Download Exercises on manipulating means and standard deviations

Exercises on manipulating means and
standard deviations
Lily Yen
Due January 12th, 2015
1. Show that for any data set {xi }ni=1 ,
a.
n
X
(xi − x̄) = 0.
i=1

b. s2 =
n
X
1 
1
x2i −
n − 1 i=1
n
n
X
!2 
xi
.
i=1
P
P
Solution
a. Consider that ni=1 xi = nx̄. Therefore ni=1 (xi − x̄) =
Pn 1. P
n
i=1 xi −
i=1 x̄ = nx̄ − nx̄ = 0.
b. Since
n
X
(xi − x̄)2 =
i=1
n
X
(x2i − 2xi x̄ + x̄2 ) =
i=1
=
n
X
=
i=1
=
n
X
x2i − 2x̄
n
X
n
X
xi +
i=1
n
X
x̄2
i=1
xi + nx̄2
i=1
2
x2i −
n
x2i −
i=1
it follows that s2 =
x2i − 2x̄
i=1
i=1
n
X
n
X
1
n−1
1
n
n
X
!2
xi
i=1
n
X
1
+
n
n
X
!2
xi
i=1
!2
xi
,
i=1
Pn
2
i=1 (xi − x̄) =
1
1
n−1
P
n
i=1
x2i −
1
n
P
2
( ni=1 xi ) .
2. Prove that if a data set {xi }ni=1 has mean x̄ and standard deviation sx ,
then
a. if yi = xi + h, then ȳ = x̄ + h and sy = sx ;
b. if yi = cxi , then ȳ = cx̄ and sy = csx ;
c. if zi =
xi −x̄
,
sx
Solution 2.
and
then z̄ = 0 and sz = 1.
a. ȳ =
1
n
Pn
i=1
yi =
1
n
Pn
i=1 (xi +h)
n
s2y
=
1
n
Pn
i=1
xi + n1 n h = x̄+h
n
1 X
1 X
(yi − ȳ)2 =
(xi + h − (x̄ + h))2
=
n − 1 i=1
n − 1 i=1
n
1 X
=
(xi − x̄)2 = s2x ,
n − 1 i=1
so sy = sx .
P
b. ȳ = n1 ni=1 yi =
1
n
Pn
i=1
cxi = c n1
Pn
n
s2y
i=1
xi = cx̄, and
n
1 X
1 X
=
(yi − ȳ)2 =
(cxi − cx̄)2
n − 1 i=1
n − 1 i=1
n
=
n
1 X 2
1 X
c (xi − x̄)2 = c2
(xi − x̄)2 = c2 s2x ,
n − 1 i=1
n − 1 i=1
so sy = csx .
c. This follows immediately from the previous two parts with h = −x̄ and
c = s1x .
3. A sample of 20 resistors yielded a mean value of 44.6 Ω and a standard
deviation of 1.3 Ω.
a. If one more resistor of 52 Ω is added to the sample, what are the mean
and standard deviation of the 21 resistors?
b. If the original sample of 20 resistors is combined with a sample of 10
resistors that had a mean of 48.3 Ω and standard deviation of 1.9 Ω,
calculate the mean and standard deviation of the combined sample of
30 resistors.
2
Solution
Pn3.
1
i=1
n
a. Consider first the original n = 20 resistors. Since x̄ =
xi ,
n
X
xi = nx̄ = 20 · 44.6 Ω = 892 Ω.
i=1
2
Moreover, since s =
n
X
x2i
i=1
1
n−1
P
n
i=1
x2i
1
= (n − 1)s +
n
2
= 19(1.3 Ω)2 +
−
1
n
Pn
2
( i=1 xi ) , it follows that
n
X
!2
xi
i=1
1
(892 Ω)2 = 39 815.31 Ω2 .
20
P
P
2
Adding the 21st resistor, 21
944 Ω and 21
i=1 = 892 Ω + 52 Ω = P
i=1 xi =
21
2
2
1
1
39 815.31 Ω +(52 Ω)2 = 42 519.31 Ω . Thus x̄ = 21 i=1 xi = 21 ·944 Ω =
45.0 Ω and

!2 
21
21
X
X
1 
1
1
1
2
2
2
2
s =
x −
42 519.31 Ω − (944 Ω)
xi  =
20 i=1 i 21 i=1
20
21
= 4.21 Ω2 ,
so s = 2.05 Ω.
b. As in the previous part, for the original 20 resistors,
20
X
xi = 892 Ω and
i=1
20
X
x2i = 39 815.31 Ω2 .
i=1
In a similar manner, for the sample of 10 resistors,
10
X
xi = 10 · 48.3 Ω = 483 Ω
i=1
and
10
X
i=1
x2i = 9(1.9 Ω)2 +
1
(483 Ω)2 = 23 361.39 Ω2 .
10
3
Therefore, for the combined 30 resistors,
30
X
xi = 892 Ω + 483 Ω = 1375 Ω
i=1
and
30
X
x2i = 39 815.31 Ω2 + 23 361.39 Ω2 = 63 176.7 Ω2 .
i=1
P30
1
1
Hence x̄ = 30
i=1 xi = 30 · 1375 Ω = 45.8 Ω and

!2 
30
30
1 X
1
1 X 2
1
2
2
2
x −
xi  =
63 176.7 Ω − (1375 Ω)
s =
29 i=1 i 30 i=1
29
30
= 5.37 Ω2 ,
so s = 2.32 Ω.
4