Stat 426 Winter 2015
Homework 2 Solutions
Takeaways from this homework.
Q1a: How do we find of mode for a discrete distribution?
Q2a: Not only the range of values of the random variable, but also the range of possible
values of its parameters affects prob
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STAT 426
Lecture 34
Fall 2012
Arash A. Amini
September 13, 2012
1 / 35
Announcements
My oce hours:
Tue 4 5p in 470 West Hall,
Wed 12 1p in 438 West Hall
Yingchuans oce hours:
Wed 2:30 3:30p in 274 West Hall
Fri 9:30 10:30a in 274 West Hall
Final exam: Wed
Stat 426 : Homework 4.
Moulinath Banerjee
November 8, 2015
Announcement: The homework carries a total of 60 points (10 6).
(1) A level 1 C.I. for the normal mean , when is assumed known, is given by:
X n q(1 2 , N (0, 1) , X n q(1 , N (0, 1) ,
n
n
where
Stat 426 Exam 1 Solution F15
Problem 1 (Show all reasoning)
15 points
Consider a simplified model for weather forecasting. Let = cfw_ and let =
( ). Further, suppose that the probability that the weather (either rain or no rain) tomorrow
will be the same
Homework 6 Stat 426 Fall 2015 Solutions
Problem 1 (Question 3, Chapter 6). Let be the average of a sample of 16 independent normal
random variables with mean 0 and variance 1. Determine c such that P(| < c) = .5. Use: qnorm(p)
to compute quantiles. H
Homework 5 Stat 426 Solutions
Review Lecture 6. Practice problems: Q 1, 7, 8, 9, 13, 15 Chap 5
1. Let " , $ , be a sequence of independent random variables with ( ) = (
"
"
and ( ) = ($ . Show that if as , 4 ( and 6 4 ($ 0,
(5
(5
"
4
Homework 4
Review lectures 4 & 5, Sec 2.3, 3.7, 4.5
Review question 67, Chapter 2; question 69, Chapter 3; 79, 81, 82, 83, 84 Chap 4
1. Let f (x) = x1, x1, > 0, and0otherwise
.
a. Find F (x)
x
x
F (x) = y1dy = y|1 = 1 x
1
b. How can we generate random va
Stat 426 W2015
Homework 1 Solution
Problem 1
A lie detector test has a 90% detection rate if a person is lying, and a 5% detection rate if a
person is not lying. Suppose that 1% of all individuals in a cert
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Solutions for HW10
Important note: Unless otherwise stated, x = (x1 , . . . , xn ) and X = (X1 , . . . , Xn ).
8.16(d) The joint density is
f (x| ) =
where T (x) =
i
n
1
1
exp
2n n
n
i=1
|xi |. Hence T =
|xi | =
i=1
1
1
exp T (x)
2n n
|Xi | is sucient by
Solutions for HW5
5.10 Let Xi = 1cfw_six turns up in the ith trial and let S =
1
1
100 6 16.67 and var(S ) = 100 1 (1 6 ) 13.89.
6
100
i=1
Xi . We have E(S ) =
S 16.67
P(15 < S < 20) P(0.45 <
< 0.89) (0.89) (0.45) 0.49.
13.89
Let Yi be the value rolled i