STAT 3021 Fall 2011, Midterm II
Sample
1.
(20 points ) A six-sided dice is rolled until the rst time T that a six turns up.
a. (5 pts )
What is the probability distribution for T ?
b. (5 pts )
Find the mean and the variance of T .
c. (5 pts )
Find P (T >
Quiz 1, Stat 3021
N amen, . i . _ m n
September 18, 2014
Read each problem carefully. Be clear and justify your answers for all the questions. The break-
down of the points for each part are provided within the parentheses. For this test you may use
one s
QUIZ 1
STAT 3021
Fall 2015
This paper contains 60 points. Show the details of your calculations for
full credit.
Name:
Student ID Number: Solutions
1
1. Suppose your commute to the university involves taking the light rail
to come to the Minneapolis campu
I30 {baptist 4: Random Variables and Expectation
§m§alems
Five men and 5 women are ranked according to their scores on an examination.
Assume that no two scores are alike and all 10! possible rankings are equally likely.
Let X denote the highest ranking
Quiz 3, Stat 3021
Name; _, r V. r- _ ._ W, T,
October 16, 2014
Read each problem carefully. Be clear and justify your answers for all the questions. The break
down of the points for each part are provided within the parentheses. For this test you may use
Quiz '6, Stat 3021
/
April 24, 2015
Read each problem carefully. Be clear and justify your answers for all the questions. The break~
down of the points for each part are provided Within theparentheses. For this test you may use
one sheet(both sides) of
Quiz 27 Stat 3021
Name: in m
October 2, 2014
Read each problem carefully. Be clear and justify your answers for all the questions. The break
down of the points for each part are provided within the parentheses. For this test you may use
one sheet(both sid
Quiz 2; Stat 3021
Name: i , ' g _
February 20, 2015
Read each problem carefully. Be clear and justify your answers for all the questions. The break-
down of the points for each part are provided within the parentheses. For this test you may use
one sheet(
Final Exam, Stat 3021
Name:
May 12, 2015
.
Read each problem carefully. Be clear and justify your answers for all the questions. The breakdown of the points for each part are provided within the parentheses. For this test you may use
two sheets(both sides
Quiz 1, Stat 3021
Name:
February 6, 2015
Read each problem carefully. Be clear and justify your answers for all the questions. The break
down of the points for each part are provided Within the parentheses. For this test you may use
one sheet(both sides
Quiz 4, Stat 3021
Name:
March 27, 2015
.
Read each problem carefully. Be clear and justify your answers for all the questions. The breakdown of the points for each part are provided within the parentheses. For this test you may use
one sheet(both sides) o
Quiz 3, Stat 3021
Name:
March 6, 2015
Read each problem carefully. Be clear and justify your answers for all the questions. The break
down of the points for each part are provided within the parentheses. For this test you may use
one sheet(both sides) o
Problem set 2
September 15, 2014
Problem numbers refer to the 9th edition of the text book.
1. Problem 2.90
2. Problem 2.100
3. Problem 2.120
4. Problem 3.5 (Note: the probability distribution of a discrete r.v. is the probability mass function
of the r.v
/
STAT 302101 ,1! Test One Spring 2012
a: + 1 .
1. (25 points) Suppose X is a random variable with p.m.f. f(ac) = 10g) < a: > 1f :5 : 123 ' 9
0, otherwise.
For your convenience, here are the values of f.
x 1 2 3 4 5 . 6 7 8 9
f(ac) 0.301 0.176 0.125 0.0
Hypothesis test for population mean:
Assumptions
Hypotheses
Random sample; the population is approximately normal or the sample size is large.
H0: = 0
H0: = 0
H0: = 0
Ha : < 0
Ha: > 0
Ha: 0
x 0
s/n
t =
Test statistic
P ( T n1 <t )
p-value
P ( T n1 >t )
If
Conditional Probability
Probability of event B, given event A = P (B|A).
P (B|A) =
P (A B)
P (A)
o the number of successes in n independent trialsXis
n
px q nx ,
x = 0, 1, 2, . . . , n
b(x; n, p) =
x
When both k (# of successes) and N k (# of failures) ar
Properties of Confidence inteavais 275
in general, the upper and lower confidence limits resultfrom replacing each < in
(7.6) by = and solving for 6. In the insulating fluid example just considered,
212x,- = 34.170 gives A = 34.170/(22xi) as the upper con
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80 {hatter 3: Elements of Probability
,P(A,AJ-) = 1 P(Af UAJ.)
= 1 [P(Af) + P(A]-) P(AfA;)]
= 1 - (1 p,-)k ~- (1 pj)/e + P{no coupon is type 1' or typej}
=1<1z>i>e (1Pj)k +(1~p,-p,-)
where the final equality follows because each of the k coupons is, indep