180
11 Statistics of Counting
σ u
u
2 =
σ x
x
2 +
σ y
y
2
(11.5)
2. For a case when u = x/y
The standard deviation (σ u ) is given by the Eq. (11.6)
σ u
u
2 =
σ x
x
2 +
σ y
y
2
(11.6)
Thus, in either of these two cases, the standard deviation (u) is calculated in the
same fashion. We can take an example to explain these two calculations.
If the activity of two sources is 15826 and 6721 cpm, what would be the standard
deviation of the ratio of these two activities?
Counts for the source 1, i.e., x = 15826 cpm. Hence, its approximate standard deviation
σ x =
√
15826 = 125.80
Likewise, counts for the source 2 (i.e., y) being 6721 cpm, its approximate standard
deviation
σ y =
√
6721 = 81.98
Ratio activity
u =
15826
6721
= 2.35
The standard deviation as per the Eq. (11.6) is
σ u
u
2 =
125.8
15826
2
+
81.98
6721
2
= 0.02015
σ u
u
= 0.14
Thus,
σ u = 0.14 × 2.35
= 0.33
The final result should be expressed as 2.35 ± 0.33 cpm.
11 Statistics of Counting
σ u
u
2 =
σ x
x
2 +
σ y
y
2
(11.5)
2. For a case when u = x/y
The standard deviation (σ u ) is given by the Eq. (11.6)
σ u
u
2 =
σ x
x
2 +
σ y
y
2
(11.6)
Thus, in either of these two cases, the standard deviation (u) is calculated in the
same fashion. We can take an example to explain these two calculations.
If the activity of two sources is 15826 and 6721 cpm, what would be the standard
deviation of the ratio of these two activities?
Counts for the source 1, i.e., x = 15826 cpm. Hence, its approximate standard deviation
σ x =
√
15826 = 125.80
Likewise, counts for the source 2 (i.e., y) being 6721 cpm, its approximate standard
deviation
σ y =
√
6721 = 81.98
Ratio activity
u =
15826
6721
= 2.35
The standard deviation as per the Eq. (11.6) is
σ u
u
2 =
125.8
15826
2
+
81.98
6721
2
= 0.02015
σ u
u
= 0.14
Thus,
σ u = 0.14 × 2.35
= 0.33
The final result should be expressed as 2.35 ± 0.33 cpm.