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practice exam statistics questions
Typology: Exercises
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(revised 6/18/21)
Hi Class!
This test was originally given in my “live” 6644 class during the Fall 2017 semester. The
students had 75 minutes, and were allowed to bring a cheat sheet (both sides). On our
Test 1, I’ll be more generous with time and cheat sheets.
Also, all of our test questions will be multiple choice, in spite of what you see below.
Dave
Solution: X ∼ Geom(1/18), so E[X] = 18.
Solution: Exp(2).
Solution: Yes.
Solution: True.
Solution:
E
0
x
4 x 3 dx =
0
4 x 2 dx = 4 / 3.
This implies that
variable eX^?
Solution: Let Y = e X
. The c.d.f. of Y is
G(y) = P(Y ≤ y) = P(e X ≤ y) = P(X ≤ n(y)) =
(^) n(y)
0
3 x 2 dx = [n(y)] 3 .
This implies that the p.d.f. of Y is
g(y) =
d
dy
G(y) =
3[n(y)] 2
y
, 1 ≤ y ≤ e.
(a) Find E[X].
(b) Find Cov(X, Y ).
Solution: We have
fX (x) =
R
f (x, y) dy =
(^) x
0
8 xy dy = 4 x 3 , 0 < x < 1 ,
so that
E[X] =
R
xfX (x) dx =
0
4 x 4 dx = 4 / 5.
Similarly,
fY (y) =
R
f (x, y) dx =
y
8 xy dx = 4(y − y 3 ), 0 < y < 1 ,
Solution: Florence Henderson. (Sadly, Florence passed away last November. )
bility 0.6. Let X be the number of games until the Hawks achieve their first win. Find the smallest x such that P(X ≤ x) ≥ 0 .9.
Solution: The number of games until the first win is X ∼ Geom(0.6), so that P(X = x) = q x−^1 p = (0.4)x−^1 (0.6). It can also be shown that the c.d.f. is
F (x) = P(X ≤ x) = 1 − q x = 1 − (0.4) x
(though you don’t need to know this if you do a trial-and-error argument to find the smallest x). Now, F (x) = 1 − (0.4)x^ ≥ 0. 9 iff 0. 1 ≥ (0.4)x^ , which is achieved by x = 3.
sample mean of the Xi’s is approximately normal for large enough n?
Solution: CLT.
0.5. (This is a simple random walk.) Find the approximate probability that the sum
i=1 Xi^ will^ be^ at^ least^ 10.
Solution: Note that E[Xi] = 0 and Var(Xi) = E[X 2 i ]^ −^ (E[Xi])
2 = 1. Then the Central Limit Theorem implies that
i=1 Xi^ ≈^ Nor(0,^ 100),^ and^ so
i=
Xi > 10) ≈ P
(a) Using X 0 = 1, calculate the first pseudo-random number U 1.
Solution: We immediately have X 1 = 4, so that U 1 = 0 .5.
(b) Using X 0 = 1, calculate the pseudo-random number U 801.
Solution: If X 0 = 1, then we get X 1 = 4, X 2 = 5, X 3 = 0, and X 4 = 1, so that the thing repeats every 4 tries. Thus, X 801 = X 1 = 4, so that U 1 = 0 .5.
Solution: Exp(1/3).
Solution: Note that − 3 n(U 1 U 2 ) = − 3 n(U 1 ) − 3 n(U 2 ). Thus, Erlang 2 (1/3) (or gamma).
− 2 n(U 1 )cos(2πU 2 )?
Solution: Nor(2,1).
0
[1 + cos(πx)]dx.
The following numbers are a Unif(0,1) sample:
Use the Monte Carlo method from class to approximate the integral via the estimator I¯ 4.
Suppose it takes Joey 3 minutes to prepare each chocolate product. Further suppose that he charges $2/chocolate. Unfortunately, the customers are unruly and each customer causes $0.50 in damage for every minute the customer has to wait in line.
(a) When does the first customer leave?
(b) What is the average number of customers in the system during the first 20 minutes?
(c) How much money will Joey make or lose with the above 4 customers?
Solution: Consider the following table.
cust intrarrl arrl time serv start serve time depart wait sys time 1 8 8 8 18 26 0 18 2 2 10 38 6 44 28 34 3 5 15 29 9 38 14 23 4 2 17 26 3 29 9 12
(a) The first customer leaves at time 26.
(b) Let Xi denote the amount of time Customer i spends in the system during the time interval [0,20]. In particular, X 1 = 20 − 8 = 12, X 2 = 20 − 10 = 10, X 3 = 5, and X 4 = 3. The average number of customers in the system during the first 20 minutes is
total customer time
20
(c) Joey makes 2(6 + 2 + 3 + 1) − 0 .5(0 + 28 + 14 + 9) = −$1.50.
easy Module 5 Arena questions, similar to those you encountered on your lesson assessments and weekly HWs.