8
The Probable Error of a Mean
Hence
y
=
N
n
s
2
π
σ
e
−
n
s
2
z
2
2
σ
2
{\displaystyle y={\frac {N{\sqrt {n}}s}{{\sqrt {2\pi }}\sigma }}e^{-{\frac {ns^{2}z^{2}}{2\sigma ^{2}}}}}
is the equation representing the distribution of
z
{\displaystyle z}
n
{\displaystyle n}
s
{\displaystyle s}
Now the chance that
s
{\displaystyle s}
s
{\displaystyle s}
s
+
d
s
{\displaystyle s+ds}
∫
s
s
+
d
s
C
σ
n
−
1
s
n
−
2
e
−
n
s
2
2
σ
2
d
s
∫
0
∞
C
σ
n
−
1
s
n
−
2
e
−
n
s
2
2
σ
2
d
s
{\displaystyle {\frac {\int _{s}^{s+ds}{\frac {C}{\sigma ^{n-1}}}s^{n-2}e^{-{\frac {ns^{2}}{2\sigma ^{2}}}}ds}{\int _{0}^{\infty }{\frac {C}{\sigma ^{n-1}}}s^{n-2}e^{-{\frac {ns^{2}}{2\sigma ^{2}}}}ds}}}
which represents the
N
{\displaystyle N}
Hence the distribution of
z
{\displaystyle z}
s
{\displaystyle s}
s
{\displaystyle s}
s
+
d
s
{\displaystyle s+ds}
y
=
∫
s
s
+
d
s
C
σ
n
n
2
π
s
n
−
1
e
−
n
s
2
(
1
+
z
2
)
2
σ
2
d
s
∫
0
∞
C
σ
n
−
1
s
n
−
2
e
−
n
s
2
2
σ
2
d
s
=
n
2
π
∫
s
s
+
d
s
s
n
−
1
e
−
n
s
2
(
1
+
z
2
)
2
σ
2
d
s
σ
∫
0
∞
s
n
−
2
e
−
n
s
2
2
σ
2
d
s
{\displaystyle y={\frac {\int _{s}^{s+ds}{\frac {C}{\sigma ^{n}}}{\sqrt {\frac {n}{2\pi }}}s^{n-1}e^{-{\frac {ns^{2}(1+z^{2})}{2\sigma ^{2}}}}ds}{\int _{0}^{\infty }{\frac {C}{\sigma ^{n-1}}}s^{n-2}e^{-{\frac {ns^{2}}{2\sigma ^{2}}}}ds}}={\frac {{\sqrt {\frac {n}{2\pi }}}\int _{s}^{s+ds}s^{n-1}e^{-{\frac {ns^{2}(1+z^{2})}{2\sigma ^{2}}}}ds}{\sigma \int _{0}^{\infty }s^{n-2}e^{-{\frac {ns^{2}}{2\sigma ^{2}}}}ds}}}
and summing for all values of
s
{\displaystyle s}
z
{\displaystyle z}
y
=
n
2
π
σ
∫
0
∞
s
n
−
1
e
−
n
s
2
(
1
+
z
2
)
2
σ
2
d
s
∫
0
∞
s
n
−
2
e
−
n
s
2
2
σ
2
d
s
{\displaystyle y={\frac {\sqrt {\frac {n}{2\pi }}}{\sigma }}{\frac {\int _{0}^{\infty }s^{n-1}e^{-{\frac {ns^{2}(1+z^{2})}{2\sigma ^{2}}}}ds}{\int _{0}^{\infty }s^{n-2}e^{-{\frac {ns^{2}}{2\sigma ^{2}}}}ds}}}
By what we have already proved this reduces to
y
=
1
2
n
−
2
n
−
3
⋅
n
−
4
n
−
5
.
.
.
5
4
⋅
3
2
(
1
+
z
2
)
−
n
2
{\displaystyle y={\frac {1}{2}}{\frac {n-2}{n-3}}\cdot {\frac {n-4}{n-5}}...{\frac {5}{4}}\cdot {\frac {3}{2}}(1+z^{2})^{-{\frac {n}{2}}}}
n
{\displaystyle n}
and to
y
=
1
π
n
−
2
n
−
3
⋅
n
−
4
n
−
5
.
.
.
4
3
⋅
2
1
(
1
+
z
2
)
−
n
2
{\displaystyle y={\frac {1}{\pi }}{\frac {n-2}{n-3}}\cdot {\frac {n-4}{n-5}}...{\frac {4}{3}}\cdot {\frac {2}{1}}(1+z^{2})^{-{\frac {n}{2}}}}
n
{\displaystyle n}
Since this equation is independent of
σ
{\displaystyle \sigma }
Section IV.
Some Properties of the Standard Deviation Frequency Curve.
By a similar method to that adopted for finding the constant we may find the mean and moments: thus the mean is at
I
n
−
1
I
n
−
2
{\displaystyle {\frac {I_{n-1}}{I_{n-2}}}}
(
n
−
2
)
(
n
−
3
)
(
n
−
4
)
(
n
−
5
)
.
.
.
2
1
2
π
σ
n
{\displaystyle {\frac {(n-2)}{(n-3)}}{\frac {(n-4)}{(n-5)}}...{\frac {2}{1}}{\sqrt {\frac {2}{\pi }}}{\frac {\sigma }{\sqrt {n}}}}
n
{\displaystyle n}
or
(
n
−
2
)
(
n
−
3
)
(
n
−
4
)
(
n
−
5
)
.
.
.
3
2
π
2
σ
n
{\displaystyle {\frac {(n-2)}{(n-3)}}{\frac {(n-4)}{(n-5)}}...{\frac {3}{2}}{\sqrt {\frac {\pi }{2}}}{\frac {\sigma }{\sqrt {n}}}}
n
{\displaystyle n}
