Homework 7 Solutions
Chapter 16.34
Let X represent the amount of money someone will pay for their dog, and Y
the amount for a cat. We're told that E(X) = 100 and E(Y) = 120.
a) The question asks to compute E(X-Y) = E(X) - E(Y) = 100-120= -$20
b) Var(X-Y) = Var(X) + Var(Y) because X and Y can reasonably be assumed to
be independent. (There's no reason that the cost of treating a randomly
selected dog should affect the cost of treating a randomly selected cat.
Var(X-Y) = 302 + 352 = 2125
SD(X-Y) = Sqrt(2125) = $46.10
We would expect the cat to cost about $20 more, give-or-take about $46.
c) If the dog costs more than the cat, then X > Y or, using a bit
of algebra, X-Y > 0.
So we want P(X - Y > 0). Let's define W = X-Y. What do we
knwo about W?
First, the probability distribution of W can be described fairly well by
the Normal model. Second, E(W) = -20 and SD(W) = 46.10. So to
find P(W >0) we need to first convert this standard normal variables.
If Z is a standard normal variable, then
P(W > 0) = P(Z > (0 - -20)/46.10) = P(Z > .433)
(in other words, 0 is .433 standard deviations above the mean.)
Using the normal table, P(Z > .433) = 1 - P(Z < .433) = 1 - .667 =
.332
*New Problem: (a) Flip a fair coin until the first head appears. Let
X represent the number of coinflips it takes before the first head appears.
Write a probability model for X.
X has values 1, 2, 3, 4, .... up to infinity. Of course, X is more
likely to be a small number than a really big one.
Let's figure out the probabilities: P(X = 1) = 1/2 because the
only way this can happen is if heads is on the first flip.
P(X =2) = (1/2) * (1/2) = 1/4, because the only this can happen
is if the first flip is tails AND the second is heads. Since flips
of a coin are independent, P(first is tailsAND second is heads= P(first is
heads) * P(second is tails) = 1/2 * 1/2.
P(X = 3) = P(first is tails and second is tails and third is heads) = (1/2)*(1/2)*(1/2)
= (1/8)
See a pattern? P(X = 4) = (1/2)4
P(X = 5) = (1/2)5
and so on...
We could write this as a formula: P(X = k) = (1/2)k
where k = 1,2,3,4,.....
But for full credit, its enough to say what the pattern is and mention that
it continues on this way. (So, for example, say that it's .5 raised
to whatever power x is equal to.)
(b) Roll a fair, six-sided die until the first 6-spot appears.
Let Y represent the number of rolls it takes. Write a probability
model for Y.
Very similar problem. But now P(Y = 1) = (1/6). P(Y = 2) = P(first
is not a 6 AND second is a six) = (5/6) * (1/6)
P(Y = 3) = (5/6)2 *(1/6)
P(Y=4) = (5/6)3 * (1/6)
The pattern is now a little different. Now we raise (5/6) to the power
that's one less than the value of x, and multiply by (1/6). In terms
of a formula: P(X = k) = (5/6)(k-1)*(1/6)