Multivariate data. Regression and Correlation The Scatter Plot.
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Transcript of Multivariate data. Regression and Correlation The Scatter Plot.
Multivariate data
Regression and Correlation
0
20
40
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40 60 80 100 120 140
The Scatter Plot
Pearson’s correlation coefficient
xy
xx yy
Sr
S S
n
x
xxxS
n
iin
ii
n
iixx
2
1
1
2
1
2
n
yx
yx
n
ii
n
iin
iii
11
1
n
y
yyyS
n
iin
ii
n
iiyy
2
1
1
2
1
2
n
iiixy yyxxS
1
where
Where for each case i, di = ri – si = difference in the rank of xi and the rank of yi.
1
61
21
2
nn
dn
ii
Spearman’s rank correlation coefficient
Simple Linear Regression
Fitting straight lines to data
The Least Squares Line The Regression Line
• When data is correlated it falls roughly about a straight line.
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80
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120
140
160
40 60 80 100 120 140
In this situation wants to:
• Find the equation of the straight line through the data that yields the best fit.
The equation of any straight line:
is of the form:
Y = a + bX
b = the slope of the line
a = the intercept of the line
For any equation of a straight line
Y = a + b X
The predicted value of Y when X = xi (ith case)
can be computed:
ˆi iy a bx
The error in the prediction is given by:
ˆi i i i ir y y y a bx
This is called the residual for the ith case.
The residuals
iiiii bxayyyr ˆ
,ˆ,,ˆ,ˆ 222111 nnn yyryyryyr
n
iii
n
iii
n
ii bxayyyrRSS
1
2
1
2
1
2 ˆ
can be computed for each case in the sample,
The residual sum of squares (RSS)
is a measure of the goodness of fit of the line
Y = a + bX to the data
The optimal choice of a and b will result in the residual sum of squares
n
iii
n
iii
n
ii bxayyyrRSS
1
2
1
2
1
2 ˆ
attaining a minimum.
If this is the case than the line:
Y = a + bX
is called the Least Squares Line
Then the slope of the least squares line can be shown to be:
n
ii
n
iii
xx
xy
xx
yyxx
S
Sb
1
2
1
and the intercept of the least squares line can be shown to be:
xS
Syxbya
xx
xy
Computing the residual sum of squares for the least squares line
n
iii
n
iii
n
ii bxayyyrRSS
1
2
1
2
1
2 ˆ
Once a and b have been determined this can be computed using the far right hand side.
This can also be computed using the values of Sxx, Syy and Sxy.
For the Least Squares Line2xy
yyxx
SRSS S
S
The following data showed the per capita consumption of cigarettes per month (X) in various countries in 1930, and the death rates from lung cancer for men in 1950. TABLE : Per capita consumption of cigarettes per month (Xi) in n = 11 countries in 1930, and the death rates, Yi (per 100,000), from lung cancer for men in 1950.
Country (i) Xi Yi
Australia 48 18Canada 50 15Denmark 38 17Finland 110 35Great Britain 110 46Holland 49 24Iceland 23 6Norway 25 9Sweden 30 11Switzerland 51 25USA 130 20
Iceland
NorwaySweden
DenmarkCanada
Australia
HollandSwitzerland
Great Britain
Finland
USA
0
5
10
15
20
25
30
35
40
45
50
0 20 40 60 80 100 120 140
Per capita consumption of cigarettes
deat
h ra
tes
from
lung
can
cer
(195
0)
Iceland
NorwaySweden
DenmarkCanada
Australia
HollandSwitzerland
Great Britain
Finland
USA
0
5
10
15
20
25
30
35
40
45
50
0 20 40 60 80 100 120 140
Per capita consumption of cigarettes
deat
h ra
tes
from
lung
can
cer
(195
0)
404,541
2
n
iix
914,161
n
iii yx
018,61
2
n
iiy
Fitting the Least Squares Line
6641
n
iix
2261
n
iiy
55.1432211
66454404
2
xxS
73.1374
11
2266018
2
yyS
82.3271
11
22666416914 xyS
Fitting the Least Squares Line - continued
First compute the following three quantities:
Computing Estimate of Slope and Intercept
288.055.14322
82.3271
xx
xy
S
Sb
756.611
664288.0
11
226
xbya
2 23271.81811374.72727-
14322.545xy
yyxx
SRSS S
S
627.3196
Computing the Residual Sum of Squares
Iceland
NorwaySweden
DenmarkCanada
Australia
HollandSwitzerland
Great Britain
Finland
USA
0
5
10
15
20
25
30
35
40
45
50
0 20 40 60 80 100 120 140
Per capita consumption of cigarettes
deat
h ra
tes
from
lung
can
cer
(195
0)
Y = 6.756 + (0.228)X
Interpretation of the slope and intercept
1. Intercept – value of Y at X = 0.– Predicted death rate from lung cancer
(6.756) for men in 1950 in Counties with no smoking in 1930 (X = 0).
2. Slope – rate of increase in Y per unit increase in X.
– Death rate from lung cancer for men in 1950 increases 0.228 units for each increase of 1 cigarette per capita consumption in 1930.
Relationship between correlation and Linear Regression
1. Pearsons correlation.
• Takes values between –1 and +1
n
ii
n
ii
n
iii
yyxx
xy
yyxx
yyxx
SS
Sr
1
2
1
2
1
2. Least squares Line Y = a + bX– Minimises the Residual Sum of Squares:
– The Sum of Squares that measures the variability in Y that is unexplained by X.
– This can also be denoted by:
SSunexplained
n
iii
n
iii
n
ii bxayyyrRSS
1
2
1
2
1
2 ˆ
Some other Sum of Squares:
– The Sum of Squares that measures the total variability in Y (ignoring X).
n
iiTotal yySS
1
2
– The Sum of Squares that measures the total variability in Y that is explained by X.
n
iiExplained yySS
1
2ˆ
It can be shown:
(Total variability in Y) = (variability in Y explained by X) + (variability in Y unexplained by X)
n
iii
n
ii
n
ii yyyyyy
1
2
1
2
1
2 ˆˆ
lainedUnExplainedTotal SSSSSS exp
It can also be shown:
= proportion variability in Y unexplained by X.
= the coefficient of determination
n
ii
n
ii
yy
yyr
1
2
1
2
2
ˆ
Further:
= proportion variability in Y that is unexplained by X.
n
ii
n
iii
yy
yyr
1
2
1
2
2
ˆ1
Example
TABLE : Per capita consumption of cigarettes per month (Xi) in n = 11 countries in 1930, and the death rates, Yi (per 100,000), from lung cancer for men in 1950.
Country (i) Xi Yi
Australia 48 18Canada 50 15Denmark 38 17Finland 110 35Great Britain 110 46Holland 49 24Iceland 23 6Norway 25 9Sweden 30 11Switzerland 51 25USA 130 20
Computing r and r2
737.0
73.137455.14322
82.3271
yyxx
xy
SS
Sr
544.0737.0 22 r
54.4% of the variability in Y (death rate due to lung Cancer (1950) is explained by X (per capita cigarette smoking in 1930)
Categorical Data
Techniques for summarizing, displaying and graphing
The frequency tableThe bar graph
Suppose we have collected data on a categorical variable X having k categories – 1, 2, … , k.
To construct the frequency table we simply count for each category (i) of X, the number of cases falling in that category (fi)
To plot the bar graph we simply draw a bar of height fi above each category (i) of X.
Example
In this example data has been collected for n = 34,188 subjects.
• The purpose of the study was to determine the relationship between the use of Antidepressants, Mood medication, Anxiety medication, Stimulants and Sleeping pills.
• In addition the study interested in examining the effects of the independent variables (gender, age, income, education and role) on both individual use of the medications and the multiple use of the medications.
The variables were: 1. Antidepressant use, 2. Mood medication use, 3. Anxiety medication use, 4. Stimulant use and 5. Sleeping pills use.6. gender, 7. age, 8. income, 9. education and 10. Role –
i. Parent, worker, partnerii. Parent, partneriii. Parent, workeriv. worker, partner
v. worker onlyvi. Parent onlyvii. Partner onlyviii. No roles
Frequency Table for Age
Age - (G)
5349 15.7 15.7 15.7
6758 19.8 19.8 35.5
6420 18.8 18.8 54.3
5528 16.2 16.2 70.5
4400 12.9 12.9 83.4
5663 16.6 16.6 100.0
34118 100.0 100.0
20-29
30-39
40-49
50-59
60-69
70+
Total
ValidFrequency Percent Valid Percent
CumulativePercent
20-29 30-39 40-49 50-59 60-69 70+
Age - (G)
0
1,000
2,000
3,000
4,000
5,000
6,000
7,000
Co
un
t
Bar Graph for Age
Frequency Table for Role
role
6614 19.4 24.5 24.5
1068 3.1 4.0 28.5
1351 4.0 5.0 33.5
5427 15.9 20.1 53.6
5711 16.7 21.2 74.7
456 1.3 1.7 76.4
3262 9.6 12.1 88.5
3097 9.1 11.5 100.0
26986 79.1 100.0
7132 20.9
34118 100.0
parent, partner, worker
parent, partner
parent, worker
partner, worker
worker only
parent only
partner only
no roles
Total
Valid
SystemMissing
Total
Frequency Percent Valid PercentCumulative
Percent
parent, partner, worker
parent, partnerparent, worker
partner, workerworker only
parent onlypartner only
no roles
role
0
1,000
2,000
3,000
4,000
5,000
6,000
7,000
Co
un
t
Bar Graph for Role
The two way frequency table
The 2 statistic
Techniques for examining dependence amongst two categorical
variables
Situation
• We have two categorical variables R and C.
• The number of categories of R is r.
• The number of categories of C is c.
• We observe n subjects from the population and count
xij = the number of subjects for which R = i and
C = j.
• R = rows, C = columns
Example
Both Systolic Blood pressure (C) and Serum Chlosterol (R) were meansured for a sample of n = 1237 subjects.
The categories for Blood Pressure are:
<126 127-146 147-166 167+
The categories for Chlosterol are:
<200 200-219 220-259 260+
Table: two-way frequency
Serum Cholesterol
Systolic Blood pressure <127 127-146 147-166 167+ Total
< 200 117 121 47 22 307200-219 85 98 43 20 246220-259 115 209 68 43 439
260+ 67 99 46 33 245
Total 388 527 204 118 1237
Example
This comes from the drug use data.
The two variables are:
1. Age (C) and
2. Antidepressant Use (R)
measured for a sample of n = 33,957 subjects.
Two-way Frequency Table
Took anti-depressants - 12 mo * Age - (G) Crosstabulation
Count
322 523 570 522 265 249 2451
5007 6201 5822 4982 4114 5380 31506
5329 6724 6392 5504 4379 5629 33957
YES
NO
Took anti-depressants- 12 mo
Total
20-29 30-39 40-49 50-59 60-69 70+
Age - (G)
Total
Age - (G)
20-29 30-39 40-49 50-59 60-69 70+6.04% 7.78% 8.92% 9.48% 6.05% 4.42%
Percentage antidepressant use vs Age
Antidepressant Use vs Age
0.0%
5.0%
10.0%
20-29 30-39 40-49 50-59 60-69 70+
The 2 statistic for measuring dependence
amongst two categorical variables
DefineTotal row
1
thc
jiji ixR
1
column Totalc
thj ij
i
C x j
n
CRE ji
ij
= Expected frequency in the (i,j) th cell in the case of independence.
Columns
1 2 3 4 5 Total
1 x11 x12 x13 x14 x15 R1
2 x21 x22 x23 x24 x25 R2
3 x31 x32 x33 x34 x35 R3
4 x41 x42 x43 x44 x45 R4
Total C1 C2 C3 C4 C5 N
Total row 1
thc
jiji ixR
1
column Totalc
thj ij
i
C x j
Columns
1 2 3 4 5 Total
1 E11 E12 E13 E14 E15 R1
2 E21 E22 E23 E24 E25 R2
3 E31 E32 E33 E34 E35 R3
4 E41 E42 E43 E44 E45 R4
Total C1 C2 C3 C4 C5 n
n
CRE ji
ij
Justification if i jij
R CE
n then ij j
i
E C
R n
1 2 3 4 5 Total
1 E11 E12 E13 E14 E15 R1
2 E21 E22 E23 E24 E25 R2
3 E31 E32 E33 E34 E35 R3
4 E41 E42 E43 E44 E45 R4
Total C1 C2 C3 C4 C5 n
Proportion in column j for row i
overall proportion in column j
and if i jij
R CE
n then ij i
j
E R
C n
1 2 3 4 5 Total
1 E11 E12 E13 E14 E15 R1
2 E21 E22 E23 E24 E25 R2
3 E31 E32 E33 E34 E35 R3
4 E41 E42 E43 E44 E45 R4
Total C1 C2 C3 C4 C5 n
Proportion in row i for column j
overall proportion in row i
The 2 statistic
r
i
c
j ij
ijij
E
Ex
1 1
2
2
Eij= Expected frequency in the (i,j) th cell in the case of independence.
xij= observed frequency in the (i,j) th cell
Example: studying the relationship between Systolic Blood pressure and Serum Cholesterol
In this example we are interested in whether Systolic Blood pressure and Serum Cholesterol are related or whether they are independent.
Both were measured for a sample of n = 1237 cases
Serum Cholesterol
Systolic Blood pressure <127 127-146 147-166 167+ Total
< 200 117 121 47 22 307200-219 85 98 43 20 246220-259 115 209 68 43 439
260+ 67 99 46 33 245
Total 388 527 204 118 1237
Observed frequencies
Serum Cholesterol
Systolic Blood pressure <127 127-146 147-166 167+ Total
< 200 96.29 130.79 50.63 29.29 307200-219 77.16 104.8 40.47 23.47 246220-259 137.70 187.03 72.40 41.88 439
260+ 76.85 104.38 40.04 23.37 245
Total 388 527 204 118 1237
Expected frequencies
In the case of independence the distribution across a row is the same for each rowThe distribution down a column is the same for each column
Table Expected frequencies, Observed frequencies, Standardized Residuals
Serum Systolic Blood pressure
Cholesterol <127 127-146 147-166 167+ Total <200 96.29 130.79 50.63 29.29 307 (117) (121) (47) (22) 2.11 -0.86 -0.51 -1.35 200-219 77.16 104.80 40.47 23.47 246 (85) (98) (43) (20) 0.86 -0.66 0.38 -0.72 220-259 137.70 187.03 72.40 41.88 439 (119) (209) (68) (43) -1.59 1.61 -0.52 0.17 260+ 76.85 104.38 40.04 23.37 245 (67) (99) (46) (33) -1.12 -0.53 0.88 1.99 Total 388 527 204 118 1237
2 = 20.85
ij
ijijij
E
Exr
Standardized residuals
ij
ijijij
E
Exr
85.20
1 1
2
1 1
2
2
r
i
c
jij
r
i
c
j ij
ijij rE
Ex
The 2 statistic
Example
This comes from the drug use data.
The two variables are:
1. Role (C) and
2. Antidepressant Use (R)
measured for a sample of n = 33,957 subjects.
Two-way Frequency Table
Percentage antidepressant use vs Role
Took anti-depressants - 12 mo * role Crosstabulation
Count
344 101 201 275 455 63 224 414 2077
6268 967 1150 5150 5249 392 3036 2679 24891
6612 1068 1351 5425 5704 455 3260 3093 26968
YES
NO
Took anti-depressants- 12 mo
Total
parent,partner,worker
parent,partner parent, worker
partner,worker worker only parent only partner only no roles
role
Total
Role parent, partner, worker
parent, partner
parent, worker
partner, worker
worker only parent only
partner only no roles
5.20% 9.46% 14.88% 5.07% 7.98% 13.85% 6.87% 13.39%
Antidepressant Use vs Role
0.0%
5.0%
10.0%
15.0%
20.0%
parent,partner,worker
parent,partner
parent,worker
partner,worker
workeronly
parentonly
partneronly
no roles
2 = 381.961
Calculation of 2
1 2 3 4 5 6 7 8 Total
YES 344 101 201 275 455 63 224 414 2077NO 6268 967 1150 5150 5249 392 3036 2679 24891
Total 6612 1068 1351 5425 5704 455 3260 3093 26968
The Raw data
Expected frequencies1 2 3 4 5 6 7 8 Total (R i )
YES 509.24 82.25 104.05 417.82 439.31 35.04 251.08 238.21 2077NO 6102.76 985.75 1246.95 5007.18 5264.69 419.96 3008.92 2854.79 24891
Total (C j ) 6612 1068 1351 5425 5704 455 3260 3093 26968
ij
ijijij
E
Exr
i jij
R CE
n
The Residuals
The calculation of 2
ij
ijijij
E
Exr
1 2 3 4 5 6 7 8
YES -7.32 2.07 9.50 -6.99 0.75 4.72 -1.71 11.39NO 2.12 -0.60 -2.75 2.02 -0.22 -1.36 0.49 -3.29
2
2 2 381.961ij ij
iji j i j ij
x Er
E
Probability Theory
Modelling random phenomena
Some counting formulae
Permutations
the number of ways that you can order n objects is:
n! = n(n-1)(n-2)(n-3)…(3)(2)(1)
Example:
the number of ways you can order the three letters A, B, and C is 3! = 3(2)(1) = 6
ABC ACB BAC BCA CAB CBA
Permutations
the number of ways that you can choose k objects from n objects in a specific order:
Example:
the number of ways you choose two letters from the four letters A, B, D, C in a specific order is
)1()1()!(
!
knnn
kn
nPkn
12)3)(4(!2
!4
)!24(
!424
P
AB BA AC CA AD DA
BC CB BD DB CD DC
Combinations
the number of ways that you can choose k objects from n objects (order irrelevant) is:
)1()1(
)1()1(
)!(!
!
kk
knnn
knk
n
k
nCkn
Example:
the number of ways you choose two letters from the four letters A, B, D, C
{A,B} {A,C} {A,D} {B,C} {B,D}{C,D}
62
12
)1)(2(
)3)(4(
!2!2
!4
)!24(!2
!4
2
424
C
Example:
Suppose we have a committee of 10 people and we want to choose a sub-committee of 3 people. How many ways can this be done
45)1)(2)(3(
)3)(9)(10(
!7!3
!10
3
10310
C
Example: Random sampling
Suppose we have a club of N =1000 persons and we want to choose sample of k = 250 of these individuals to determine there opinion on a given issue. How many ways can this be performed?
The choice of the sample is called random sampling if all of the choices has the same probability of being selected
2422501000 10823.4
!750!250
!1000
250
1000
C
Important Note:
0! is always defined to be 1.
Also
are called Binomial Coefficients
)!(!
!
knk
n
k
nCkn
Reason:
The Binomial Theorem
nyx
0222
111
00 yxCyxCyxCyxC n
nnn
nn
nn
n
022110
210yx
n
nyx
nyx
nyx
n nnnn
Binomial Coefficients can also be calculated using Pascal’s triangle
11 1
1 2 11 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
Random Variables
Probability distributions
Definition:
A random variable X is a number whose value is determined by the outcome of a random experiment (random phenomena)
Examples1. A die is rolled and X = number of spots
showing on the upper face.2. Two dice are rolled and X = Total number
of spots showing on the two upper faces.3. A coin is tossed n = 100 times and
X = number of times the coin toss resulted in a head.
4. A person is selected at random from a population and
X = weight of that individual.
5. A sample of n = 100 individuals are selected at random from a population (i.e. all samples of n = 100 have the same probability of being selected) .
X = the average weight of the 100 individuals.
In all of these examples X fits the definition of a random variable, namely:– a number whose value is determined by the
outcome of a random experiment (random phenomena)
Probability distribution of a Random Variable
Random variables are either
• Discrete– Integer valued – The set of possible values for X are integers
• Continuous– The set of possible values for X are all real
numbers – Range over a continuum.
Examples
• Discrete
– A die is rolled and X = number of spots showing on the upper face.
– Two dice are rolled and X = Total number of spots showing on the two upper faces.
– A coin is tossed n = 100 times and X = number of times the coin toss resulted in a head.
Examples
• Continuous– A person is selected at random from a
population and X = weight of that individual.– A sample of n = 100 individuals are selected
at random from a population (i.e. all samples of n = 100 have the same probability of being selected) . X = the average weight of the 100 individuals.
The probability distribution of a discrete random variable is describe by its :
probability function p(x).
p(x) = the probability that X takes on the value x.
Examples
• Discrete
– A die is rolled and X = number of spots showing on the upper face.
– Two dice are rolled and X = Total number of spots showing on the two upper faces.
x 1 2 3 4 5 6
p(x) 1/6 1/6 1/6 1/6 1/6 1/6
x 2 3 4 5 6 7 8 9 10 11 12p(x) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
Graphs
To plot a graph of p(x), draw bars of height p(x) above each value of x.
Rolling a die
0
1 2 3 4 5 6
Rolling two dice
0
Note:1. 0 p(x) 1
2.
3.
x
xp 1
b
ax
xpbXaP )(
The probability distribution of a continuous random variable is described by its :
probability density curve f(x).
i.e. a curve which has the following properties :• 1. f(x) is always positive.
• 2. The total are under the curve f(x) is one.
• 3. The area under the curve f(x) between a and b is the probability that X lies between the two values.
0
0.005
0.01
0.015
0.02
0.025
0 20 40 60 80 100 120
f(x)
An Important discrete distribution
The Binomial distribution
Suppose we have an experiment with two outcomes – Success(S) and Failure(F).
Let p denote the probability of S (Success).
In this case q=1-p denotes the probability of Failure(F).
Now suppose this experiment is repeated n times independently.
Let X denote the number of successes occuring in the n repititions.
Then X is a random variable.
It’s possible values are
0, 1, 2, 3, 4, … , (n – 2), (n – 1), n
and p(x) for any of the above values of x is given by:
xnxxnx qpx
npp
x
nxp
1
X is said to have the Binomial distribution with parameters n and p.
Summary:
X is said to have the Binomial distribution with parameters n and p.
1. X is the number of successes occuring in the n repititions of a Success-Failure Experiment.
2. The probability of success is p.
3. xnx pp
x
nxp
1
Examples:
1. A coin is tossed n = 5 times. X is the number of heads occuring in the 5 tosses of the coin. In this case p = ½ and
3215
215
21
21
555
xxxxp xx
x 0 1 2 3 4 5
p(x)321
325
325
321
3210
3210
Random Variables
Numerical Quantities whose values are determine by the outcome of a
random experiment
Discrete Random VariablesDiscrete Random Variable: A random variable usually assuming an integer value.
• a discrete random variable assumes values that are isolated points along the real line. That is neighbouring values are not “possible values” for a discrete random variable
Note: Usually associated with counting• The number of times a head occurs in 10 tosses of a coin
• The number of auto accidents occurring on a weekend
• The size of a family
Continuous Random Variables
Continuous Random Variable: A quantitative random variable that can vary over a continuum
• A continuous random variable can assume any value along a line interval, including every possible value between any two points on the line
Note: Usually associated with a measurement• Blood Pressure
• Weight gain
• Height
Probability Distributionsof a Discrete Random Variable
Probability Distribution & Function
Probability Distribution: A mathematical description of how probabilities are distributed with each of the possible values of a random variable.
Notes: The probability distribution allows one to determine probabilities
of events related to the values of a random variable. The probability distribution may be presented in the form of a
table, chart, formula.
Probability Function: A rule that assigns probabilities to the values of the random variable
x 0 1 2 3
p(x) 6/14 4/14 3/14 1/14
ExampleIn baseball the number of individuals, X, on base when a home run is hit ranges in value from 0 to 3. The probability distribution is known and is given below:
P X( )the random variable equals 2 p ( ) 23
14
Note: This chart implies the only values x takes on are 0, 1, 2, and 3. If the random variable X is observed repeatedly the probabilities,
p(x), represents the proportion times the value x appears in that sequence.
2least at is variablerandom the XP 32 pp 14
4
14
1
14
3
A Bar Graph
No. of persons on base when a home run is hit
0.429
0.286
0.214
0.071
0.000
0.100
0.200
0.300
0.400
0.500
0 1 2 3
# on base
p(x)
Comments:Every probability function must satisfy:
1)(0 xp
1. The probability assigned to each value of the random variable must be between 0 and 1, inclusive:
x
xp
1)(
2. The sum of the probabilities assigned to all the values of the random variable must equal 1:
b
ax
xpbXaP )(3.
)()1()( bpapap
Mean and Variance of aDiscrete Probability Distribution
• Describe the center and spread of a probability distribution
• The mean (denoted by greek letter (mu)), measures the centre of the distribution.
• The variance (2) and the standard deviation () measure the spread of the distribution.
is the greek letter for s.
Mean of a Discrete Random Variable• The mean, , of a discrete random variable x is found by
multiplying each possible value of x by its own probability and then adding all the products together:
Notes: The mean is a weighted average of the values of X.
x
xxp
kk xpxxpxxpx 2211
The mean is the long-run average value of the random variable.
The mean is centre of gravity of the probability distribution of the random variable
-
0.1
0.2
0.3
1 2 3 4 5 6 7 8 9 10 11
2
Variance and Standard DeviationVariance of a Discrete Random Variable: Variance, 2, of a discrete random variable x is found by multiplying each possible value of the squared deviation from the mean, (x )2, by its own probability and then adding all the products together:
Standard Deviation of a Discrete Random Variable: The positive square root of the variance:
x
xpx 22
2
2
xx
xxpxpx
22 x
xpx
ExampleThe number of individuals, X, on base when a home run is hit ranges in value from 0 to 3.
x p (x ) xp(x) x 2 x 2 p(x)
0 0.429 0.000 0 0.0001 0.286 0.286 1 0.2862 0.214 0.429 4 0.8573 0.071 0.214 9 0.643
Total 1.000 0.929 1.786
)(xp )(xxp )(2 xpx
• Computing the mean:
Note: • 0.929 is the long-run average value of the random variable • 0.929 is the centre of gravity value of the probability
distribution of the random variable
929.0x
xxp
• Computing the variance:
x
xpx 22
2
2
xx
xxpxpx
923.0929.786.1 2
• Computing the standard deviation:
2
961.0923.0
The Binomial distribution1. We have an experiment with two outcomes
– Success(S) and Failure(F).
2. Let p denote the probability of S (Success).
3. In this case q=1-p denotes the probability of Failure(F).
4. This experiment is repeated n times independently.
5. X denote the number of successes occuring in the n repititions.
The possible values of X are
0, 1, 2, 3, 4, … , (n – 2), (n – 1), n
and p(x) for any of the above values of x is given by:
xnxxnx qpx
npp
x
nxp
1
X is said to have the Binomial distribution with parameters n and p.
Summary:
X is said to have the Binomial distribution with parameters n and p.
1. X is the number of successes occurring in the n repetitions of a Success-Failure Experiment.
2. The probability of success is p.
3. The probability function
xnx ppx
nxp
1
Example:
1. A coin is tossed n = 5 times. X is the number of heads occurring in the 5 tosses of the coin. In this case p = ½ and
3215
215
21
21
555
xxxxp xx
x 0 1 2 3 4 5
p(x)321
325
325
321
3210
3210
0.0
0.1
0.2
0.3
0.4
1 2 3 4 5 6
number of heads
p(x
)
Computing the summary parameters for the distribution – , 2,
x p (x ) xp(x) x 2 x 2 p(x)
0 0.03125 0.000 0 0.0001 0.15625 0.156 1 0.1562 0.31250 0.625 4 1.2503 0.31250 0.938 9 2.8134 0.15625 0.625 16 2.5005 0.03125 0.156 25 0.781
Total 1.000 2.500 7.500
)(xp )(xxp )(2 xpx
• Computing the mean: 5.2
x
xxp
• Computing the variance:
x
xpx 22
2
2
xx
xxpxpx
25.15.25.7 2
• Computing the standard deviation:
2
118.125.1
Example:
• A surgeon performs a difficult operation n = 10 times.
• X is the number of times that the operation is a success.
• The success rate for the operation is 80%. In this case p = 0.80 and
• X has a Binomial distribution with n = 10 and p = 0.80.
xx
xxp
1020.080.0
10
x 0 1 2 3 4 5p (x ) 0.0000 0.0000 0.0001 0.0008 0.0055 0.0264
x 6 7 8 9 10p (x ) 0.0881 0.2013 0.3020 0.2684 0.1074
Computing p(x) for x = 1, 2, 3, … , 10
The Graph
-
0.1
0.2
0.3
0.4
0 1 2 3 4 5 6 7 8 9 10
Number of successes, x
p(x
)
Computing the summary parameters for the distribution – , 2,
)(xxp )(2 xpx
x p (x ) xp(x) x 2 x 2 p(x)
0 0.0000 0.000 0 0.0001 0.0000 0.000 1 0.0002 0.0001 0.000 4 0.0003 0.0008 0.002 9 0.0074 0.0055 0.022 16 0.0885 0.0264 0.132 25 0.6616 0.0881 0.528 36 3.1717 0.2013 1.409 49 9.8658 0.3020 2.416 64 19.3279 0.2684 2.416 81 21.743
10 0.1074 1.074 100 10.737Total 1.000 8.000 65.600
• Computing the mean: 0.8
x
xxp
• Computing the variance:
x
xpx 22
2
2
xx
xxpxpx
60.10.86.65 2
• Computing the standard deviation:
2 118.125.1