NOBEL PRIZE 1939-ERNEST LAWRENCE
WHAT IS A PARTICLE ACCELERATOR?
Any device that
accelerates charged
particles to very high
speeds using electric
and/or magnetic fields.
This piece of the accelerator helped in the development of the atom bomb.
Built in 19
Cell Organelles
Dr Apekshita Singh
Asst Prof
AIB
AUUP
CHLOROPLAST
Chloroplasts were first described as
chloroplyllkornern
(chlorophyll
granules) by German botanist Hugo von
Mohl in 1837, though because of their
large size they were seen long back by
Nehem
Cell Organelles
Dr Apekshita Singh
Asst Prof
AIB
AUUP
PROKARYOTIC CELL
E
U
K
A
R
Y
O
T
I
C
C
E
L
L
NUCLEUS
Every eukaryotic cell possesses one or
more nuclei during their life cycle at some
stage of their existence.
Eg-RBCs with nucleus when young and
lat
Eukaryotic cell
The term, eukaryote was first of all used by Hans Ris in early 1960s, which means
organisms, which possess true nucleus (Gr. Eu-true, Karyon-nucleus). The cells are
characterized by the presence of nuclear membrane, which separates genetic
2011 Douglas Theobald A formal test of the theory of universal common ancestry. . . . Takahiro Yonezawa and Masami Hasegawa Some Problems in Proving the Existence of the Universal Common Ancestor of Life on Earth.
. . .
http:/www.
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
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
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
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 3 Section 3-1 3-1. 3-2. 3-3. 3-4. 3-5. The range of X is
cfw_0,1,2,.,1000
The range of X is cfw_0,12,.,50 , The range of X is cfw_0,12,.,99999 , The range of X is cfw_0,12,3,4,5 , The range of X is 1,2,.,491 . Because 490 parts are conforming, a n
CHAPTER 3 Section 3-1 3-1. 3-2. 3-3. 3-4. 3-5. The range of X is
cfw_0,1,2,.,1000
The range of X is cfw_0,12,.,50 , The range of X is cfw_0,12,.,99999 , The range of X is cfw_0,12,3,4,5 , The range of X is 1,2,.,491 . Because 490 parts are conforming, a n
125 375 5 0 2.3453E8 = = 0.00092 f X (5) = 2.5524 E11 500 5
b)
x f(x)
3-150.
0 1 2 3 4 5 6 7 8 9 10 0.0546 0.1866 0.2837 0.2528 0.1463 0.0574 0.0155 0.0028 0.0003 0.0000 0.0000
Let X denote the number of totes in the sample that exceed the moisture conte
P( A | B) =
2-121.
(0.38)(0.53) P( A B) P( A) P( B | A) = = = 39.3821% P( B) P( A) P( B | A) + P ( A' ) P( B | A' ) (0.38)(0.53) + (0.62)(0.5)
Let G denote a product that received a good review. Let H, M, and P denote products that were high, moderate, an
P( A | B) =
2-121.
(0.38)(0.53) P( A B) P( A) P( B | A) = = = 39.3821% P( B) P( A) P( B | A) + P ( A' ) P( B | A' ) (0.38)(0.53) + (0.62)(0.5)
Let G denote a product that received a good review. Let H, M, and P denote products that were high, moderate, an