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Selected Statistics Examples from
Lectures
Matlab: Histograms
>> y = [1 3 5 8 2 4 6 7 8 3 2 9 4 3 6 7 4 1
535896246156987534529
659416785429679253196
843679134752985745436
793169567321578531975
3 4 7 9 1]’;
>> mean(y)  ans = 5.1589
16
14
12
10
8
6
4
2
0
1
2
3
4
5
6
7
8
>> var(y)  ans = 6.1726
>> std(y)  ans = 2.4845
>> hist(y,9)  histogram plot with 9 bins
>> n = hist(y,9)  store result in vector n
>> x = [2 4 6 8]’
>> n = hist(y,x)  create histogram with
bin centers specified by vector x
35
30
25
20
15
10
5
0
2
4
6
8
9
Probability Examples

Probability that at least one coin will turn heads
up from five tossed coins
»
»
»
»

Number of outcomes: 25 = 32
Probability of each outcome: 1/32
complement set
Probability of no heads: P(AC) = 1/32
Probability at least one head: P(A) = 1-P(AC) = 31/32
Probability of getting an odd number or a number
less than 4 from a single dice toss
»
»
»
»
Probability of odd number: P(A) = 3/6
Probability of number less than 4: P(B) = 3/6
Probability of both: P( A  B)  2 / 6
Probability of either:
P( A  B)  P( A)  P( B)  P( A  B)  3 / 6  3 / 6  2 / 6  2 / 3
Permutation Examples

Process for manufacturing polymer thin films
» Compute the probability that the first 6 films will be too thin
and the next 4 films will be too thick if the thickness is a
random variable with the mean equal to the desired thickness
» n1 = 6, n2 = 4
» Probability
1
6!4!
P

 0.005  0.5%
n! n1!n2 ! 10!

Encryption cipher
» Letters arranged in five-letter words: n = 26, k = 5
» Total number of different words: nk = 265 = 11,881,376
» Total number of different words containing each letter no
n!
26!
more than once:

 7,893,600
(n  k )! (26  5)!
Combination Examples

Effect of repetitions
» Three letters a, b, c taken two at a time (n = 3, k = 2)
» Combinations without repetition
 n   3
3!
     
 3 ab ac bc
 k   2  2!(3  2)!
» Combinations with repetitions
ab ac bc
 n  k  1  4 
4!

    
6
aa bb cc
 k   2  2!(4  2)!

500 thin films taken 5 at a time
» Combinations without repetitions
 n   500 
500!
   
 
 255,244,686,600
 k   5  5!(500  5)!
Matlab: Permutations and Combinations
>> perms([2 4 6])  all possible permutations of 2, 4, 6
6 4 2
6 2 4
4 6 2
4 2 6
2 4 6
2 6 4
>> randperm(6)  returns one possible permutation of 1-6
5 1 2 3 4 6
>> nchoosek(5,4)  number of combinations of 5 things taken 4 at a
time without repetitions
ans = 5
>> nchoosek(2:2:10,4)  all possible combinations of 2, 4, 6, 8, 10
taken 4 at a time without repetitions
2 4 6 8
2 4 6 10
2 4 8 10
2 6 8 10
4 6 8 10
Continuous Distribution Example

Probability density function
0.75(1  x 2 )  1  x  1
f ( x)  
0
otherwise





1
f (v)dv   0.75(1  v 2 )dv 1
1
Cumulative distribution function
x
x

1
F ( x)   f (v)dv   0.75(1  v 2 )dv 0.5  0.75 x  0.25 x 3
x  1
0


F ( x)  0.5  0.75 x  0.25 x 3

1


1  x  1
x 1
Probability of events
P(0.5  x  0.5)  F (0.5)  F (0.5)  
0.5
 0.5
0.75(1  v 2 )dv  68.75%
P( X  x)  F ( x)  0.95  0.5  0.75 x  0.25 x 3  x  0.73
Matlab: Normal Distribution

Normal distribution: normpdf(x,mu,sigma)
» normpdf(8,10,2)  ans = 0.1210
» normpdf(9,10,2)  ans = 0.1760
» normpdf(8,10,4)  ans = 0.0880

Normal cumulative distribution:
normcdf(x,mu,sigma)
» normcdf(8,10,2)  ans = 0.1587
» normcdf(12,10,2)  ans = 0.8413

Inverse normal cumulative distribution:
norminv(p,mu,sigma)
» norminv([0.025 0.975],10,2)  ans = 6.0801 13.9199

Random number from normal distribution:
normrnd(mu,sigma,v)
» normrnd(10,2,[1 5])  ans = 9.1349
10.2507 10.5754 7.7071
6.6688
Matlab: Normal Distribution Example


The temperature of a bioreactor follows a
normal distribution with an average temperature
of 30oC and a standard deviation of 1oC. What
percentage of the reactor operating time will the
temperature be within +/-0.5oC of the average?
Calculate probability at 29.5oC and 30.5oC, then
calculate the difference:
0.4
» p=normcdf([29.5 30.5],30,1)
p = [0.3085 0.6915]
» p(2) – p(1)
0.3829
0.3
0.2
0.1
0
25
26
27
28
29
30
31
Temperature

The reactor temperature will be within +/- 0.5oC
of the average ~38% of the operating time
32
33
34
35
Discrete Distribution Example

Matlab function: unicdf(x,n),
>> x = (0:6);
>> y = unidcdf(x,6);
>> stairs(x,y)
Poisson Distribution Example

Probability of a defective thin polymer film p = 0.01

What is the probability of more than 2 defects in a
lot of 100 samples?


Binomial distribution: m = np = (100)(0.01) = 1
Since p <<1, can use Poisson distribution to
approximate the solution.
2
2
P( X  2)  F (2)   f ( x j )  
m
xj
e
m
e

1 10
0!
 
x j!
P( X  2)  1  P( X  2)  8.03%
j 0
j 0
11
1!
12
2!
  0.9197
Matlab: Maximum Likelihood


In a chemical vapor deposition process to make a
solar cell, there is an 87% probability that a
sufficient amount of silicon will be deposited in
each cell. Estimate the maximum likelihood of
success in 5000 cells.
s = binornd(n,p) – randomly generate the number of
positive outcomes given n trials each with
probability p of success
» s = binornd(5000,0.87)
s=4338

phat = binofit(s,n) – returns the maximum likelihood
estimate given s successful outcomes in n trials
» phat = binofit(s,5000)
phat = 0.8676
Mean Confidence Interval Example

Measurements of polymer molecular weight (scaled
by 10-5)
1.25

1.36 1.22 1.19 1.33 1.12 1.27 1.27 1.31 1.26
Confidence interval
  0.95 m  10  1  9 F (c)  0.975 c  2.26
x  1.26 s 2  0.0049
(2.26) 0.0049
k
 0.050 CONF0.95 1.21  m  1.31
10
Variance Confidence Interval Example

Measurements of polymer molecular weight (scaled
by 10-5)
1.25

1.36 1.22 1.19 1.33 1.12 1.27 1.27 1.31 1.26
Confidence interval
  0.95 m  9 F (c1 )  0.025 c1  2.70 F (c2 )  0.975 c2  19.02
(9)(0.0049)
(9)(0.0049)
s  0.0049 k1 
 0.0163 k 2 
 0.0023
2.70
19.02
2


CONF0.95 0.002   2  0.016
Matlab: Confidence Intervals
>> [muhat,sigmahat,muci,sigmaci]
= normfit(data,alpha)
data: vector or matrix of data
 alpha: confidence level = 1-alpha
 muhat: estimated mean
 sigmahat: estimated standard deviation
 muci: confidence interval on the mean
 sigmaci: confidence interval on the standard deviation
>> [muhat,sigmahat,muci,sigmaci] = normfit([1.25 1.36
1.22 1.19 1.33 1.12 1.27 1.27 1.31 1.26],0.05)
muhat = 1.2580
sigmahat = 0.0697
muci = 1.2081
1.3079
sigmaci = 0.0480
0.1273

Mean Hypothesis Test Example





Measurements of polymer molecular weight
x  1.258 s 2  0.0049
Hypothesis: m0 = 1.3 instead of the alternative m1 < m0
Significance level: a = 0.10
Degrees of freedom: m = 9
Critical value
P(T  c~) m0  1  a  0.90  c~  1.38
P(T  c) m0  a  0.10  c  c~  1.38

Sample t
1.258  1.3
t
 1.897  c  1.38
0.0049 10

Reject hypothesis
Polymerization Reactor Control
Monomers
Catalysts
Hydrogen
Solvent
Polymer
On-line
measurements
Lab
measurements

Monomer Content
Viscosity
Lab measurements
» Viscosity and monomer content of polymer every four hours
» Three monomer content measurements: 23%, 18%, 22%

Mean content control
» Expected mean and known variance: m0 = 20%, 2 = 1
» Control limits:
LCL  m0  2.58
2
1
2
 20  2.58
 18.5% UCL  m0  2.58
 21.5%
n
3
n
» Sample mean: 21%  process is in control
Goodness of Fit Example

Maximum likelihood estimates
1 n
mˆ   x j  x  364.7
n j 1
1 n
2
ˆ   ( x j  x ) 2  712.9
n j 1
Goodness of Fit Example cont.
Matlab: Goodness of Fit Example

Find maximum likelihood estimates for µ and σ of a normal
distribution with function mle
>> data=[320 … 360];
>> phat = mle(data)
phat =

364.7
26.7
Analyze dataset for goodness of fit using function chi2gof –
usage: [h,p,stats]=chi2gof(x,’edges’,edges) for data in vector x
divided into intervals with the specified edges
»
»
»
»
If h = 1, reject hypothesis at 5% significance
If h = 0, accept hypothesis at 5% significance
p: probability of observing given statistic
stats: includes chi-square statistic and degrees of freedom
>> [h,p,stats]=chi2gof(data,’edges’,[-inf,325:10:405,inf]);
>> h = 0
>> p = 0.8990
>> chi2stat = 2.8440
>> df = 7