Download Probability and Statistics in Engineering

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

History of statistics wikipedia, lookup

Bootstrapping (statistics) wikipedia, lookup

Time series wikipedia, lookup

Misuse of statistics wikipedia, lookup

Transcript
Probability and
Statistics in
Engineering
Philip Bedient, Ph.D.
Probability: Basic Ideas
 Terminology:




Trial: each time you repeat an
experiment
Outcome: result of an experiment
Random experiment: one with random
outcomes (cannot be predicted exactly)
Relative frequency: how many times a
specific outcome occurs within the entire
experiment.
Statistics: Basic Ideas

Statistics is the area of science that deals with
collection, organization, analysis, and
interpretation of data.
 It also deals with methods and techniques that
can be used to draw conclusions about the
characteristics of a large number of data points-commonly called a population- By using a smaller subset of the entire data.
For Example…





You work in a cell phone factory and are asked
to remove cell phones at random off of the
assembly line and turn it on and off.
Each time you remove a cell phone and turn it
on and off, you are conducting a random
experiment.
Each time you pick up a phone is a trial and the
result is called an outcome.
If you check 200 phones, and you find 5 bad
phones, then
relative frequency of failure = 5/200 = 0.025
Statistics in Engineering


Engineers apply physical
and chemical laws and
mathematics to design,
develop, test, and
supervise various
products and services.
Engineers perform tests
to learn how things
behave under stress, and
at what point they might
fail.
Statistics in Engineering
 As
engineers perform experiments, they
collect data that can be used to explain
relationships better and to reveal
information about the quality of products
and services they provide.
Frequency Distribution:
Scores for an engineering class are as follows: 58, 95, 80,
75, 68, 97, 60, 85, 75, 88, 90, 78, 62, 83, 73, 70, 70, 85,
65, 75, 53, 62, 56, 72, 79
To better assess the success of the class, we make a
frequency chart:
Now the information can be better analyzed.
For example, 3 students did poorly, and 3 did
exceptionally well. We know that 9 students
were in the average range of 70-79. We can also
show this data in a freq. histogram (PDF).
Divide each no. by 26
Cumulative Frequency


The data can be further organized by calculating the
cumulative frequency (CDF).
The cumulative frequency shows the cumulative number
of students with scores up to and including those in the
given range. Usually we normalize the data - divide 26.
Measures of Central Tendency &
Variation

Systematic errors, also called fixed errors, are
errors associated with using an inaccurate
instrument.


These errors can be detected and avoided by properly
calibrating instruments
Random errors are generated by a number of
unpredictable variations in a given measurement
situation.

Mechanical vibrations of instruments or variations in
line voltage friction or humidity could lead to random
fluctuations in observations.


When analyzing data, the mean alone cannot signal
possible mistakes. There are a number of ways to define
the dispersion or spread of data.
You can compute how much each number deviates from
the mean, add up all the deviations, and then take their
average as shown in the table below.

As exemplified in Table 19.4, the sum of deviations
from the mean for any given sample is always zero.
This can be verified by considering the following:
n
1
x   xi
n i1

di  (xi  x )
Where xi represents data points, x is the average, n
is the number of data points, and d, represents the
deviation from
the average.
n
n
n
n
d   x   x
d
i1
i1
i
i
i1
i1
i
 nx  nx  0
Therefore the average of the deviations from the
mean of the data set cannot be used to measure
the spread of agiven data set.
Instead we calculate the average of the absolute
values of deviations. (This is shown in the third
column of table 19.4 in your textbook)
For group A the mean deviation is 290, and Group
B is 820. We can conclude that Group B is more
scattered than A.
Variance
 Another
way of measuring the data is by
calculating the variance.
 Instead of taking the absolute values of
each deviation, you can just square the
deviation and find the means.
 (n-1) makes estimate unbiased
n
v

i1
(x i  x )
n 1
2
 Taking
the square root of the variance
which results in the standard deviation.
n
s
 The

i1
(x i  x )
2
n 1
standard deviation can also provide
information about the relative spread of a
data set.


The mean for a grouped distribution is calculated
from:
(xf )

x
n

Where
x = midpoints of a given range
f = 
frequency of occurrence of data in the range
n = f = total number of data points
The standard deviation for a grouped distribution is
calculated from:
2
(x  x ) f

s
n 1

Normal Distribution

We could use the probability distribution from the figures
below to predict what might happen in the future. (i.e.
next year’s students’ performance)
Normal Distribution

Any probability distribution with a bell-shaped
curve is called a normal distribution.
 The detailed shape of a normal distribution
curve is determined by its mean and standard
deviation values.
THE NORMAL CURVE

zi = (xi - x) / s
Using Table 19.11, approx. 68% of the data will
fall in the interval of -s to s, one std deviation
 ~ 95% of the data falls between -2s to 2s, and
approx all of the data points lie between -3s to 3s
 For a standard normal distribution, 68% of the
data fall in the interval of z = -1 to z = 1.
AREAS UNDER THE NORMAL CURVE


z = -2 and z = 2 (two standard deviations below and
above the mean) each represent 0.4772 of the total area
under the curve.
99.7% or almost all of the data points lie between -3s
and 3s.
Analysis of Two Histograms
Graph A is class distribution of numbers 1-10
Graph B is class distribution of semester credits
Data for A = 5.64 +/- 2.6 (much greater spread than B)
Data for B = 15.7 +/- 1.96 (smaller spread)
Skew of A = -0.16 and Skew B = 0.146
CV of A = 0.461 and CV of B = 0.125 (CV = SD/Mean)
Frequency B
Frequency A
9
8
7
6
5
4
3
2
1
0
7
6
5
4
3
2
1
0
2
3
4
5
6
7
8
9
10
12
13
14
15
16
17
18
19
20