Download Exam 2 w Solutions (2011) – Intro to Probability and

Math 447 - November 17, 2011 - Test 2 Solutions
Name:
Read these instructions carefully: The points assigned are not a guide to the difficulty
of the problems. If the question is multiple choice, there is a penalty for wrong answers, so
that your expected score from guessing at random is zero. No partial credit is possible on
multiple-choice and other no-work-required questions.
You must show your work to obtain full credit.
1. (12 points) A soft-drink machine can be regulated so that it discharges an average of µ
ounces per cup. If the ounces of fill are normally distributed with standard deviation 0.3
ounce, give the setting for µ so that 16-ounce cups will overflow only 1% of the time. You
must show your work to obtain full credit.
Let Y be the number of ounces of soda discharged by the machine. We are given that Y is
normally distributed with mean µ and σ = 0.3. We wish to find µ so that P (Y > 16) = 0.01.
Observe that Y > 16 is equivalent to
Y −µ
16 − µ
>
σ
σ
and that Z = (Y − µ)/σ is a standard normal random variable. To have P (Z > z) = 0.01
we must have (from the table) Z ≈ 2.33. So if we select µ so that
16 − µ
= 2.33
σ
we will have the required condition. Solving, it follows that µ ≈ 15.3.
2. (3 points) Let Z be a standard normal random variable, i.e. Z ∼ N (0, 1). Which of the
following numbers is closest to P (Z 2 > 4)?
(a)
(b)
(c)
(d)
0.025
0.475
0.95
0.05
Observe that Z 2 > 4 is equivalent to |Z| > 2. By the “95% rule”, this probability is
approximately 5%. So the answer is (d).
3. (3 points) Let Y be a normal random variable with mean 2 and variance 9. What is the
median value of Y ? (No explanation required.)
The normal random variable is symmetric about its mean, i.e. 50% of the probability
density is below the mean and 50% above. So the mean and median are the same and the
answer is 2.
4. (8 points) Annual household incomes in a city have approximately a gamma distribution
with parameters α = 30 and β = 1000.
a. (4 points) Find the mean and variance of these incomes. (No explanation required.)
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Using the results obtained for the mean and variance of the gamma distribution, we have
that the mean is αβ = $30, 000 and that the variance is αβ 2 = 30, 000, 000 (in units of
dollars-squared).
b. (4 points) Would you expect to find many incomes in excess of $40,000 in this city? Why
(or why not)?
√
Note that the standard deviation of these incomes is 30 · 1000 ≈ $5, 500. So an income
2 standard deviations above the mean is $41,000. Using the 5% rule, and noting that the
gamma distribution is approximately symmetric for these parameter values and this distance
from the mean, we should expect about 2.5% of the incomes to be above $41,000. Above
$40,000, we should expect more, but still a small percentage, not more than 5%. (In fact it
is about 4.3%; you could get an approximate value by noting that $40,000 is more than 1.8
standard deviations from the mean and using the table for the normal distribution.)
If you got this far, you will get full credit. You can argue that less than 5% is not many.
Or you can argue that by a city typically has a large population (say 100,000) and that one
will thus be able to find thousands of incomes at the required level, and that “thousands” is
“many”.
5. A machine used to fill cereal boxes dispenses, on average, µ ounces per box. The manufacturer wants the actual ounces dispensed Y to be within 1 ounce of µ at least 75% of the
time.
a. (4 points) Let σ be the standard deviation of Y , and state Tchebysheff’s theorem for Y .
(Either formulation is acceptable.)
If Y is a random variable with mean µ and standard deviation σ, then
P (|Y − µ| <
1
1
) ≥ 1 − 2.
kσ
k
Note that the sentence beginning “If Y ” is part of the theorem: it is the hypothesis. Not
every random variable has a standard deviation, or even a mean!
b. (4 points) Use your answer from a. to determine the largest value of σ that can be
tolerated if the manufacturer’s objectives are to be met.
1
= 1 It follows that k = 2
We need to choose k so that 1 − k12 = 0.75, and σ so that kσ
and σ = 1/2.
6. (3 points) A random variable Y has cumulative distribution function F given by the
formula:

0
y<0


√

y + 0.2 0 ≤ y < 14
F (y) =
1

0.5 + y
≤ y < 12

4


1
y ≥ 12
Which of the following numbers is closest to P (Y = 14 )?
(a) 0.25
(b) 0.02
(c) 0.2
(d) 0
Observe that
√
1
1
1
P (Y = ) = lim F ( ) − F ( − x) = 0.5 + 0.25 − 0.25 + 0.2 = 0.05
x→0
4
4
4
2
so that the answer is (b).
7. (3 points) Let Y1 and Y2 be random variables whose joint distribution function is given
by
(
6y1 y2 0 ≤ y1 ≤ y2 ≤ 1
f (y1 , y2 ) =
0
otherwise
Are Y1 and Y2 independent? (Just a yes or no answer; no reason required.)
No: note the region of definition.
8. (20 points) A point (Y1 , Y2 ) is chosen at random (according to the uniform distribution)
from the triangle with vertices (−1, 0), (1, 0), and (0, 1). That is, Y1 and Y2 are random
variable whose joint distribution is defined by the previous sentence.
a. (4 points) Write the definition of the conditional density function f (y1 | Y2 = y2 ).
f (y1 , y2 )
,
f2 (y2 )
f (y1 | Y2 = y2 ) =
where
Z
∞
f (y1 , y2 )dy1 = 2 − 2|y2 |
f2 (y2 ) =
−∞
and f (y1 , y2 ) is 1 if (y1 , y2 ) is inside the triangle and 0 outside. It follows that
f (y1 | Y2 = y2 ) =
1
2 − |y2 |
b. (6 points) Determine the conditional density function f (y1 | Y2 = y2 ). Indicate clearly the
values of y2 for which the conditional density function is defined.
Note that P (Y2 ≥ 1) = P (Y2 < 0) = 0, so the conditional density function is not defined
for y2 ≥ 1 or y2 < 0. For 0 ≤ y2 < 1 we have:
f (y1 | Y2 = y2 ) =
1
2 − 2|y2 |
| Y2 = 31 ) in terms of the conditional density function.
Z 2/3
Z 2/3
1
1
3
P (Y1 ≥ | Y2 = ) =
f (y1 | Y2 = 1/3)dy1 =
dy1
2
3
1/2
1/2 4
c. (4 points) Write P (Y1 ≥
1
2
d. (6 points) Find P (Y1 ≥ 12 | Y2 = 13 ).
You can do the integral from the previous part or use geometry to get the answer: 1/8.
9. (8 points) Suppose a random variable Y has a probability density function given by
(
ky 3 e−y/2 y ≥ 0
f (y) =
0
y<0
a. (6 points) Find the value of k that makes f a probability density function. You must give
a numerical answer.
Observe that f is a multiple of the density function for the gamma distribution with
parameters α = 4, β = 2, and so the constant k must be the same as it is in that density
function to make the integral of f (over the whole real line) exactly 1. Thus we have
k=
1
β α Γ(α)
=
1
24 Γ(4)
3
=
24 (4
1
1
==
− 1)!
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b. (2 points) Does Y have a χ2 distribution? (Just a yes or no answer; no explanation
required.)
Yes. The definition of the χ2 -distribution with ν degrees of freedom is Γ(ν/2, 2), so take
ν = 8.
0.4
0.0
f (x)
0.8
In each of the following problems you are asked to identify the probability distribution
giving rise to the graph preceding the question.
−0.5
0.0
0.5
1.0
1.5
x
10. (4 points) From which of the following distributions does the probability density function
graphed above arise? The near-vertical lines in the graph are meant to be vertical, or better
yet nonexistent—I’m still having trouble getting R to display graphs in the desired form.
(a) the beta distribution with parameters α = 5 and β = 10
(b) the uniform distribution on the interval (0, 2)
(c) the gamma distribution with parameters α = 10 and β = 10
(d) the beta distribution with parameters α = 1 and β = 1
(e) the χ2 distribution with 4 degrees of freedom.
Answer: (d)
4
4
3
2
0
1
g (x)
−0.2
0.2
0.6
1.0
x
0.06
0.00
f (x)
11. (4 points) From which of the following distributions does the probability density function
graphed above arise?
(a) the beta distribution with parameters α = 2 and β = 3
(b) the beta distribution with parameters α = 5 and β = 5
(c) the normal distribution with parameters µ = 0.2 and σ = 0.1
(d) the beta distribution with parameters α = 10 and β = 15
(e) the normal distribution with parameters µ = 0.4 and σ = 0.5
Answer: (d)
0
5
10
15
20
x
12. (4 points) From which of the following distributions does the probability density function
graphed above arise?
(a) the gamma distribution with parameters α = 2 and β = 1
5
(b) the beta distribution with parameters α = 0.5 and β = 1
(c) the uniform distribution on [0, 20]
(d) the gamma distribution with parameters α = 10 and β = 10
(e) the chi-square distribution with 8 degrees of freedom
Answer: (e)
13. (10 points) Two gamblers A and B each toss a coin until the first head appears. The
coin they use is biased, so that the probability of heads is p and the probability of tails is
q = 1 − p. Let Y1 be the number of times A tosses the coin and Y2 be the number of times
B tosses the coin. You may assume in what follows that Y1 and Y2 are independent and
you may take as given that the mean and variance of the geometric random variable with
probability of success p are 1/p and q/p2 respectively.
a. (3 points) Find E[Y1 − Y2 ].
E[Y1 − Y2 ] = E[Y1 ] − E[Y2 ] =
1 1
− =0
p p
b. (7 points) Find V [Y1 − Y2 ].
Observe that knowing −Y2 is equivalent to knowing Y2 , so Y1 and −Y2 are independent.
Therefore, the variances add and we have
V [Y1 − Y2 ] = V [Y1 + (−Y2 )] = V [Y1 ] + V [−Y2 ]
We showed V [aY ] = a2 V [Y ] in a homework problem, and using this we have:
V [Y1 ] + (−1)2 V [Y2 ] = V [Y1 ] + V [Y2 ] =
2q
.
p2
Of course this also follows from the theorem on the bilinearity of covariance in Section
5.8.
14. (10 points) Let Z be a normal random variable with mean 0 and variance 1.
a. (2 points) Write an expression for E[Z 4 ] as an integral.
Z ∞
1 −z2
4
z 4 · √ e 2 dz
E[Z ] =
2π
−∞
b. (8 points) Find E[Z 4 ]. Suggestion: find a way to use what you know about the gamma
function.
The answer is 3. There are at least 3 ways you could have arrived at this answer. The
first two are require very specific pieces of knowledge, but once you recall the relvant facts,
the work is very easy. The suggestion uses only things that you are absolutely supposed to
know, but requires more work.
Method 1: By theorem 7.2 in your text, which was discussed in class, Z 2 has the χ2 [1]
distribution, which is the same as the Γ( 12 , 2) distribution, so using the equation
E[Z 4 ] = E[(Z 2 )2 ] = V [Z 2 ] + E[Z 2 ]2
and the known mean and variance of the gamma distribution we get 3.
2
Method 2: Take the known moment generating function (E t /2 ) of Z, take 4 derivatives,
and evaluate at t = 0. Of course this requires that you remember the moment generating
function!
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Method 3: Work directly with the integral from part(a), and change it so that we can use
what we know about the gamma function to evaluate it.
First observe that the function to be integrated is even, so we may write
Z ∞
1 −z2
4
z 4 · √ e 2 dz
E[Z ] = 2
2π
0
Change variables in the integral above, setting y = z 2 , dy = 2zdz, and we get
Z ∞
Z ∞
1 −y dy
1 −y dy
4
2
E[Z ] = 2
y ·√ e2 √ =2
y2 · √ e 2
y
2z
2π
2π
0
0
√
Note that z = y and cancel the 2s to get:
Z ∞
Z ∞
−y
5
5
1
1 −y
4
−1
y 2 · √ e 2 dy = √
y 2 −1 · e 2 dy
E[Z ] =
2π
2π 0
0
This integral can now be computed using what we know about the gamma function and the
density function for the gamma distribution.
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