Download Econ 250 Fall 2010 Due at November 16 Assignment 2: Binomial

Econ 250 Fall 2010
Due at November 16
Assignment 2: Binomial Distribution, Continuous Random Variables and
Sampling
1. Suppose a firm wishes to raise funds and there are a large number of independent
financial lenders who might lend from 0 to $10 million dollars each. The total amount
raised follows a uniform distribution from 0 to $10n million dollars, where n is the
number of lenders. If the firm is to have at least 80% chance of raising at least $100
million 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].
1
The density function is: f (x) = 10n
The probability of raising more than $100 is at least 80%
P (X > 100) = 1 − P (X < 100) = 1 −
100
≥ 80%
10n
⇒ 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 a
cold during the cold and flu season. The probability of a success is 0.4. Let X be the
number of successes in 10 people.
(a) What is the probability that exactly 4 people have the flu?
P (X = 4) = (
10
)(0.4)4 (0.6)6 = 0.2508
4
(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] = σ 2 = np(1 − p) = 2.4
1
(d) Approximate your answer in (b) using the normal approximation with and without
continuity corrections.
Without continuity correction:
2−4
5−4
X −µ
P (2 ≤ X ≤ 5) ≈ P ( √
≤ √ )
≤
σ
2.4
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:
5.5 − 4
X −µ
1.5 − 4
≤ √
≤
)
P (2 ≤ X ≤ 5) ≈ P ( √
σ
2.4
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
2
0
.13
.16
.07
1
2
.08 .02
.14 .23
.04 .13
Solution:
(a) What is the covariance and correlation between the X and Y? Are these variables
independent? Explain.
X
E[X] = µX =
xP (x) = 1.01
X
E[Y ] = µY =
yP (y) = 1.02
X
V [X] = σ X 2 =
(X − µX )2 P (x) = 0.4699
X
V [Y ] = σ Y 2 =
(Y − µY )2 P (y) = 0.7396
XX
Cov(X, Y ) = E(XY ) − µX µY =
xyP (x, y) − µX µY = 0.1698
X
Y
0.1698
Cov(X, Y )
√
=√
= 0.2880
Corr =
σX σY
0.4699 0.7396
2
(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] = 42 V [X] + 22 V [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
0.23
F(1,Y)
0.29
0.51
0.76
F(2,Y)
0.36
0.62
1
4. Suppose we know the number of sales X by any sales person follows a normal distribution 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)?
64 − 61.7
62.5 − 61.7
<Z<
) = P (0.1538 < Z < 0.4423)
5.2
5.2
= F (0.44) − F (0.15) = 0.67 − 0.5596 = 0.1104
P (62.5 < X < 64) = P (
(b) What is the P (62.5 < X̄3 < 64) where X̄3 is the average sales from three (independent) sales persons?
62.5 − 61.7
64 − 61.7
√
√ ) = P (0.2665 < Z < 0.7661)
<Z<
5.2/ 3
5.2/ 3
= F (0.77) − F (0.27) = 0.7794 − 0.6064 = 0.173
P (62.5 < X̄3 < 64) = P (
(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?
59 − 61.7
< Z < Zk ) = P (−0.52 < Z < Zk )
5.2
= F (Zk ) − (1 − F (0.52)) = 0.54
⇒ F (Zk ) = 0.54 + 1 − F (0.52) = 1.54 − 0.6985 = 0.8415
⇒ Zk ≈ 1
P (59 < X < k) = P (
Thus
k = Zk σ + 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 as
deliveries. During the course of a month, over-the-counter sales have a mean of $96,780
with 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 sales
and 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
p
p
V [S] = σ s 2 = 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)?
240, 400 − 326400
222, 900 − 326400
<Z< √
)
P (222, 900 < S < 240, 400) = P ( √
10
5.613 × 10
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.
150 + 0
a+b
=
= 75
2
2
(b − a)2
1502
=
=
= 1875
12
12
µX =
σ2
(b) Find the value that leaves .05 in the lower tail and also the value that leaves .05
in the upper tail.
XL − 0
⇒ XL = 7.5
150
150 − XU
0.05 =
⇒ XU = 142.5
150
0.05 =
4
(c) Suppose that you do not know that the variable is uniform but are given the mean
and 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 distributed
with 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] = σ C 2 = V [C1 ] + V [C2 ] = 12 + 22 = 34
(b) What is the probability that the difference from picking 1 student from each class
is between -1 and 1?
−1 − (−5)
1 − (−5)
√
<Z< √
)
34
34
= P (0.6860 < Z < 1.0290) = F (1.03) − F (0.69) = 0.0936
P (−1 < C < 1) = P (
(c) To give the class the same mean as the second class, the professor adds 5 to all
grades. Explain why this does not leave the two classes equivalent. Which class
would you prefer to be in?
8. A car company says their car gets a mean of 45km per liter with a standard deviation
of 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?
1
47 − 45
) = P (Z ≥ ) = 1 − F (0.333) = 0.3707
6
3
(b) If we have 4 cars and take the average, calculate the P (44 < X̄4 < 46).
P (X ≥ 47) = P (Z ≥
44 − 45
46 − 45
√ <Z<
√ )
6/ 4
6/ 4
= P (−0.3333 < Z < 0.3333) = F (0.33) − (1 − F (0.33))
= 2(0.6293) − 1 = 0.2586
P (44 < X̄4 < 46) = P (
5
9. Suppose you wish to drive across a country that is 2625 km wide and you intend to
rent a series of cars from Rent-A-Wreck. The distance that the first car they give you
is normally distributed with a mean distance of 1500km and a variance of 500km. Each
subsequent 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 on
Consider the first car
1
>
P (X1 > 2625) = P ( X1σ−µ
1
2625−1500
√
)
500
⇒ P (Z > 50.3) ≈ 0
Consider the distance traveled by the first car and then the second car
√ 2 −(µ1 +µ2 ) >
P (X1 + X2 > 2625) = P ( X1 +X
2
2
σ 1 +σ 2
2625−(1500+750)
√
)
500+125
⇒ P (Z > 15) ≈ 0
Now the third car:
P (X1 + X2 + X3 > 2625)
√2 +X3 −(µ1 +µ2 +µ3 ) >
= P ( X1 +X
2
2
2
σ 1 +σ 2 +σ 3
2625−(1500+750+375)
√
)
500+125+31.5
⇒ P (Z > 0) = 0.5
So
n=3
Expected cost of the trip is $300.
Farthest trip possible T (geometric series)
T∞ = X1 + X2 + X3 + ... =
1
× X1
1 − 0.5
E[T∞ ] = 2 × 1500 = 3000
6