Notes on Probability Distributions and Expected Value


A Probability Distribution refers to the Distribution of Probabilities corresponding to each Outcome or Value taken by r.v. X:

X

X1

X2

X3

Xn

P(X)

P(X1)

P(X2)

P(X3)

P(Xn)



(X: L1, and P: L2)


2. Mean, μ or Expected Value, E(X) = ∑X·P(X)

S.d., σ = √[∑(X – µ)2·P(X)]

Variance, V(X) = σ2


3.
Linear Combination of r.v.:

a. E(k) = k

b. E(kX) = kE(X)

c. E(kX ± lY ± a) = kE(X) ± lE(Y) ± a

d. V(k) = 0

e. V(kX) = k2V(X)

f. V(kX ± lY ± a) = k2V(X) + l2V(Y)

g. E(k1X1 ± k2X2 ± … knXn + a) = k1E(X1) ± k2E(X2) ± …knE(Xn) + a

h. For σ(k1X1 ± k2X2 ± … knXn + a), assuming independence, first find V(k1X1 ± k2X2 ± … knXn + a) = k12V(X1) + k22V(X2) + … + kn2V(Xn)




Problem. ​Steven and Beth visit a record store frequently together. The Number of CDs each buys has the following probability distributions:

S

0

1

2

3

4

P(S)

0.20

0.15

0.35

0.05

0.25



B

0

1

2

P(B)

0.15

0.35

0.50



a) What is the probability that neither Steven not Beth purchases a CD on their trip?


b) What is the probability that Steven and Beth purchase exactly 3 CDs in all?


c) What is the probability that Steven and Beth purchase at most 2 CDs in all?

d) Find the expected Number of CDs Steven purchases. Find the s.d. of the Number of CDs Steven purchases.


e) Find the expected Number of CDs Beth purchases. Find the s.d. of the Number of CDs Beth purchases.


f) If each CD that Steven purchases costs $11.99 and each CD that Beth purchases costs $16.99, define a random variable to denote the Total expenditure per visit, T. Find the Mean and s.d. of the Total Expenditure per visit, T.


g) On 2 successive trips, find the probability that Steven shall purchase at most 1 CD?

Solution.
a) P(S = 0 and B = 0) = P(S = 0)·P(B = 0) = 0.20·0.15 = 0.03


b) P(S + B = 3) = P(S = 1)·P(B = 2) + P(S = 2)·P(B = 1) + P(S = 3)·P(B = 0) = 0.205

c) P(S + B < 2) = P(S = 0)·P(B = 0) + P(S = 0)·P(B = 1) + P(S = 0)·P(B = 2) + P(S = 1 1)·P(B = 0) + P(S = 1)·P(B = 1) + P(S = 2)·P(B = 0) = 0.3275

d) E(S) = μ = ∑S∙P(S) = 0(0.2) + ...4(0.25) = 2 using 1-Var Stats L1, L2

σ(S) = √∑(S – μ)2∙P(S) = √(0 – 2)2∙(0.2) + ...(4 – 2)2∙(0.25) = 1.4142 using 1-Var Stats L1, L2


e) E(B) = μ = ∑B∙P(B) = 0(0.2) + ...4(0.25) = 1.35 using 1-Var Stats L1, L2

σ(B) = √∑(B – μ)2∙P(B) = √(0 – 1.35)2∙(0.15) + ...(2 – 1.35)2∙(0.5) = 0.7263 using 1-Var Stats L1, L2


f) Let Total, T = 11.99S + 16.99B
E(T) = 11.99∙E(S) + 16.99∙E(B) = 11.99∙2 + 16.991.35 = $46.9165
and
Variance, V(T) = 11.992∙V(S) + 16.992∙V(B) = 11.992∙4.41422 + 16.992∙0.72632 = 439.787, since it's safe to assume independence of expenditures.

σ(T) = = √V(T) = √439.787 = $20.9711

g) P(S1 + S2 < 1) = P(S1 = 0)·P(S2 = 0) + P(S1 = 1)·P(B = 0) + P(S1 = 1)·P(S2 = 0) = 0.1

Problem. Pete's Jeep Tours offers a popular all-day trip in a tourist area. The number of passengers, X, on a randomly selected day has the following probability distribution:


X

2

3

4

5

P(X)

0.15

0.35

0.3

0.20


Pete charges $150 per passenger, and it costs him $100/day to buy permits, gas and a ferry pass for each all-day trip. He is open for business every day of the week.

a) On 2 successive days, what is the probability that Pete takes exactly 7 tourists on tour?

b) On 2 successive days, what is the probability that Pete takes 9 or more tourists on tour?

c) Calculate the Mean and s.d. of the number of tourists Pete's Jeep Tours takes on an all-day trip.


d) Calculate the mean and s.d. of the daily profit for Pete.


e) Pete’s sister Erin starts a similar business Erin's Adventures in another part of the country. She discovers that the number of passengers, Y, on her all-day tours have the probability distribution:


Y

2

3

4

P(Y)

0.35

0.45

0.20


Erin charges $175 per passenger and it costs $120 per day for her to buy permits, gas and a ferry pass for each all-day trip.


e) Assuming the number of passengers that Pete and Erin take are independent of each other, on any day
1. what is the probability that Pete and Erin together have fewer than 6 passengers?
2. what is the probability that Pete and Erin together have exactly 8 passengers?
3. what is the probability that Pete and Erin have the same number of passengers?
4. what is the probability that Pete has fewer passengers than Erin?

f) Assuming the number of passengers that Pete and Erin take are independent of each other, determine the probability distribution of the Difference [Pete – Erin] in number of passengers on any day.
Tip! Make a probability table showing values of D = X – Y vs. P(D). What are all the values – Smallest? Largest? – D can take? Scan both tables to determine the relevant outcomes for each value of D and then multiply the corresponding probabilities. Caution! if you dont do this carefully and consider ALL possible outcomes, the probabilities shant add up to 100%.

g) Calculate the expected daily profit of Erin, and the s.d. of the profit
Tip! Define a new r.v.


h) Determine the probability distribution of the total number of tourists Erin would have on 2 successive days, assuming independence of the outcomes.
Tip! As in j, make a probability distribution table showing values of S = Y1 + Y2 vs. P(S). What are all the values – Smallest? Largest? – S can take? Scan the Y distribution to determine the relevant outcomes for each value of S and then multiply the corresponding probabilities. Show as little work as you want, but just be careful and consider ALL possibilities for each value of S.
Caution! if you dont do this carefully and consider ALL possible outcomes, the probabilities shant add up to 100%.


i) What is the IQR of the distribution in part h? Show work.


j) For the distribution in h, what percentile does 5 passengers correspond to?

Solution.

a) P(X1 + X2 = 7) = P(X1 = 2)·P(X2 = 5) + P(X1 = 5)·P(X2 = 2) + P(X1 = 3)·P(X2 = 4) + P(X1 = 4)·P(X2 = 3) = 0.27

b) P(X1 + X2 > 9) = P(X1 = 4)·P(X2 = 5) + P(X1 = 5)·P(X2 = 4) + P(X1 = 5)·P(X2 = 5) = 0.16

c) E(X) = Show steps! ∑X·P(X) = Show work! 2·0.15 + …5·0.2 = 3.55, using 1-Var Stats L1, L2

σ(X) = Show steps! √∑(X – μX)2·P(X) = Show work! √(2 – 3.55)2·0.15 + … (5 – 3.55)2·0.2 = 0.9734, using 1-Var Stats L1, L2

d) If B denotes the daily profit for Pete, E(B) = Show steps! 150·E(X) – 100 = Show work! 150·3.55 – 100 = $432.50

Operate on Variances! V(B) = Show steps! 1502 V(B) = Show work! 1502·0.97342 = 21318.9201

σ(B) = Show steps! √V(B) = Show work! √21318.9201 = $146.01

e)

1. P(X + Y < 6) = Show steps! P(X = 2)·P(Y = 2) + P(X = 3)·P(Y = 2) +

P(X = 2)·P(Y = 3), given independence. Substitute and finish.
Answer. 0.2425.

2. P(X + Y = 8) = Show steps! P(X = 4)·P(Y = 4) + P(X = 5)·P(Y = 3), given independence. Substitute and finish.
Answer. 0.15

3. P(X = Y) = Show steps! P(X = 2)·P(Y = 2) + P(X = 3)·P(Y = 3) + P(X = 4)·P(Y = 4), given independence. Substitute and finish.
Answer. 0.27

4. P(X < Y) = Show steps! P(X = 2)·P(Y = 3) + P(X = 2)·P(Y = 4) + P(X = 3)·P(Y = 4), given independence. Substitute and finish.
Answer. 0.1675

f)

D = X – Y

-2:

2, 4

-1:

2, 3; 3, 4

0:

2, 2; 3, 3; 4, 4

1:

3, 2; 4, 3; 5; 4

2:

4, 2; 5, 3

3:

5, 2

P(D)

0.03

0.1375

0.27

0.2975

0.195

0.07


g) If L is a variable denoting Erin’s daily profit, L = $175Y – 120
E(L) = Show steps! 175·E(Y) – 100 = Show work! 175·2.85 – 120 = $378.75


Operate on Variances! V(L) = Show steps! 1752 V(Y) = Show work! 1752·0.72632

= 16155.04551

σ(L) = Show steps! √V(L) = Show work! √16155.04551 = $127.1025

h) If S = Y1 + Y2, then we need:

S =

Y1 + Y2

4:

2, 2

5:

2, 3; 3, 2

6:

2, 4; 4, 2; 3, 3

7:

3, 4; 4, 3

8:

4, 4

P(S)

0.1225

0.315

0.3425

0.18

0.04


i) Show steps! P(S < Q1) = 0.25 → Q1 = 5
Show steps! P(S < Q3) = 0.75 → Q1 = 6
IQR = Show steps! Q3 – Q1 = 6 – 5 = 1

j) P(S < 5) = Show steps! P(S = 4) + P(S = 5) = Show work! 0.1225 + 0.315 = 0.4375

PROBLEM. A box contains ten $1 bills, five $2, three $5 bills, one $10 bill, and one $100 bill. In a game, the player is charged $20 to select one bill and wins the bill chosen.

a. Find the expected value and s.d. of the one’s gain.

b. Is the game fair? Explain.

Solution.

If is a r.v. denoting gain, Outcomes: drawing $1, $2, $5, $10 and $100.

Value taken by X [adjusted for the cost of the game]: -$19, -$18, -$15, -$10, $80

P(outcomes): 10/20, 5/20, 3/20, 1/20, 1/20


E(X) = μ = ΣX∙P(X) = $-19(10/20) + ...$80(1/20) = -$12.75, using 1-Var Stats L1, L2

In the long run, one may expect to lose $12.75 per game.


b) No, the game is not fair, since in the long one may expect to lose $12.75 per game: the expectation is non-zero.