MAE 3241: AERODYNAMICS AND
FLIGHT MECHANICS
Flow Over Rotating Cylinders and Applications
February 23, 2011
Mechanical and Aerospace Engineering Department
Florida Institute of Technology
D. R. Kirk
INVISCID VS. VISCOUS FLOWS
Theoretical: Beautifully
beha
CHAPTER 4
Section 4-2
4-1.
a) P (1 < X ) = e x dx = ( e x )
1
1
2.5
b) P (1 < X < 2.5) =
e
x
= e 1 = 0.3679
dx = ( e x )
1
2 .5
1
= e 1 e 2.5 = 0.2858
3
x
c) P( X = 3) = e dx = 0
3
4
4
x
x
4
d) P( X < 4) = e dx = (e ) = 1 e = 0.9817
0
0
x
x
e) P(3 X ) = e
X np
3.5 6.48
P ( X 4) P
np(1 p )
64800(0.0001)(0.9999)
= P( Z 1.17) = 1 0.1210 = 0.8790
b)
4-150
Using the normal approximation to the binomial with X being the number of people who will be
seated. Then X ~Bin(200, 0.9).
X np
np (1 p )
185.5 180
3-118.
a) Let X denote the failures in 8 hours. Then, X has a Poisson distribution with = 0.16.
P( X = 0) = e 0.16 = 0.8521
b) Let Y denote the number of failure in 24 hours. Then, Y has a Poisson distribution with = 0.48.
P(Y 1) = 1 P(Y = 0) = 1 e 48 = 0
Section 3-9 3-107.
e 4 4 0 = e 4 = 0.0183 0! b) P( X 2) = P( X = 0) + P( X = 1) + P( X = 2)
a) P( X = 0) =
e 4 41 e 4 42 + 1! 2! = 0.2381 = e 4 +
e 4 4 4 = 01954 . 4! e 4 4 8 = 0.0298 d) P( X = 8) = 8!
c) P ( X = 4 ) = 3-108 a) P ( X = 0) = e 0.4 = 0.6703
3-96.
Let X denote a geometric random variable with parameter p. Let q = 1-p.
1 1 E ( X ) = x (1 p ) x 1 p = p xq x 1 = p 2 = x =1 x =1 p p
V ( X ) = ( x 1 ) 2 (1 p) x 1 p = px 2 2 x + p
x =1 x =1
1 p
(
1 p
)(1 p)
x 1
= p x 2 q x 1 2 xq x 1 +
x =1
x =1
q
3-79.
The probability is 0.651 that at least one sample from the next five will contain more than one defective Let X denote the passengers with tickets that do not show up for the flight. Then, X is binomial with n = 125 and p = 0.1.
a) P( X 5) = 1 P( X
3-60.
E (cX ) = cxf ( x) = c xf ( x) = cE ( X ) , V (cX ) = (cx c ) f ( x) = c 2 ( x ) 2 f ( x) = cV ( X )
2 x x x x
Section 3-6 3-61. A binomial distribution is based on independent trials with two outcomes and a constant probability of success on each t
50.25
4-7
a) P( X > 50) =
2.0dx = 2 x
50.25
50
= 0.5
50
50.25
b) P( X > x) = 0.90 =
2.0dx = 2 x
50.25
x
= 100.5 2 x
x
Then, 2x = 99.6 and x = 49.8.
74.8
4-8.
a) P( X < 74.8) =
1.25dx = 1.25x
74.8
74.6
= 0.25
74.6
b) P(X < 74.8 or X > 75.2) = P(X < 74.8
2
4-123 Let X ~N(, ), then Y = e follows a lognormal distribution with mean and variance
X
2. By definition, FY(y) = P(Y y) = P(eX < y) = P(X < log y) = FX(log y) =
Since Y = eX and X ~ N(, 2), we can show that
fY (Y ) =
log y
.
1
f X (log y )
y
log y
= 1 / 105 = 10 5 error per bit = 1 error per 105 bits.
1 0
1 1
1 2
P(Y 3) = 1 P(Y 2) = 1 e 0!1 + e 1!1 + e 2!1 = 0.0803
[
4-102
]
= 20 r = 100
a) E ( X ) = r / = 100 / 20 = 5 minutes
b) 4 min - 2.5 min=1.5 min
c) Let Y be the number of calls before 15 s
e) P(Y = 5) = (1 0.2019) 4 0.2019 = 0.0819 .
f) E(Y) = 1/0.2019 = 4.95.
4-93.
a) P( X > ) =
e
1 x /
dx = e x /
b) P ( X > 2 ) = e x /
c) P ( X > 3 ) = e x /
3
= e 1 = 0.3679
= e 2 = 0.1353
2
= e 3 = 0.0498
d) The results do not depend on .
4-94.
E ( X
Then, Y is a binomial random variable with n=365 and p=0.0222.
E(Y) = 8.10 and V(Y) = 365(0.0222)(0.9778) = 7.92
15 8.10
P (Y > 15) P Z
= P ( Z 2.45) = 1 P ( Z < 2.45)
7.92
= 1 0.9929 = 0.0071
Section 4-8
0
4-76.
x
a) P( X 0) = e dx = 0
0
2
1
2
1
0
0
Section 4-7
4-65.
a) E(X) = 200(0.4) = 80, V(X) = 200(0.4)(0.6) = 48 and
Then, P( X 70) P Z
X
=
48 .
70.5 80
= P( Z 1.37) = 0.08534
48
b)
89.5 80
70.5 80
P (70 < X < 90) P
<Z
= P ( 1.37 < Z 1.37)
48
48
= 0.91466 - 0.08534 = 0.82931
c)
80.5 80
4-38.
Let X denote the changed weight.
Var(X) = 42/12
Stdev(X) = 1.1547
4-39. (a) Let X be the time (in minutes) between arrival and 8:30 am.
f ( x) =
1
,
90
for 0 x 90
So the CDF is F ( x) =
x
,
90
for 0 x 90
(b) E ( X ) = 45
Var(X) =902/12=675
(c ) The
Therefore, V ( X ) = (5 5.1) 2 = 0.01 .
10 ( x 5 )
dx = e 10( x 5)
b) P( X > 5.1) = 10e
5.1
5.1
= e 10( 5.15) = 0.3679
Section 4-5
4-32.
a) E(X) = (5.5+1.5)/2 = 3.5,
V (X ) =
(5.5 1.5) 2
= 1.333, and x = 1.333 = 1.155 .
12
2.5
b) P( X < 2.5) =
0.25dx = 0
3-38.
The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; pmf: f1/8) = 0.2, f(1/4) = 0.7, f(3/8) = 0.1 a) P(X1/18) = 0 b) P(X1/4) = 0.9 c) P(X5/16) = 0.9 d) P(X>1/4) = 0.1 e) P(X1/2) = 1
Section 3-4 3-39. Mean and Va
CHAPTER 2
Section 2-1 2-1. 2-2. Let "a", "b" denote a part above, below the specification S = cfw_aaa, aab, aba, abb, baa, bab, bba, bbb
Let "e" denote a bit in error Let "o" denote a bit not in error ("o" denotes okay)
eeee, eoee, oeee, ooee, eeeo, eoeo,
(d) P ( A' B ) =
63 + 35 + 15 = 0.113 1000 0.032 P( A B) (e) P ( A | B ) = = = 0.2207 (25 + 63 + 15 + 7 + 35) / 1000 P( B) 5 (f) P = = 0.005 1000
2-167.
(a) Let A denote that a part conforms to specifications and let B denote for a part a simple component
Q27-1
A gold-leaf electroscope is given a net negative charge and the leaf rises to
some angle. A source of ultraviolet (UV) light shines on the electroscope.
The leaf will.
B) rise to a higher angle
C) fall to a lower angle
UV source
B
C
Q27-2.
A photoel
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Or March MAE 3241: AERODYNAMICS AND FLIGHT MECHANIC
(b) There are 36 experiments that use all three steps. The probability the best result uses all three steps is 36/52 = 0.6923. (a) No, it will not change. With k amounts in the first step the number of experiments is k + 3k + 9k = 13k. The number of exper