1.3 Solutions
79
M x (t) = Ee
xt
=
∞
x=0
e
xt B(x)
=
∞
x=0
N
r
( pe
t )
r (1 − p)
N −r
= ( pe
t
+ 1 − p)
N
N
r
p
r q
N −r
= ( pe
t
+ 1 − p)
N
μ
0
n =
∂
n M x (t)
∂t n | t=0
Therefore μ
0
1 =
∂ M
∂t
| t=0 = N pe
t (q + pe
t )
N −1
| t=0 = N p
Thus the mean = N p
(c) μ
0
2 =
∂
2 M
∂t 2 = [N (N − 1) p
2 e
t (q + pe
t )
N −2
+ N pe
t (q + pe
t )
N −1 ] t=0
= N (N − 1) p
2
+ N p
But μ 2 = μ
0
2 − (μ
0
1 )
2
= N (N − 1) p
2
+ N p − N
2 p
2
= N p − N p
2
= N p(1 − p) = N pq
or σ =
N pq
1.95 Total counting rate/minute, m 1 = 245
Background rate/minute, m 2 = 49
Counting rate of source, m = m 1 − m 2 = 196
m 1 =
n 1
t 1
±
√
n 1
t 1
; m 2 =
n 2
t 2
±
√
n 2
t 2
; Net count m = m 1 − m 2
σ = (σ
2
1 + σ
2
2 )
1/2
=
n 1
t
2
1
+
n 2
t
2
2
1/2
=
m 1
t 1
+
m 2
t 2
1/2
σ =
m 1
t 1
+
m 2
t 2
1/2
=
49
100
+
245
20
1/2
= 3.57
Percentage S.D. =
σ
m
× 100 =
3.57
196
× 100 = 1.8%
1.96 (a) B(x) =
N !
x!(N − x)!
p
x q
N −x
Using Sterling’s theorem