Econ 250 Fall 2010

Due at November 16

Assignment 2: Binomial Distribution, Continuous Random Variables andSampling

1. Suppose a firm wishes to raise funds and there are a large number of independentfinancial lenders who might lend from 0 to $10 million dollars each. The total amountraised follows a uniform distribution from 0 to $10n million dollars, where n is thenumber of lenders. If the firm is to have at least 80% chance of raising at least $100million dollars what is the minimum number of lenders that should be contacted?

Solution: This is a uniform distribution question. Let X be the total amount raised.X is distributed uniformly on [0,10n].The density function is: f(x) = 1

10n

The probability of raising more than $100 is at least 80%

P (X > 100) = 1− P (X < 100) = 1− 100

10n≥ 80%

⇒ n ≥ 50

At least 50 lenders should be contracted.

2. The probability that a person catches a cold during the cold and flu season is 0.4.Assume that 10 people are chosen at random.

Solution: This is a binomial question. A success occurs when a person catches acold during the cold and flu season. The probability of a success is 0.4. Let X be thenumber of successes in 10 people.

(a) What is the probability that exactly 4 people have the flu?

P (X = 4) = (104

)(0.4)4(0.6)6 = 0.2508

(b) What is the probability that between 2 and 5 people inclusive have the flu?

P (2 ≤ X ≤ 5) = P (X ≤ 5)− P (X ≤ 1) = 0.834− 0.046 = 0.788

(c) What is the expected number of people with the flu and what is its variance?

E[X] = µ = np = 10× 0.4 = 4

V [X] = σ2 = np(1− p) = 2.4

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(d) Approximate your answer in (b) using the normal approximation with and withoutcontinuity corrections.

Without continuity correction:

P (2 ≤ X ≤ 5) ≈ P (2− 4√

2.4≤ X − µ

σ≤ 5− 4√

2.4)

= P (−1.2910 ≤ Z ≤ 0.6455)

= F (0.6455)− (1− F (1.2910))

= 0.7406− (1− 0.9015) = 0.6421

With continuity correction:

P (2 ≤ X ≤ 5) ≈ P (1.5− 4√

2.4≤ X − µ

σ≤ 5.5− 4√

2.4)

= P (−1.6137 ≤ Z ≤ 0.9682)

= F (0.9682)− (1− F (1.6137))

= 0.834− (1− 0.9463) = 0.7803

3. The following table displays the joint probability distribution of X and Y.

X/Y 0 1 20 .13 .08 .021 .16 .14 .232 .07 .04 .13

Solution:

(a) What is the covariance and correlation between the X and Y? Are these variablesindependent? Explain.

E[X] = µX =∑

xP (x) = 1.01

E[Y ] = µY =∑

yP (y) = 1.02

V [X] = σX2 =

∑(X − µX)2P (x) = 0.4699

V [Y ] = σY2 =

∑(Y − µY )2P (y) = 0.7396

Cov(X, Y ) = E(XY )− µXµY =∑X

∑Y

xyP (x, y)− µXµY = 0.1698

Corr =Cov(X, Y )

σXσY=

0.1698√0.4699

√0.7396

= 0.2880

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(b) Calculate the mean and variance of D

D = 2 + 4X − 2Y

E(D) = 2 + 4E(X)− 2E(Y ) = 2 + 4(1.01)− 2(1.02) = 4

V [D] = 42V [X] + 22V [Y ]− 2(4)(2)Cov(X, Y )

= 16(0.4699) + 4(0.7396)− 16(0.1698) = 7.76

(c) Write out the joint cumulative distribution.

F(X,Y) F(X,0) F(X,1) F(X,2)F(0,Y) 0.13 0.21 0.23F(1,Y) 0.29 0.51 0.76F(2,Y) 0.36 0.62 1

4. Suppose we know the number of sales X by any sales person follows a normal distri-bution with a mean of 61.7 and a standard deviation of 5.2.

Solution: X ∼ N(61.7, 5.22)

(a) What is P (62.5 < X < 64)?

P (62.5 < X < 64) = P (62.5− 61.7

5.2< Z <

64− 61.7

5.2) = P (0.1538 < Z < 0.4423)

= F (0.44)− F (0.15) = 0.67− 0.5596 = 0.1104

(b) What is the P (62.5 < X̄3 < 64) where X̄3 is the average sales from three (inde-pendent) sales persons?

P (62.5 < X̄3 < 64) = P (62.5− 61.7

5.2/√

3< Z <

64− 61.7

5.2/√

3) = P (0.2665 < Z < 0.7661)

= F (0.77)− F (0.27) = 0.7794− 0.6064 = 0.173

(c) What is the value of k such that P (X > k) = 0.63?

F (−0.33) ≈ 1− 0.63

X = Zσ + µ = −0.33(5.2) + 61.7 = 59.984

(d) What is the value of k such that P (59 < X < k) = 0.54?

P (59 < X < k) = P (59− 61.7

5.2< Z < Zk) = P (−0.52 < Z < Zk)

= F (Zk)− (1− F (0.52)) = 0.54

⇒ F (Zk) = 0.54 + 1− F (0.52) = 1.54− 0.6985 = 0.8415

⇒ Zk ≈ 1

Thusk = Zkσ + µ = 1(5.2) + 61.7 = 66.9

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5. Sales at a local electrical wholesaler consist of both over-the-counter sales as well asdeliveries. During the course of a month, over-the-counter sales have a mean of $96,780with a standard deviation of $12,520. Over the same time period, deliveries average$229,620 with a standard deviation of $234,100. Assume that over-the-counter salesand deliveries have a correlation of .2.

Solution:Let SC denote over-the-counter sales and SD denote deliveries.SC ∼ N(96780, 125202) and SD ∼ N(229620, 2341002)Corr(SC , SD) = 0.2

(a) What is the mean, variance and distribution of all sales S?All sales S = SC + SD

E[S] = µS = E[SC ] + E[SD] = 96780 + 229620 = 326400

V [S] = σs2 = V [SC ] + V [SD] + 2(Corr)(

√V (SC))(

√V (SD))

= 125202 + 2341002 + 2(0.2)(12520)(234100) = 5.613× 1010

Now we can see that S ∼ N(326400, 5.613× 1010)

(b) What is the P (222, 900 < S < 240, 400)?

P (222, 900 < S < 240, 400) = P (222, 900− 326400√

5.613× 1010< Z <

240, 400− 326400√5.613× 1010

)

= P (−0.4369 < Z < −0.3630)

= F (0.4369)− F (0.363) = 0.0294

6. Suppose X is uniform distribution over the interval 0 to 150.

(a) Find the mean and variance.

µX =a+ b

2=

150 + 0

2= 75

σ2 =(b− a)2

12=

1502

12= 1875

(b) Find the value that leaves .05 in the lower tail and also the value that leaves .05in the upper tail.

0.05 =XL − 0

150⇒ XL = 7.5

0.05 =150−XU

150⇒ XU = 142.5

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(c) Suppose that you do not know that the variable is uniform but are given the meanand variance from (a). Calculate the same magnitudes for (b).

F (ZL = −1.645) = 0.05⇒ XL = (−1.645)√

1875 + 75 = 3.7694

1− F (ZU = 1.645) = 0.05⇒ XU = (1.645)√

1875 + 75 = 146.2306

(d) Draw the two distributions to explain these results.

7. Two classes of statistics have grades that are normally and independently distributedwith C1 ∼ N(75, 12) and C2 ∼ N(80, 22).

(a) What is the expected difference and its variance?C = C1 − C2

The expected difference and its variance:

E[C] = µC = E[C1]− E[C2] = 75− 80 = −5

V [C] = σC2 = V [C1] + V [C2] = 12 + 22 = 34

(b) What is the probability that the difference from picking 1 student from each classis between -1 and 1?

P (−1 < C < 1) = P (−1− (−5)√

34< Z <

1− (−5)√34

)

= P (0.6860 < Z < 1.0290) = F (1.03)− F (0.69) = 0.0936

(c) To give the class the same mean as the second class, the professor adds 5 to allgrades. Explain why this does not leave the two classes equivalent. Which classwould you prefer to be in?

8. A car company says their car gets a mean of 45km per liter with a standard deviationof 6. Suppose we assume a normal distribution.

Solution:

(a) Suppose some sales representative claims you will get at least 47 80% of the time,what can you tell him?

P (X ≥ 47) = P (Z ≥ 47− 45

6) = P (Z ≥ 1

3) = 1− F (0.333) = 0.3707

(b) If we have 4 cars and take the average, calculate the P (44 < X̄4 < 46).

P (44 < X̄4 < 46) = P (44− 45

6/√

4< Z <

46− 45

6/√

4)

= P (−0.3333 < Z < 0.3333) = F (0.33)− (1− F (0.33))

= 2(0.6293)− 1 = 0.2586

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9. Suppose you wish to drive across a country that is 2625 km wide and you intend torent a series of cars from Rent-A-Wreck. The distance that the first car they give youis normally distributed with a mean distance of 1500km and a variance of 500km. Eachsubsequent car you rent gets 50% less km on average than the previous one with a 75%reduction in the variance.

(a) Try to formulize this problem.

(b) What is the probability that the trip can be done using exactly 2 cars?

(c) What is the probability that you do the trip with more than 3 cars?

(d) If each car costs $100 what is the expected cost of the trip?

(e) Approximate the expected length of the farthest trip that can be taken.

Let Xi be the distance travelled by car i

X1 ∼ N(1500, 500)

X2 ∼ N(750, 125)

X3 ∼ N(375, 31.5)

...and so onConsider the first car

P (X1 > 2625) = P (X1−µ1σ1

> 2625−1500√500

) ⇒ P (Z > 50.3) ≈ 0

Consider the distance traveled by the first car and then the second car

P (X1 +X2 > 2625) = P (X1+X2−(µ1+µ2)√σ1

2+σ22

> 2625−(1500+750)√500+125

) ⇒ P (Z > 15) ≈ 0

Now the third car:

P (X1 +X2 +X3 > 2625)

= P (X1+X2+X3−(µ1+µ2+µ3)√σ1

2+σ22+σ3

2> 2625−(1500+750+375)√

500+125+31.5) ⇒ P (Z > 0) = 0.5

Son = 3Expected cost of the trip is $300.

Farthest trip possible T (geometric series)

T∞ = X1 +X2 +X3 + ... =1

1− 0.5×X1

E[T∞] = 2× 1500 = 3000

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