An instructor has given a short quiz consisting of two parts. For a randomly selected students, let X = the number of
points earned on the first part and Y=the number of points earned on the second part. Suppose that the joint pmf of
X and Y is given in the accompanying table.
y
0
5
10
15
Total
x
0
0.02
0.06
0.02
0.10
0.20
5
0.04
0.15
0.20
0.10
0.49
10
0.01
0.15
0.14
0.01
0.31
Total
0.07
0.36
0.36
0.21
1
Use this information to answer the following 5 questions.
1. Which of the following is the expected number of points earned on the first part?
(a) 5.55 =E(X)=0(0.2)+5(0.49)+10(0.31)
(b) 8.55
(c) 13.55
(d) 18.55
(e) 21.55
2.
Which of the following is the probability of maximum number of points earned in two parts, max(x,y) being 5
pts.?
(a) 0.06
(b) 0.10
(c) 0.15
(d) 0.21
(e) 0.25 =P(max(x,y)=5)=p(0,5)+p(5,0)+p(5,5)
3.
Which of the following is the probability of total number of points earned in two parts, x+y being 5 pts.?
(a) 0.06
(b) 0.10 =P(x+y=5)=p(0,5)+p(5,0)
(c) 0.15
(d) 0.21
(e) 0.25
4.
Given that students did not earn any points on the first part, what is the probability that they will earn 10 points
on the second part?
(a) 0.01
(b) 0.02
(c) 0.10 =P(Y=10|X=0)=p(0,10)/P(X=0)=0.02/0.2
(d) 0.12
(e) 0.20
5.
Are X and Y independent?
(a) Yes
(b) No because p(0,0)=0.02 P(X=0)P(Y=0)=0.2(0.07)=0.014
The tensile strength of paper is modeled by a normal distribution with a mean of 35 pounds per square inch and a
standard deviation of 2 pounds per square inch. Answer the following 2 questions using this information.
6. What is the probability that the strength of a paper is less than 40 lb/in 2?
(a) 0.0062
(b) 0.25
(c) 0.5364
(d) 0.75
(e) 0.9938 = P(X<40)=P(Z<(40-35)/2)=P(Z<2.5) where X: tensile strength ~ N(=35,2=4)
7.
If the specifications require the tensile strength to exceed 30 lb/in2, what proportion of the samples is scrapped?
(a) 0.0062 = P(X30)=P(Z (30-35)/2)=P(Z-2.5) where X: tensile strength ~ N(=35,2=4)
(b) 0.25
(c) 0.5364
(d) 0.75
(e) 0.9938
The probability density function of the net weight in pounds of packaged chemical herbicide, X is
f(x)=2 for 49.75  x  50.25 pounds. (Hint: Did you notice that X~U[49.75,50.25])
Answer the following 10 questions using this information.
8.
Which of the following is the probability that a package weighs more than 50 pounds?
(a) 0
(b) 0.25
(c) 0.50 =P(X<50)=2(50.25-50)
(d) 0.75
(e) 1
9.
How much chemical is contained in 90% of all packages?
(a) 49.9
(b) 50
(c) 50.1
(d) 50.2 =x* where P(Xx*)=2(x*-49.75)=0.90
(e) 50.25
10. Which of the following is P(X  50.3)?
(a) 0
(b) 0.25
(c) 0.50
(d) 0.75
(e) 1 which is the total area under the curve
11. Which of the following is the median net weight in pounds of packaged chemical herbicide?
(a) 49.9
(b) 50 it is already shown for question 8 but P(Xmedian)=2(median-49.75)=0.50
(c) 50.1
(d) 50.2
(e) 50.25
12. Which of the following is the probability of the package net weight exceeding the mean value by more than 2
standard deviations?
(a) 0 =P(X>+2)=P(X>50.29) where =50 and 2=(50.25-49.75)2/12 in uniform distribution
(b) 0.25
(c) 0.50
(d) 0.75
(e) 1
13. If there is a random sample of 25 packages, which of the following is the expected value of the average package
net weight?
(a) 49.9


_


(b) 50 = E  X   E ( X ) proven in handout 5
(c) 50.1
(d) 50.2
(e) 50.25
14. If there is a random sample of 25 packages, which of the following is the variance of the average package net
weight?


_


(a) 0.0008 = Var X  =Var(X)/n=((50.25-49.75)2/12)/25 proven in handout 5
(b) 0.0042
(c) 0.0208
(d) 0.025
(e) 0.25
15. If there is a random sample of 25 packages, which of the following is the expected value of the sum of the net
weights of any two packages?
(a) 0
(b) 2
(c) 50
(d) 100 =E(Xi+Xj)=E(Xi)+E(Xj)=50+50
(e) 1250
16. If there is a random sample of 25 packages, which of the following is the variance of the difference between the
net weights of any two packages?
(a) I don’t know what the covariance is and I need too know it to answer the question
(b) 0
(c) 0.0417 = Var(Xi-Xj)=Var(Xi)+Var(Xj)= (50.25-49.75)2/6 where X’s are random sample (implies indep.)
(d) 0.0834
(e) 0.1203
17. If there is a random sample of 25 packages, which of the following is the joint pdf of any two packages net
weights?
(a) 2
(b) 4 = f(xi)f(xj)=2(2) because they are independent
(c) 8
(d) 12
(e) 16
Let X be a continuous random variable with the pdf, f(x)=k(x-1) for 2<x<4 . Answer the following 2 questions
using this information.
18. Which of the following k makes f(x) a legitimate pdf?
4
(a) 0.25
where the total area under the curve is
 k ( x  1)dx  1 then 4k=1
2
(b) 0.50
(c) 1
(d) 2
(e) 4
19. Which of the following is the expected value of X?
(a) 0.0790k
(b) 6k
4
(c) 12.6667k =E(X)=
 xk ( x  1)dx
2
(d) 70k
(e) 76k
20. Let Z be a standard normal random variable. Which of the following is P(0  Z  0.5)?
(a) 0
(b) 0.1915 =P(Z 0.5)-P(Z<0)=0.6915-0.5
(c) 0.3915
(d) 0.5
(e) 0.6915
21. Let Z be a standard normal random variable. Which of the following c value satisfies P(-c  Z  c)=0.984?
(a) 2.145
(b) 2.41 because of the symmetry P(Z<-c)=P(Z>c)=0.008
(c) 2.8
(d) 2.95
(e) 3.01
Suppose the counts recorded by a Geiger counter follow a Poisson process with an average of two counts per
minute. Answer the following 4 questions using this information.
22. What is the probability that there are no counts in a 30-second interval?
(a) 0.1353
(b) 0.2835
(c) 0.3679
(d) 0.4355
(e) 0.6719
X: counts ~ Poisson (=2 per minute or 1 per 30 second or 2/60 per second)
Y: time between counts ~exponential((=2 per minute or 1 per 30 second or 2/60 per second)
P(X=0)=
e 110
 e 1 or P(Y>30 second)= e  ( 2 / 60)( 30)  e 1
0!
23. Which of the following is the probability that the time until the next count being less than 10 seconds?
(a) 0.1353
(b) 0.2835 =P(Y<10)= 1  e
(c) 0.3679
(d) 0.4355
(e) 0.6719
 ( 2 / 60)(10)
 e 1 / 3
24. Which of the following is the expected time between counts?
(a) 0.25 minutes
(b) 0.5 minutes because 2 counts per minute
(c) 1 minutes
(d) 2 minutes
(e) 4 minutes
25. Which of the following is the expected count per hour?
(a) 2
(b) 12
(c) 20
(d) 60
(e) 120 because 2 counts per minute then 2(60)=120 counts per hour