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Math 411 Solutions to Final Exam December 7, 2001
- (15) Define the following terms:
(a) The distribution function of a random variable.
Let X be a random variable. Then the distribution function F (x) of X equals F (x) = P {X ≤ x} (b) The covariance of two random variables.
The covariance of the random variables X and Y is cov (X, Y ) = E [(X − E [X]) (Y − E [Y ])]
(c) The conditional expectation of X given Y , E [X | Y ]. E [X | Y ] =
x
xpX|Y (x|y) ,
where pX|Y (x|y) =
p (x, y) pY (y). This formula assumes that^ X^ and^ Y^ are discrete random variables.
- (5) An urn contains 15 red balls and 5 blue balls. If two balls are picked from the urn without replacement, what is the probability that they are both blue? ( 5 2
19 ≈^0.^0526
- (15) Stores A, B, and C have 50, 75, and 100 employees and , respectively, 50, 60, 70 percent of these are women. Resignations are equally likely among all employees, regardless of sex. One employee resigns, and this is a woman. What is the probability that she works in store C? The easiest way to work this problem is to say that there are a total of 140 women, 70 of whom work in store C. Thus, the probility that a women comes from store C is 70 140
=^1
The harder way is to use Bayes’ ideas
P {C|w} = P^ {C, w} P {w}
= P^ {w|C}^ P^ {C} P {w|A} P {A} + P {w|B} P {B} + P {w|C} P {C}
= (7/10) (4/9) (1/2) (2/9) + (3/5) (3/9) + (7/10) (4/9)
=
- (15) The table below gives the values of the joint probability mass function of two discrete random variables X and Y. Note that the values of X are arranged vertically and those of Y are horizontal. That is, p (3, −1) = 0, where X = 3 and Y = −1.
X \ Y − 1 0 1 2 1 1/10 1/10 1/20 1/ 2 1/50 1/10 0 1/ 3 0 1/50 1/5 1/ 4 1/20 1/10 1/25 0
(a) p (2 | Y = 1) =? p (2 | Y = 1) = p^ (2,^ 1) pY (1)
(b) E [X | Y = 1] =? E [X | Y = 1] = 1 pX|Y (1|1) + 2pX|Y (2|1) + 3pX|Y (3|1) + 4pX|Y (4|1)
=
29 / 100 + 2 (0) + 3^
29 / 100 + 4^
- (15) In a game a fair die is rolled, and the amount of money won or lost is determined by the number rolled. The amounts are given in the table below Number 1 2 3 4 5 6 Winnings? 3 1 − 8 1 5
Let X be the random variable, which denotes the amount of money won or lost.
(a) If a one is rolled, what should the winnings be so that the expected value of X is −1. (House advantage) E [X] =
6 (c^ + 3 + 1^ −^ 8 + 1 + 5) =^ −^1
implies that c = −8, where c represents the winings if a one is rolled. (b) Using the value you found in part a determine the variance of X. Var (X) = E
[
X^2
]
− (E [X])^2
6 (64 + 9 + 1 + 64 + 1 + 25)^ −^ (−1)
2
- (20) One of the more remarkable things about normal random variables is that sums of independent ones are also normal. As a first step to this we saw that if X is a normally distributed random variable with mean μ and variance σ^2 , then the random variable Y = aX + b is also normal with mean aμ + b and variance a^2 σ^2.
(a) What is the density function of a normally distributed random variable with mean 5 and variance 36? 1 6
2 π
e
− (x − 5)^2 72
(b) Verify that Y = aX + b is also normal distributed with mean aμ + b and variance a^2 σ^2 , where μ is the mean of X and σ^2 is the variance of X. The problem is not to show that the mean are variance are what is expected, but to show that Y is normally distributed. To that end let G (y) denote the distribution function of Y , then we have
G (y) = P {Y ≤ y} = P {aX + b ≤ y}
= P
X ≤ y^ −a^ b
assume that a > 0
σ
2 π
∫ y^ −^ b a − ∞
e
− (x^ −^ μ)
2 2 σ^2 dx
To get the density function, g (y) of Y , we differentiate the distribution function of Y
g (y) =
aσ
2 π
e
− ((y^ −^ b)^ /a^ −^ μ)
2 2 σ^2
aσ
2 π
e
− (y^ −^ (aμ^ +^ b))
2 2 a^2 σ^2
Thus, we see that Y is normally distributed with mean aμ + b and variance a^2 σ^2.
- (15) Suppose that a coin is flipped 3200 times and the probability that it comes up heads is 1/4. Find, by using a normally distributed random variable, an approximation to the probability that there are between 740 and 750 heads. If X is the random variable which counts the number of heads flipped then the expected value of X and its variance equal
E [X] = 3200
var (X) = 3200
Let N denote a random variable which is normally distribute with parameters 800 and 600. Then
P { 740 ≤ X ≤ 750 } ≈ P { 739. 5 ≤ N ≤ 750. 5 }
= P
≤ N^ −^800
≤ 750.^5 −^800
≈ P {− 2. 469 ≤ Z ≤ − 2. 0208 }
= P { 2. 02 ≤ Z ≤ 2. 47 }
- (15) A model proposed for college football supposes that when two teams with roughly the same record play each other, then the number of points scored in a half by the home team minus the number scored by the visiting team is approximately a normal random variable with mean 5 and variance
- In addition the model supposes that the point differentials for the two halves are independent. Assuming this model
(a) What is the probability that the home team wins? Let H 1 denote the score differential in the first half and H 2 the score differential in the second half. Then H 1 + H 2 is a normally distributed random variable with mean 10 and variance 16. The probability that the home team wins is then approximated by
P
H 1 + H 2 >
= P
{ H
1 +^ H 2 −^10
4 >^ −^
= P {Z > − 2. 375 } = P {Z < 2. 375 }
(b) Given that the home team is behind by 3 points at the half, what is the conditional probability that the home team wins?
P
H 1 + H 2 > 1
| − 3. 5 < H 1 < − 2. 5
P
H 1 + H 2 > 1
, − 3. 5 < H 1 < − 2. 5
P {− 3. 5 < H 1 < − 2. 5 }
The computation of these probabilities was done using Maple.
