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HELIA
BITE / Applied Mathematics
FORMULAS
30.4.2017
1 (7)
Statistics
Median from grouped data:
n
 FMd 1
2

 c Md
f Md
Md  LMd
LMd = lower class boundary of median class
n = total number of observations
FMd-1 = cumulative frequency of the class preceding median class
fMd = frequency of median class
cMd = width of median class
Quartiles (Q1 ja Q3), deciles (Dn) and percentiles (Pn%) can be defined by applying the same principle.
Arithmetic mean (=average) from ungrouped and grouped data:
x
x
i
x
n
xi = single observation
n = total number of observations
fx
i
i
n
fi = frequency of class i
xi = midpoint of class i
n = total number of observations
Range and Range-Width (R):
[min ; max]
R = max - min
Standard deviation from ungrouped data (with and without arithmetic mean):
s
 x
i
x
2

2
s
n 1
From grouped data:
 f x 

2
s
fx
i
2
i
i
n 1
 x 
x  n
i
2
n
i
i
n 1
HELIA
BITE / Applied Mathematics
FORMULAS
30.4.2017
Variance (s2) and proportional variance (V):
s2 = s2 (square of standard deviation)
V
s
x
Linear regression:
y = a + bx
b
n  xi y i   xi  y i
a
n xi   xi 
2
2
y
n
i
b
x
i
n
Pearson’s coefficient of correlation (rxy):
rxy 
n   x
n   xi y i   xi  y i
2
i

  xi   n   yi   yi 
2
2
2

Spearman’s coefficient of rank correlation (rs):
rs  1 
6   di
2
n(n 2  1)
di = difference between ranks of i
n = total number of ranked items
Probability
Theoretical probability
P( A) 
k
n
k = number of ways as event can occur
n = total number of different equally likely events that can occur
P(-A) = 1 – P(A)
-A = “Not A” = complement of A
P(A and B) = P(A) × P(B|A)
P(B|A) = P(B) when A does occur
2 (7)
HELIA
BITE / Applied Mathematics
FORMULAS
30.4.2017
3 (7)
Independent events, A  B :
(1) P(A and B) = P(A) × P(B)
P(A or B) = P(A) + P(B) – P(A and B) 
(2) P(A or B) = P(A) + P(B) - P(A) × P(B)
Mutually exclusive events: [P(A and B) =0]
P(A or B) = P(A) + P(B)
Permutations and combinations:
Factorial notation:
n! = n × (n-1) × (n-2) × … × 3 × 2 × 1
0! = 1
In how many different ways can one put n items in order?
n!
Permutation:
If there are n items and k are to be placed in order, the number of different ways in which this can be
done is:
n
Pk 
n!
(n  k )!
Combination:
If there are n items and k are to be placed irrespective of order, the number of different ways in
which this can be done is:
n
n
n!
C k    
 k  k!  (n  k )!
Probability distributions:
Binomial distribution:
n
P(k )     p k  q n  k
k 
p = the probability of ‘success’
q = 1 - p = the probability of ’failure’
n = number of trials
k = number of ‘successes’
HELIA
BITE / Applied Mathematics
Poisson distribution:
FORMULAS
30.4.2017
P(k ) 
 k  e 
k!
 = mean number of successes [ = np ]
n = number of trials
k = number of ‘successes’
e = exponential constant = 2,71828
Exponential distribution: P(>T) = e-aT
P(<T) = 1 - e-aT
T = waiting time for the next ‘failure’
a = constant failure rate (= failures per time unit)
e = exponential constant = 2,71828
Normal distribution:


Standard normal distribution: x  0 ,   1
See z statistic in full on the separate sheet
If x ~N(,), then z 
x

~ N(0,1)
Normal estimate of binomial and Poisson distributions:
x  np (  ) ,   npq
Estimate of populations mean:
 is within the values x  cr 
s
n
 = population mean
x = sample mean
cr = critical factor based on the significance level (see below)
s = sample standard deviation
n = total number of observations in sample
level of accuracy
factor (cr)
95 % 99 %
1,96 2,58
99,9 %
3,30
4 (7)
HELIA
BITE / Applied Mathematics
FORMULAS
30.4.2017
Estimation of the percentage value in population:
pˆ qˆ
n
p is within values pˆ  cr 
p = percentage value in population

p = percentage value in sample (in decimal format)


q= 1- p
n = total number of observations in sample
cr = critical factor based on the significance level (see the previous page)
Statistical testing
Critical values of the z-test:
two-tailed test
one-tailed test
5%
1,960
1,645
1%
2,576
2,326
0,1 %
3,291
3,090
Critical values of the t-test:

See the separate sheet. Degrees of freedom f = n – 1
n = total number of observations in sample
Mean test
Test variable:
z
x
t

n
x

n
x = sample mean
 = comparison mean
 = population (or sample) standard deviation
n = total number of observations in sample
z -test, if n  30
t -test, if n < 30
Percentage value test (only if n > 30)
z
x  np 0
np o q 0
tai
z
P  p0
p0 q0
n
x = number of ’successes’ in sample
n = total number of observations in sample
p0 = comparison percentage value
q0 = 1 - p0
P = x/n : percentage value of ‘successes’ in sample (in decimal format)
Critical values of z-test: see above.
5 (7)
HELIA
BITE / Applied Mathematics
FORMULAS
30.4.2017
Two sample mean test (t -test):
t
x1  x 2
(n1  1) s1  (n2  1) s 2
(n1  1)  (n2  1)
2
where s 
1
1
s

n1 n 2
2
x1, s1, n1, : mean, standard deviation and number of observations in sample 1
x2, s2, n2, : mean, standard deviation and number of observations in sample 2
Degrees of freedom: f = n1 + n2 -2
Two sample percentage value test (z -test):
z
P1  P2
where
1
1 
P (1  P )    
 n1 n 2 
P
n1 P1  n2 P2
n1  n2
P1 = percentage value of ‘successes’ in sample 1 (in decimal format)
P2 = percentage value of ‘successes’ in sample 2 (in decimal format)
n1 = number of observations in sample 1
n2 = number of observations in sample 2
2 –test (Chi squared –test)
Critical values: see the separate sheet
Distribution test:
2  
(oi  ei ) 2
ei
oi = observed values
ei = expected values : ei 
row total  column total
grand total
Degrees of freedom: f = number of categories (usually = number of columns) - 1
Independency test:
 2  
i
j
(oij  eij ) 2
eij
oij = observed values
eij = expected values : eij 
row total  column total
grand total
Degrees of freedom: f = (nr. of rows -1) × (nr. of columns -1)
6 (7)
HELIA
BITE / Applied Mathematics
FORMULAS
30.4.2017
7 (7)
Queuing theory
 = average number of clients joining the queue within a time unit T
 = average number of clients that could be served within a time unit T
Time unit T can be chosen freely.
’Jam factor’ :  =


Single line system M/M/1//FIFO defaults:


(1) Number of customers in the system: n 

(2) Number of customers in queue:

(3) Time spent in the system:
w

(4) Queuing time:
wq 
 
2
nq 
 (   )
1
 

 (   )
Time from formulas (3) and (4) is presented in the same unit to which parameters  and  refer.
Intervals between to consequential arrivals follow exponential distribution
Number of arrivals / exits follow Poisson distribution
Probability for total number of customers in system:
P(n) = n(1-)
P(no queue) = P(0) + P(1)
P(queue) = 1 - P(no queue)