Probability And Statistics

Review of Probability and statistics Sea of integration in science and engineering By Oladokun Sulaiman

Slide navigator  Statistics  Probability  Application

ship and structure design  Goal base system initiative for reliability and risk assessment

1.0 Introduction Population Sample

Hybrid FSA Goal based probability emperical

Sampling a Process Process A sequence of operations that takes inputs (labor, raw materials, methods, machines, and so on) and turns them into outputs (products, services, and the like.)

Inputs

Process

Outputs

A process is in statistical control if it displays constant level and constant variation.

2.0 Descriptive Statistics

Descriptive shapes Stem and Leaf Display Frequency Histogram

1 5 12 21 (11) 17 7 1

29 8 30 1344 30 5666889 31 001233444 31 55566777889 32 0001122344 32 556788 33 3

The Normal Curve

Skewness Left Skewed

Symmetric

Right Skewed

The Mean Population X1, X2, …, XN µ

x

Population Mean N

µ=

Sample x1, x2, …, xn

∑ Xi i=1

N

Sample Mean n

∑x

x = i=1 n

i

The Sample Mean

The sample mean x is defined as n

x=

∑x i =1

n

i

x1 + x2 + ... + xn = n

and is a point estimate of the population mean .

The Median Example 2.6: Internist’s Salaries (x$1000) 127 132 138 141 144 146 152 154 165 171 177 192 241 Since n = 13 (odd,) then the median is the middlemost or 7th 1 10 0 measurement, Md=152 2 11 0 Example 2.2: The Accounts Receivable Case The value 16 occurs 9 times therefore: Mo = 16

4 7 11 18 27 (8) 30 24 19 16 13 10 8 5 3 2 1 1

12 00 13 000 14 0000 15 0000000 16 000000000 17 00000000 18 000000 19 00000 20 000 21 000 22 000 23 00 24 000 25 00 26 0 27 0 28 29 0

The Range Range = largest measurement - smallest measurement Example: Internist’s Salaries (in thousands of dollars) 127 132 138 141 144 146 152 154 165 171 177 192 241 Range = 241 - 127 = 114 ($114,000)

The Standard Deviation Population Standard Deviation, σ: Sample Standard Deviation, s:

σ= σ

2

s = s2

The Variance Population X1, X2, …, XN

σ

Sample x1, x2, …, xn

s2

2

Population Variance

Sample Variance

N

σ2 =

2 (X µ ) ∑ i i=1

N

n

s2 =

2 (x x ) ∑ i i=1

n -1

The Empirical Rule for Normal Populations If a population has mean µ and standard deviation σ and is described by a normal curve, then 68.26% of the population measurements lie within one standard deviation of the mean: [µ−σ, µ+σ] 95.44% of the population measurements lie within two standard deviations of the mean: [µ−2σ, µ+2σ] 99.73% of the population measurements lie within three standard deviations of the mean: [µ−3σ, µ+3σ]

Chebyshev’s Theorem Let µ and σ be a population’s mean and standard deviation, then for any value k>1, At least 100(1 - 1/k2 )% of the population measurements lie in the interval: [µ−kσ, µ+kσ]

2.4 Percentiles and Quartiles For a set of measurements arranged in increasing order, the pth percentile is a value such that p percent of the measurements fall at or below the value and (100-p) percent of the measurements fall at or above the value. The first quartile Q1 is the 25th percentile The second quartile (or median) Md is the 50th percentile The third quartile Q3 is the 75th percentile. The interquartile range IQR is Q3 - Q1

Example: Quartiles 20 customer satisfaction ratings:

1 3 5 5 7 8 8 8 8 8 8 9 9 9 9 9 10 10 10 10

Md = (8+8)/2 = 8 Q1 = (7+8)/2 = 7.5

Q3 = (9+9)/2 = 9

IRQ = Q3 - Q1 = 9 - 7.5 = 1.5

Describing Qualitative Data

Population and Sample Proportions Population X1, X2, …, XN

Sample x1, x2, …, xn

pˆ

p

Population Proportion

Sample Proportion n

∑x

pˆ = i=1 n

i

xi = 1 if characteristic present, 0 if not

Pareto Chart Scatter Plots

Pareto Chart of Labeling Defects

2.0 Probability If E is an experimental outcome, then P(E) denotes the probability that E will occur and Conditions

0 ≤ P( E ) ≤ 1 If E can never occur, then P(E) = 0 If E is certain to occur, then P(E) = 1 The probabilities of all the experimental outcomes must sum to 1. Interpretation: long-run relative frequency or subjective

The Sample Space

Computing Probabilities of Events An event is a set (or collection) of experimental outcomes.

The probability of an event is the sum of the probabilities of the experimental outcomes that belong to the event. If the sample space outcomes (or experimental outcomes) are all equally likely, then the probability that an event will occur is equal to the ratio the number of sample space outcomes that correspond to the event The total number of sample space outcomes

Event Relations The complement A of an event A is the set of all sample space outcomes not in A. Further, P(A) = 1 - P(A) Union of A and B, A

B

Elementary events that belong to either A or B (or both.) Intersection of A and B, A B Elementary events that belong to both A and B.

The Addition Rule for Unions The probability that A or B (the union of A and B) will occur is P(A B) = P(A) + P(B) - P(A B)

A and B are mutually exclusive if they have no sample space outcomes in common, or equivalently if P(A B) = 0

Conditional Probability The probability of an event A, given that the event B has occurred is called the “conditional probability of A given B” and is denoted as . Further, P(A | B)

P(A B) P(A|B) = P(B)

Independence of Events

Two events A and B are said to be independent if and only if: P(A|B) = P(A) or, equivalently, P(B|A) = P(B)

The Multiplication Rule for Intersections The probability that A and B (the intersection of A and B) will occur is P(A ∩ B) = P(A) P(B | A) = P(B) P(A | B)

If A and B are independent, then the probability that A and B (the intersection of A and B) will occur is P(A ∩ B) = P(A) P(B) = P(B) P(A)

Contingency Tables P(R 1 )

P(R1 C1 )

R1 R2 Total P(R2 C2 )

C1 .4 .1 .5

C2 .2 .3 .5

Total .6 .4 1.00 P(C 2 )

Application to ship wave loading LINIEAR WAVE THEORY VN l

z c x

n

z (x,t)

H

still water level d

N VN = ∂ f / ∂N = 0

IREGULAR WAVES For an irregular seaway, the amplitude and period is constantly changing from time to time and place to place. This can be illustrated as follows: Tc

Tc

Tz Hw

ζ (t)

t (time) zero line

(Lw)c

ζ (x)

(Lw)z x (space)

IRREGULARITY OF THE SEAWAY AND THE HISTOGRAM Histogram is basically derived from the record of wave elevations as shown: mean elevation ζa mean

Percent occurence

ζa mean

ζ

Ts/s Ts

-6

-4

-2

0

2

4

6

8 ζ (m) a

IRREGULARITY OF THE SEAWAY AND THE HISTOGRAM Also, another way of plotting the Histogram is to plot the cumulative distribution as by diagram below: 100 %

The significant value or the average value of 1/10 or 1/100 highest value can be obtained using the cumulative distribution diagram.

50

2

4

6

8

10 ζa (m)

experience Percentage of waves below ζa = 2m theoretical Rayleigh fits the histograms for the height (double amplitude) very well.

It

has

been

found by that the curve wave

WAVE SPECTRUM As proposed by Longuet–Higgins, the seaway can be described by a superposition of many sinusoidal waves, each having a particular frequency, amplitude, direction and phase. The wave surface is then expressed as the sum of a large number (N) of sinusoidal waves, N   A n. sin(kn x - ωnt +∈n ) ζ(t) = n=1

∑

The energy of each component wave per unit area of water surface is ½ pgA2n

WAVE SPECTRUM The total energy per unit surface area of the wave w+dw in the A 2n frequency range w to w + δw is ½ w ρg S (ω)   w+dw Energy Let us write, 2 ρg An S(w) δω = ½ w

∑

∑

δω

ω

Where S(w) is the wave spectrum or wave energy spectrum.

WAVE SPECTRUM Now, N 2 2 { } E A sin ( k x w t + ∈ ) E [ ζ (t) ]= n n n n

∑

n=1

[

Since mean value of sin2 is ½,   σ2 = E[ζ2(t)] - {E [ ζ(t) ] }2  

σ

2

= E∞[ζ2(t)] S(ω).dω = o

∫

= mo

]

1 N 2 An = 2 n=1

∑

WAVE SPECTRUM Some commonly quoted spectra.   i) Energy Spectrum Ordinate = A2

∴S

2 ½A

 

ii) Amplitude Spectrum   Ordinate = A2

∞

S(oω= ).dω = ∫m o

∴SA2 = 2 mo

 

iii) Height Spectrum   Ordinate = H2

∴SH2 = 8 mo

 

iv) Double-Height Spectrum Ordinate = (2H)2 ∴S(2H)2 = 32 mo  

Commonly used energy spectra, S(w) = S

½

A2

STANDARD WAVE SPECTRUM Darbyshire Wave Spectrum (1963)  

1/ 2

S(w)

=

  (ω - ωo ) A. exp( -   )  0.054 (ω- ωo +0. 265) ω - ωo +0.265>o 0, otherwise

A V ωo

= = =

0.186 X 10 –3.V4 (units are m2) wind speed (m/s) 6.284 (1.94 V ½ + 2.5 x 10 –7.V4)-1 rad/s

2

STANDARD WAVE SPECTRUM British Towing Tank Wave Spectrum   1/ 2 2   S(w) = (ω - ωo ) A. exp( -   )  0.065 (ω- ωo +0. 265) ω - ωo +0.265>o 0, otherwise   A = 21.5 (0.0625 V – 0.442)2 m2 V = wind speed (m/s) ωo = 6.142 (0.1545 V + 7.389) –1 (rad/s)

STANDARD WAVE SPECTRUM Pierson – Moskowitz Wave Spectrum. 1963 (North Atlantic) 4 -3  8.1x10  g    2   g exp . 0 . 74     6 ω  V.ω    S(w) =   V

=

wind speed (m/s) at height of 19.5m

 

In general form, S(ω) = When ω = ωmax = (0.8B) ∴S (ωmax) = A (0.8 B)

¼

–5/4 .

exp(-5/4)

 

Note:

A

=

0.0081 g2 & B

=

0.74

STANDARD WAVE SPECTRUM International Towing Tank Conf. (I.T.T.C) Wave Spectrum -     H1/ 3  Assumes only significant wave ht.  is known A 4 . exp( B / ω ). s Then use S(ω) =ω  - 2  H1/ 3   A = 8.1 x 10 –3 g2, B = 3.11/  -1 /2   5/2  H1/ 3  0.25 H1/ 3 . exp - 5   4 S(ωmax) = at ωmax =1.26

( )

( )

STANDARD WAVE SPECTRUM International Ship Structure Congress (I.S.S.C.) Wave Spectrum -

-

H1/ 3 T1/ 3 Assumes and 4 are known. A . exp( - B / ω ).   s ω 2 S(ω) =   691  mo    H , B = , T = 2 π 1 / 3     1/ 3   4 m 1   T   1/ 3

Where, A = 173  

S(ωmax) = 0.065 (H1/3)2 .T1/3.exp (-5/4) 4.85/T1/3

at

ωmax =

 

Note:

I.T.T.C. and I.S.S.C. have maxima at

STANDARD WAVE SPECTRUM If no wave data is available but wind speed is known, use the following table: Wind Speed (knots) 20 30 40 50

Sig. Wave Ht. (m) P.M 2.2 5.0 8.8 13.8

I.T.T.C 3.1 5.1 8.1 11.0

Scott (1968) 3.5 5.2 7.5 9.6

STANDARD WAVE SPECTRUM Joint North Sea Wave Project (JONSWAP) Wave Spectrum       



2  − ( ω − ω ) m ax 4 exp For Limited fetch and shallow water condition.  A 2 2 2  − 5  ω max    2 σ ω  m ax .g . exp     . *  5

γ

 4  ω   S(ω) = ω   -0.22 S J= A 0.477 X , Xx = g.x/V10 max x ( ≅ 3.3) p S max γ = σ* = σa (≅ 0.07), ω ≤ ωmax σb (≅ 0.09), ω > ωmax x = fetch, V10 = wind speed at height of 10m  

Typical wmax = 22 X

-0.33

STANDARD WAVE SPECTRUM Bretschneider Wave Spectrum -

S(ω) =

48 .7(H1/ 3 )2 T14/ 3.ω5

exp

 -1050   4 4  T1/ 3.ω  Fully developed sea

5 4

Partially developed sea

3 2 1

SPECTRAL MOMENT AND SPECTRAL BANDWITCH e --> 0

e --> 1

S (w)

S (w)

Narrow Bandwidth

w

z (t)

Broad Bandwidth

w

z (t) t

T = 2p w

t

PROBABILITY DENSITY FUNCTION P (x)

Gaussian P (x) =

(∈→ 1)

Rayleigh 1

σ 2π

x −1/ 2   σ e

2

P ( x) =

σ

(∈→ 0 ) P ( x) =

−3σ

x

−2σ

−σ

σ

2σ

2σ

x −1 / 2   σ e

x e m0 x

−1 / 2

x2 m0

2

PROBABILITY DENSITY FUNCTION

If denotes the mean of the highest 1/n waves in a sea state, then we can show that: Mean wave height,

H1= 2.50 mo

  Significant wave height,

H1/3 = 4.00 mo

  1/10

th

highest wave,

H1/10 = 5.10 mo

1/2

½

½

PROBABILITY DENSITY FUNCTION

Example Calculation – 4.1

Using ITTC formulation, plot a wave spectrum for a wing speed of 31 knot (H1/3 = 5.6 m) ! S(ω) =  

A

A ω

4 . exp( B / ω ). s

= 8.1 x 10 –3 g2 = 8.1 x 10 = 0.779512

 

B = 3.11/(H1/3)2 = 3.11/(5.6)2 = 0.099171

–3

x (9.81)2

MOTION IN IRREGULAR SEAWAY

The basic steps in predicting the vessel motions are: 1. Choose a suitable wave spectrum S(ω) 2. Calculate the encounter frequency ωE, as a function of the wave frequency w, vessel speed V, wave speed c, and heading µ ωE = ω(1 – V.c-1 cos µ) [µ = o0 for following seas, 90o for beam seas, 180o for head seas]

MOTION IN IRREGULAR SEAWAY

3. Calculate the encounter spectrum S(ωE) = S(ω) x d ω/dωE Vessel response Wave input parameter * 4. Evaluate R.A.O = Where it is assumed that;  

a. The response to each component wave is independent of the response to the other waves.  

b. The response is a linear function of the component wave amplitudes.  

[ * a commonly used wave input parameter is the wave amplitude for translational motions or wave slope for rotational motions ]

MOTION IN IRREGULAR SEAWAY

5. Plot the response spectrum SR(ωE) = S(ωE) x (R.A.O)2 (RAO)2

S (ω)

SR (ω) mR

ω

INPUT FORCE

2 mR

ω

VESSEL (RAO)2

ω

OUTPUT RESPONSE

Probability of extreme limit

Reliability

Risk assessment

Design life approach

Conclusion and discussion 

We use statistics everyday. Perhaps it is because they are so common that there is so much confusion, cynicism and uncertainty in their use. We make measurements and record data with the intent of learning something useful about the system or factors that generate those data.



Statistics helps us in that learning process. In this day of information explosion we are confronted more and more frequently with questions; "What do all those data mean?", "What are the data trying to tell me?” Statistics is the key to unlocking the answers to many of those questions.



Not only does it assist in extracting information from data, but it also tells us how to plan experiments that will provide informative data in the first place. Rather than a "smoke screen" for ignorance, statistics is a vehicle for clarity in discovering the value in good data and for weeding out numbers with little information content.

Offshore standard