4 * 2.4^2 / 2! + e^-2.4 * 2.4^3 / 3!

= e^-2.4 [ 1 + 2.4 + (2.4)^2 / 2 + (2.4)^3 / 6 ]

= e^-2.4 * 8.584

= 0.4703

c) Expected No.of bottles of drink not filled properly () = 2.4 ~ 2 bottles

d) In poisson distribution Variance = Mean =

     variance of the number of bottle of drink are not filled in properly = 2.4 ~ 2 bottles