Download Quiz 5 2015 (Solutions)

CME 263H1: Quiz 05 – April 6th, 2015
Time:
10:30 pm to 11:15 am (45 minutes)
Location:
MP202
Total Marks: 45
Student Number:___________________________
Name:_____________________________________
Marks Distribution:
Question
Number
Q1
Q2
Q3
Total marks
of Part a
3
4
3
Total marks
of Part b
3
4
Total marks
of Part c
3
Total marks
of Part d
3
Total marks
Total marks
of Part a
Total marks
of Part b
Total marks
of Part c
Total marks
of Part d
Total marks
12
8
3
Total 23
Marks:
Question
Number
Q1
Q2
Q3
Total
Q1) Each front tire on a particular type of vehicle is supposed to be filled to a pressure of 26 psi. Suppose the actual
air pressure in each tire is a random variable—X for the right tire and Y for the left tire, with joint pdf
a.
b.
c.
d.
What is the value of K?
What is the probability that both tires are underfilled?
Determine the (marginal) distribution of air pressure in the right tire alone.
Are X and Y independent random variables?
ANS:
a.
b.
c.
d.
is obtained by substituting y for x in (d); clearly
not independent.
are
Q2) Let
represent the times necessary to perform three successive repair tasks at a certain service
facility. Suppose they are independent normal random variables with expected values
respectively.
a.
If
Calculate
b.
What is
Using the
given in part (a), calculate
ANS:
a.
b.
!"
!
= 2.581
Q3)
A study of the ability of individuals to walk in a straight line reported that accompanying data on cadence (strides
per seconds) for a sample of n – 20 randomly selected healthy men:
.95
.92
.81
.96
.93
.92
.95
1.00
.93
.78
.86
1.06
1.05
1.06
.92
.96
.85
.85
.81
.92
A normal probability plot gives substantial support to the assumption that the population distribution of
cadence is approximately normal. A descriptive summary of the data is:
Variable
Cadence
a.
Mean
0.9255
Median
0.9300
StDev
0.0809
Calculate a 95% confidence interval for a population mean cadence.
ANS:
a.
N
20
A 95% C.I.:
SEMean
0.0181
EQUATIONS: