Download MATH 3160, SPRING 2013 HOMEWORK #5

MATH 3160, SPRING 2013
HOMEWORK #5—SOLUTIONS
JOHANNA FRANKLIN
This assignment will be due on Wednesday, March 6 at the beginning of class. Remember to
show your reasoning and name the classmates you worked with. Answers without work shown will
receive almost minimal credit.
(1) Suppose that two fair six-sided dice are rolled. Write down the probability mass function
and draw a graph of the cumulative distribution function for each of the random variables
described below.
(a) X, which represents the smaller of the two values that are rolled
Solution. We start by making a table that gives the value of X for each possible roll:
X 1 2 3 4 5 6
1 1 1 1 1 1 1
2 1 2 2 2 2 2
3 1 2 3 3 3 3
4 1 2 3 4 4 4
5 1 2 3 4 5 5
6 1 2 3 4 5 6
9
7
5
We can see from this table that pX (1) = 11
36 , pX (2) = 36 , pX (3) = 36 , pX (4) = 36 ,
1
3
, and pX (6) = 36
. A graph of the cumulative distribution function appears
pX (5) = 36
below:
(b) Y , which represents the absolute value of the difference of the values of the two rolls
Solution. We start by making a table that gives the value of Y for each possible roll:
1
2
FRANKLIN
Y 1 2 3 4 5 6
1 0 1 2 3 4 5
2 1 0 1 2 3 4
3 2 1 0 1 2 3
4 3 2 1 0 1 2
5 4 3 2 1 0 1
6 5 4 3 2 1 0
6
8
6
We can see from this table that pY (0) = 36
, pY (1) = 10
36 , pY (2) = 36 , pY (3) = 36 ,
4
2
pY (4) = 36 , and pY (5) = 36 . A graph of the cumulative distribution function appears
below:
(2) A card is drawn at random from a standard deck of playing cards. If it is a heart, you win
$1. If it is a diamond, you lose $2. If it is a black card, you win $3. What is the expected
value of your winnings?
Solution. The probability of drawing a heart and the probability of drawing a diamond
are both 14 , and the probability of drawing a black card is 21 . Therefore, if X is the random
variable that represents your winnings, your expected winnings (in dollars) is
1
1
1
5
E[X] = 1 · + (−2) · + 3 · = .
4
4
2
4
(3) A royal family has children until it has a boy or until it has three children, whichever comes
first. Find the expected number of boys and the expected number of girls in this family.
You may assume that the probability of a girl (and thus the probability of a boy) is 12 each
time, that the sexes of their children are independent, and that they will only have one
child at a time (so, for instance, they will never have twins).
Solution. The possible sequences of children this family might have are B (with probability
1
1
1
2 ), GB (with probability 4 ), and GGB and GGG (with probability 8 each). We let X be
a random variable representing the number of girls the family has. We can see that the
probability mass function of X is given by pX (0) = 12 , pX (1) = 14 , pX (2) = 18 , and pX (3) = 18 .
Therefore,
1
1
1
1
7
E[X] = 0 · + 1 · + 2 · + 3 · = .
2
4
8
8
8
MATH 3160, SPRING 2013
HOMEWORK #5—SOLUTIONS
3
Now we let Y be a random variable representing the number of boys the family has. The
probability mass function of Y is given by pY (0) = 18 and pY (1) = 78 , so
1
7
7
E[Y ] = 0 · + 1 · = .
8
8
8
(4) A random variable X has the following probability mass function: p(0) = 31 , p(1) = 16 ,
p(2) = 14 , p(3) = 41 . Find its expected value, variance, and standard deviation.
Solution. This is a simple matter of applying formulas:
1
1
1
34
1
E[X] = 0 · + 1 · + 2 · + 3 · = .
3
6
4
4
24
Now we calculate the variance:
V ar(X) = E[X 2 ] − (E[X])2
2
1
34
2 1
2 1
2 1
0 · +1 · +2 · +3 · −
3
6
4
4
24
2
82 34
−
24 242
812
242
root of the variance to get the standard deviation: SD(X) =
2
=
=
=
Finally,
we take the square
√
2 203
24 .
(5) Suppose X is a random variable such that E[X] = 50 and V ar(X) = 12. Calculate the
following quantities.
(a) E[X 2 ]
Solution.
E[X 2 ] = V ar(X) + (E[X])2 = 12 + 502 = 2512
(b) E[3X + 2]
Solution.
E[3X + 2] = 3E[X] + 2 = 152
(c) V ar(−X)
Solution.
V ar(−X) = (−1)2 V ar(X) = 12
(d) SD(2X)
Solution.
SD(2X) =
p
p
√
√
V ar(2X) = 22 V ar(X) = 48 = 2 12
(6) Is it possible for there to be a random variable X such that E[X] = 4 and E[X 2 ] = 11?
Why or why not?
Solution. This is not possible. If such a random variable existed, its variance would be
E[X 2 ] − (E[X])2 = 11 − 42 = −5, and a variance must be nonnegative.
4
FRANKLIN
Suggested problems: Chapter 4 Problems: 1-2, 5-8, 12-13, 17, 19, 23, 25; Chapter
4 Theoretical Exercises: 3, 7, 8