Unformatted text preview: 6.7 , 2.4
6.7
N = 13
µ =8
5.8
5.8
3 9.7 σ 2 =14 Set 2 should be viewed as
Set
the this table:
the
x f 8 − 14 6.5 8 + 14 6.5
13 Solution:
ON MODE COMP
MODE SD
6.7 DT 2.4 DT
6.7
 5.8 DT SHIFT
5.8
88+ SD
3 9.7 DT 14 ) EXE SHIFT ; 6.5 DT
14 ) EXE SHIFT ; 6.5 DT SHIFT
SHIFT
SHIFT n SSUM
EXE 17 = N
SVAR x EXE 6.4372 = µ
SVAR xσn EXE 4.841 = σ ∴ µ = 6.4372, σ = 4.8410 Simple Exercises:
Simple
1. Create 10 numbers so that the mean=143 and
1.
s.d.=15
s.d.=15 2. Data set 1: N1 = 10 µ1 = 141 σ1 =15
Data set 2: N 2 = 15 µ 2 = 156 σ 22 = 345
Data
Find the mean and s.d. of the 25 numbers. B. Linear Transformation
B.
Consider the data set x1 , x2 ,......., xN
Consider
2
µ X and σ X
Suppose their mean and variance are
respectively.
Often we are interested in a linear transformation of
them: yi = a + bxi ............(1)
them:
Thus, we have new data: y1 , y2 ,......., y N
For instance x1 , x2 ,......., xN may be temperatures in
For
centigrade. Now we like to express them in
Fahrenheit:
Fahrenheit: y = 32 + 9 x
i 5 i We like to ask:
What are the mean µ and variance
these new data y = a + bx ?
these
Y i The answers are: Proof: From (1),
From
yi = a + bxi
N ∑y
i =1 i = Na + b∑ xi
i =1 N ∑y
i =1 N i = a+b i µ =a +bµ .......... ..( 2)
Y
X
2
σ2 =b 2σX .......... ......(3)
Y
σ = b  × X .......... ....( 4)

σ
Y µY = a + bµ X
yi − µY = a + bxi − a − bµ X
N N ∑x
i =1 N
N
∴ µ Y = a + bµ X i of 2
σY ∑(y − µ
i =1 i N ∑(y − µ
i =1 i Y Y) =b ∑(y − µ
i =1 Y N
2
2
σ Y = b 2σ X 2 i =1 2 N i N ) = ∑ (a + bxi − a − bµ X ) = b
2 ) 2 N ∑ (x − µ
i =1 i )2
X N 2 = b2 ∑ (x − µ
i =1 i N X )2 2 N ∑ (x − µ
i =1 i X )2 Example 6: Salaries (in $1000 p.m.) of the
Example
Salaries
employees of Company C are given below:
employees
Salary
x
Frequency
f 5.5 6.5 8.0 total 20 50 25 5 (a) Find the mean and standard deviation of X.
(b) If everyone’s salary is revised 6% up, find the
(b)
new mean and s.d.
new
(c) If everyone is further given a travel allowance
(c)
of $400, find the new mean and s.d.
of Solution:
Solution
(a)
By use of SDMODE,
5.5 SHIFT ; 20 DT
6.5 SHIFT ; 25 DT
8 SHIFT ; 5 DT N X = 50, µ X = 6.25, σ X = 0.75 we can easily obtain (b) By (1)
(b)
by (2)
by by (4)
(c) Now yi = 0 + 1.06 xi µY = 0 + 1.06 µ X = 1.06 × 6.25 = 6.625
2
2
σ Y = 1.06 2 σ X
σ Y = 1.06  ×σ X = 1.06 × 0.75 = 0.795 wi = 0.4 + yi µW = 0.4 + µY = 0.4 + 6.625 = 7.025
2
2
σW = σY
σ W = σ Y = 0.795 Example 7:
The pass mark of an examination is 50.
The table below shows the marks (X) of 250
students who took the examination.
The examiner revised the marks by the formula
given by Y.
given
Mark
Mark
X No. of Mean SD
No.
Mean SD
students
students Revised mark
Y x ≥ 50 200 70 8 4 y = 60 + ( x − 50)
5 x < 50 50 40 6 11 y = 60 − (50 − x)
10 Find the mean and standard deviation of the 250
revised marks.
revised Solution:
Solution
For the 200 students whose old mark are x ≥ 50 ,
4
Mean(Y ) = 60 + (70 − 50) = 76
5 4
SD(Y ) = × 8 = 6.4
5 Similarly, for the 50 students with x < 50 ,
Similarly,
Mean(Y ) = 60 − 11
(50 − 40) = 49
10 SD(Y ) = 11
× 6 = 6.6
10 Hence the mean & s.d. of the 250 new marks can be
Hence
found from the frequency table:
found x 766.4= 69.6 82.4 42.4 55.6
55.6 f 100 100 25 25 76+6.4= 496.6= 49+6.6= ∴ µ = 70.6, σ = 12.5748 N
250 Simple Exercise:
Simple
Data set : x , x ,........, x
Data
Mean=12, Median=14, Mode=15 and
CV=30%.
Define new data set: yi = 80 − 4 xi i = 1,2,........,80
1 2 80 Find the mean, median, mode, s.d, CV and
Sk for the yi ' s .
Sk Solution:
Solution
RULES:
RULES
Y = a + cX
then
µY = a + cµ X
MedY = a + c ⋅ Med X
ModY = a + c ⋅ Mod X σ Y = c  ⋅σ X
and
2
2
σ Y = c 2 ⋅σ X Given and µX =12 C .V X = Med X =14 σX Mod X =15 σX
=30%
µX = 0.3
12
∴ X =3.6
σ Y = a + cX
Y = 80 − 4 X
⇒ a = 80, c = −4 µY = a + cµ X = 80 − 4 ×12 = 32
MedY = a + c ⋅ Med X = 80 − 4 ×14 = 24
ModY = a + c ⋅ Mod X = 80 − 4 ×15 = 20
σ Y = c  ⋅σ X = 4 × 3.6 = 14.4
σ
14.4
C.VY = Y =
= 45%
µY
32
3( µY − MedY ) 3(32 − 24)
SkY =
=
= 1.66667
σY
14.4...
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 Fall '13
 Statistics, Standard Deviation, Trigraph, .........

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