Download Question 1 Question 2 Question 3

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
C our ses
Jua n Luis Her r er a C or t ijo 
Mathematical Biostatistics Boot Camp
by Brian Caffo
Feedback — Week 2 Quiz
You submitted this quiz on Sun 28 Apr 2013 9:17 AM CEST (UTC +0200). You got a score of 10.00 out of 10.00.
GENERAL
Home
Question 1
Syllabus
Let
Grading Policy
Faculty
g(x) = πf1 (x) + (1 − π)f2 (x) where f1 and f2 are densities with associated means μ1 and μ2 , respectively. Here 0 ≤ π ≤ 1 . Note that g is a valid
density. What is it's associated mean?
Your Answer
1
π1
Github Repository 
Explanation
μ2
μ2
LEARNING ACTIVITIES
μ1 + μ2
Video Lectures
μ1
Homework
Quizzes
1
1−π1
μ1 +
Score
πμ1 + (1 − π)μ2
1.00

Total
1.00 / 1.00
COMMUNITY
Discussion Forums
Course Wiki 
Join a Meetup 
Question Explanation
∞
∫−∞ x{πf1 (x) + (1 − π)f2 (x)}dx
∞
∞
= π ∫−∞ xf1 (x)dx + (1 − π) ∫−∞ xf2 (x)dx
= πμ1 + (1 − π)μ2
 Help Articles
Question 2
Suppose that a density is of the form (k + 1)x k for some constant k
Your Answer
> 1 and 0 ≤ x ≤ 1 . What is the mean associated with this density?
Score
Explanation
k+1
k
k+1
k+2
k+1
k+2
k+1
k+2

Total
1.00
1.00 / 1.00
Question Explanation
1
∫0 (k + 1)xk+1 dx =
k+1
Note that this is a special case of the so-called Beta distribution. (Look it up on Wikipedia!)
k+2
Question 3
You are playing a game with a friend where you flip a coin and if it comes up heads you give her X dollars and if it comes up tails she gives you Y
dollars. The probability that the coin is heads in p (some number between
0 and 1.) What has to be true about X and Y to make so that both of your
expected total earnings is 0. (The game would then be called "fair".)
Your Answer
p
p−1
=
X
Y
p
p−1
=
Y
X
p
1−p
=
X
Y
Score
Explanation
X=Y
p
1−p
=
Y
X
Total

1.00
1.00 / 1.00
Question Explanation
p
=
Y
p
= 0 Then it must be the case that 1−p = YX Or that the ratio of the payouts has to equal the odds. So
= 2. The game is 2 to 1 against you, p = 2/3 ; she is twice as likely to win as you. Then she will have to pay out twice
Your expected earnings is −pX + (1 − p)Y
p
consider, for example, if 1−p
as much if you win to make the game fair.
Question 4
You are playing a game with a friend where you flip a coin and if it comes up heads you give her 1 dollar and if it comes up tails she gives you one
dollar. If you play the game 10 times what would be the variance of your earnings?
Your Answer
Score
Explanation
√2‾
2
1
‾‾
‾
√10
10

Total
1.00
1.00 / 1.00
Question Explanation
10
Let X1 , … , X10 be random variables that take the value −1 with probability .5 and 1 with probability. Let X = ∑ i=1 Xi be the total earnings. Consider
the fair coin. Note that E[X] = 0 . Also notice that Var(Xi ) = .5(−1)2 + .5(−1) = 1 . Then the variance of X is the sum of the variances, 10.
Question 5
When at the free-throw line for two shots, a basketball player makes at least one free throw 90% of the time. 80% of the time, the player makes the first
shot, while 70% of the time she makes both shots. Which number is closest to the conditional probability that the player makes the second shot given
that she missed the first?
Your Answer
Score
Explanation
90%
80%
10%
30%
40%
60%
50%

1.00
20%
70%
Total
1.00 / 1.00
Question Explanation
A be the event that the player makes the first shot and B be the event that she makes the second. Then, P(A ∪ B) = .9 , P(A) = .8 ,
P(B)−P(A∩B)
.8−.7
P(A ∩ B) = .70 . Then P(B) = P(A ∪ B) − P(A) + P(A ∩ B) = .8 . We want P(B|Ac) = P(B ∩ Ac )/P(Ac) =
. Thus, it is
= 50%
.2
P(Ac )
Let
Question 6
Let X1 , … , Xn1 be random variables independent of Y1 , … , Yn2 , where both groups are iid with associated population means μ1 and μ2 and population
variances σ12 and σ22 , respectively. Let X̄ and Ȳ be their sample means. What is the variance of X̄ −
Your Answer
σ12
n1
+
Score
σ22
n2

1.00
σ12 + σ22
σ12
n1
−
σ22
n2
σ12 − σ22
Total
1.00 / 1.00
Question Explanation
Var(X̄ − Ȳ) = Var(X̄) + Var(Ȳ) =
σ12
n1
+
σ22
n2
.
Ȳ?
Explanation
Question 7
Quality control experts estimate that the time (in years) until a specific electronic part from an assembly line fails follows (a specific instance of) the
Pareto density, f (x)
=
3
for
x4
1 < x < ∞ . Which option is closest to the mean failure time?
Your Answer
Score
Explanation
1 year
2 years
3 years
3.5 years
1.5 years

1.00
2.5 years
Total
1.00 / 1.00
Question Explanation
∞
E[X] = ∫1 3x−4+1 dx =
3
2
∞
x−2 ∣∣1 =
3
2
.
Question 8
Let
f be a continuous density having a finite mean and μ be any number. Suppose that f (x) = f (−x) (i.e. f is symmetric about 0). Convince yourself
that f (x − μ) is a valid density. What is its associated mean?
Your Answer
Score
μ

Explanation
1.00
1
0
It can't be ascertained.
Total
1.00 / 1.00
Question Explanation
Note that the density g(x)
∞
formally ∫−∞ xf (x − μ)dx
= f (x − μ) is symmetric about μ. This is enough information to conclude that the mean associated with f is μ. More
∞
∫−∞ (y + μ)f (y)dy via the change of variable y = x − μ
∞
∫−∞ yf (y)dy + μ
∞
Then following the homework problem ∫−∞ yf (y)dy = 0
Question 9
Suppose that f be a mean 0 density having variance 1. What is the variance associated with the density
Your Answer
Score
σ
1/σ
1
1/σ 2
σ2
Total

1.00
1.00 / 1.00
Question Explanation
First note the mean associated with g(x)
∞
Then the variance is ∫−∞ x 2 f (x/σ)/σdx
= f (x/σ)/σ is 0. (Show this for yourself.)
∞
= ∫−∞ (σy)2 f (y)dy via the change of variable y = x/sigma
∞
= σ 2 ∫−∞ yf (y)dy = σ 2 since f has associated mean 0 and variance 1 .
Question 10
If
X and Y are mean 0, variance 1 independent random variables, what is E[X 2 Y 2 ]?
g(x) = f (x/σ)/σ ?
Explanation
Your Answer
Score
Explanation
0
2
It can not be calculated.
1/2
1

Total
1.00
1.00 / 1.00
MathJax no longer loads a default configuration file; you must specify such files
x
Question Explanation
explicitly. This page seems to use the older default config/MathJax.js file, and
so needs to be updated. This is explained further at
Note 1
= Var(X) = E[X 2 ] − E[X]2 = E[X 2 ] − 0 and thus E[X 2 ] = 1 and similarly E[Y 2 ] = 1 . E[X 2 Y 2 ] = E[X 2 ]E[Y 2 ] = 1 × 1
http://www.mathjax.org/help/configuration