426.
Now,
f ( x)
e
x /10
for 0 < x and
10
x
x
FX ( x) 1/10 e x /10 dx e x /10 1 e x /10
0
0
for 0 < x.
Then,
0, x 0
FX ( x )
1 e
x /10
, x0
a) P(X<60) = F(60) =1  e6 = 1  0.002479 = 0.9975
b)
30
1/10 e x /10 dx e1.5 e3 =0.173343
15
c) P(X1>40)+
513.
Determine c such that
3 3
3
2
c xydxdy c x2
y
0 0
0
3
dy c(4.5
0
y2
2
3
0
) 81 c.
4
Therefore, c = 4/81.
a)
3 2
P ( X 2, Y 3)
4
81
3
xydxdy
4
81
0 0
4
(2) ydy 81 (2)( 9 ) 0.4444
2
0
b) P(X < 2.5) = P(X < 2.5, Y < 3) because the range of Y is from
Assignment 11 solution
214. An order for an automobile can specify either an automatic or a
standard transmission, either with or without air conditioning, and with
any one of the four colors red, blue, black, or white. Describe the set of
possible orde
Assignment 12 solutions
339.
0,
x 2
0.2 2 x 1
0.6 1 x 0
0.7
0 x 1
0.9 1 x 2
1
2x
F (x)
a) P(X 1.25) = 0.9
b) P(X 2.2) = 1
c) P(1.1 < X 1) = 0.7
d) P(X > 0) = 1 P(X 0) = 0.3
340.
0
x 1
7 1 x 2
4
F(x)
6
7 2 x 3
1
3 x
a)
b)
c)
d)
P(X < 1.5) = 4
Statistical Tools for Engineers
Spring 2016
Colisted as CHE 426 and CAE 523
COURSE OBJECTIVES

Design tests that make it possible to efficiently and effectively collect data;
Understand how to obtain information from existing data;
Make meaningful predi