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nFinal ExamSTATS 120A/281A
Sevan Koko Gulesserian
Directions: Please read.
Please write your name and student ID number on your submission.
If you do not submit your exam properly on Gradescope (mark where each question is) and on time,
you risk a few points off your total.
Read each question and parts carefully, some parts are asking two questions.
Exam is closed book. You can use a sheet of notes (8.5x11, front and back). Use a calculator but
no R or calculator applets in a browser. There is a page of useful formulas at the end of the
exam.
Each part is worth 5 points unless otherwise noted for a total 100 points.
Please show all work to get full credit (work shown is much more important than the final numeric
result). Numeric only answers will get little to no credit.
Can leave answers in unsimplified form (do not need to calculate a numeric answer, can leave answer
in combinatoric forms for example or summation form or leave integrals not evaluated, but solve
the integral).
When it says explain your work, just a quick sentence so we can see your reasoning.
If you cannot work out a problem, or are stuck, list what you would do had you not got stuck. And
if a part of a problem uses a previous part’s answer you did not finish or work out numerically, just
call it "answer to previous part" and use it in the next question part(s).
1. A technology company has two independent servers. Denote X as the downtime in hours of server
1 and denote Y as the downtime in hours of server 2. X follows a continuous exponential distribution
with parameter β > 0 and Y follows a continous exponential distribution with parameter α > 0,
where β and α are both finite.
Assume X and Y are independent of one another (thus f(x, y) = fx(x)fy(y)).
a. What is P (X = 1 and Y = 1) and P (X = Y )?
b. What is the probability that server 1 will be down longer than server 2 is down, P (X > Y )?
Show work.
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c. What is E[X(X + Y )] and var(X + Y )? Show work and reasoning.
d. Show that the moment generating function for server 1’s downtime (X) is ββ−t for β > t and
use the MGF to get the the expectation of X, E(X). Show work.
e. Derive the moment generating function for the combined down time across server 1 and server
2, X + Y . Show work.
f. Say we have a sequence Xn that follows an exponential distribution with parameter n.
Show that Xn converges in probability to 0 as n goes to infinity, that is to say show that
lim
n→∞P (|Xn − 0| > ) = 0 for some > 0 (say = 0.01).
Hint: Note that P (|Xn−0| > ) is the same as P (Xn > ) since Xn’s support being strictly positive.
2. Let X be a continuous random variable with the following probability density function f(x):
f(x) =
 34(1− x2) = 0.75− 0.75x20 −1 ≤ x ≤ 1otherwise
a. What is P (X < 0|X < 0.5)? The probability that X < 0 given X < 0.5.
b. Show that E( 1
X2
), the expectation of 1
X2
, does not exist (it is infinite).
c. Now say we have two (2) independent random variables, X1 and X2, that have density
functions f(x) defined earlier. What is the probability that only one of them (exactly one) is less
than 0 (and so the other is greater than 0) ?
3. Let X denote the amount of gasoline, in millions of gallons, a gas station buys from a supplier in
a year. Let Y be the amount of gas, in millions of gallons, sold by the gas station in a year.
Thus for example 0.2 means 200,000 gallons.
Both X and Y are continuous, and the joint probability density function is as follows:
f(x, y) =
 3x0 0 ≤ y ≤ x ≤ 1otherwise
a. Find fx(x), the marginal density of X. Show work.
b. Find fy(y), the marginal density of Y. Show work.
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c. Derive f(y|x), the conditional density of Y given X and use it to find the probability that
the gas station sells more than 500,000 gallons given it buys 800,000 gallons from the supplier,
P (Y > 0.5|X = 0.8).
d. What is E(Y ) and E(Y 2)? The expectation of Y and Y 2. Show work.
e. What is P(X + Y ≤ 1)? Show work.
f. Now say we observe this gas station for n=30 years, and record it’s yearly sale of gasoline (in
millions of gallons), Y1, Y2,...,Y30. Assume the yearly sales are independent.
Use the central limit theorem to approximate the probability that the average yearly sale of gasoline
is more than 0.465 (465,000 gallons). That is to say the probability that Y¯ = 130
30∑
i=1
Yi > 0.465.
4. A bowl contains 10 red balls, 6 white balls, and 4 blue balls. The red balls are all marked "10",
the white balls are all marked "5", and all the blue balls are all marked "0".
Two (2) balls are going to be randomly selected with replacement (the picks are independent).
That is to say, you will select a ball, record the color and number marking, then put that ball back
into the bowl and then select a second ball. Thus the picks are independent as well.
Let X be the number of red balls picked (marked "10") and let Y be the number of white balls
picked (marked "5").
a. Tabulate the joint probability mass function table (pmf values of P (X = x, Y = y)) for X and
Y. Note X = {0, 1, 2} , Y = {0, 1, 2}, and 0 ≤ X + Y ≤ 2.
b. The number markings indicate the amount of points you will receive when picking that ball.
For example a red ball gives you 10 points. So if your two picks are a red ball and a white ball, the
total you have is 15. If you pick two blue balls, your total is 0.
We are interested in T, the sum of the numbers that are marked on the two balls picked. Note that
T = 10X + 5Y + 0(2− Y −X). Show E(T)=13 and var(T)=30.5.
c. Can do this part without parts a. and b. Now suppose instead of picking just two balls,
we pick n=200 balls with replacement from the bowl ( can think of this as picking 2 balls 100 times
).
Let Xi denote the amount of points of the i-th picked ball, for i=1,2,...,200. And since it is with
replacement, the Xi’s are independent.
Use the Central Limit Theorem (CLT) to approximate the probability that the sum of the points
across the 200 picks is greater than 1410. That is P (
200∑
i=1
Xi > 1410).
Hint: From part b. we have that E(Xi) = 6.5 and var(Xi) = 15.25.
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5. The number of emails you receive in a day follows a Poisson distribution with parameter/rate
λ > 0. Emails are either junk or non-junk, and emails are independent of one another.
The probability that a single given email is junk is p, and with probability 1 − p it is non-junk
(where 0 < p < 1).
Let Y denote the number of emails you receive in a day, and let X denote the number of junk mail.
Thus Y follows a Poisson distribution with parameter λ and X given Y (X|Y) follows a binomial
distribution with parameters Y (n=Y) and p.
a. Given you receive Y = 100 emails today, what is the expect d number of junk mail you receive
in a day, E(X|Y)? Show reasoning.
b. Now say you do not know how many emails you will receive, what is the expected number of
junk mail you will receive in a day if you do not know how many emails in total you will receive
that day, E(X)? Show reasoning.
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Useful formulas.
Exponential function:
∞∑
z=0
wz
z! = e
w.
lim
n→∞
(
1 + an
)bn
= eab.
Geometric series:
Let |r| < 1, then
∞∑
k=0
rk = 11−r .
Also:
∞∑
k=1
akrk−1 = a(1− r)−2 = a
(1−r)2 .
Markov’s and Chebyshev’s inequality:
Markov: If X > 0, then
P (X ≥ t) ≤ E(X)t
Chebyshev: Let E(X) = µ and k > 0, then
P (|X − µ| > k) ≤ var(X)
k2
Variance Formula
var(X) = E(X2)− [E(X)]2.
General formula. Let a and b be constants and let X and Y be random variables:
var(aX + bY ) = a2var(X) + b2var(Y ) + abcov(X,Y ).
If X and Y are independent, then:
var(aX + bY ) = a2var(X) + b2var(Y ).
Covariance Formula
cov(X,Y)=E(XY)-E(X)E(Y).
Bernoulli Distribution:
f(x) = P (X = x) = px(1− p)1−x with SX = {0, 1} and parameter 0 < p < 1.
E(X) = p and var(X) = p(1− p)
Binomial Distribution:
f(x) = P (X = x) =
(
n
x
)
px(1− p)n−x with SX = {0, 1, 2, ..., n} where 0 < p < 1 and for positive
integer n (n=1,2,3,...).
E(X) = np and var(X) = np(1− p).
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Geometric Distribution:
f(x) = P (X = x) = (1− p)x−1p with SX = {1, 2, ...} where 0 < p < 1.
F(x)=P (X ≤ x) = 1− (1− p)x.
E(X) = 1p and var(X) =
1−p
p2
.
Poisson Distribution:
f(x) = P (X = x) = e
−λλx
x! with SX = {0, 1, 2, ..., } where 0 < λ.
E(X) = λ and var(λ).
Uniform Distribution:
f(x) = 1b−a with SX = (a, b) where a < b.
E(X) = 12(a+b) and var(X) =
1
12(b−a)2 .
Exponential Distribution:
f(x) = βe−βx with SX = [0,∞) where 0 < β.
F(x)=P (X ≤ x) = 1− e−βx.
E(X) = 1β and var(X) =
1
β2
.
Normal Distribution:
f(x) = 1√
2piσ2
e−
(x−µ)2
2σ2 with SX = (−∞,∞) where −∞ < µ <∞ and 0 < σ2 <∞.
E(X) = µ and var(X) = σ2.
Gamma Distribution:
f(x) = β
α
Γ(α)x
α−1e−βx with SX = (0,∞) where 0 < α <∞ and 0 < β <∞.
E(X) = αβ and var(X) =
α
β2
.
Beta Distribution:
f(x) = x
α−1(1−x)β−1
B(α,β) with SX = (−∞,∞) where 0 < α <∞ and 0 < β <∞.
E(X) = αα+β and var(X) =
αβ
(α+β)2(α+β+1)
.
Normal Distribution Probabilities:
If Z follows a Normal distribution with parameter µ = 0 and σ2 = 1 then we approximately have:
P (−1 < Z < 1) = 0.68
P (−2 < Z < 2) = 0.95
P (−3 < Z < 3) = 0.997
P (Z < 0) = 0.5
P (Z < 1) = 0.84
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P (Z < 2) = 0.975
P (Z < 3) = 0.99.
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