Transcript BSc/HND IETM Week 8 - Some Probability Distributions

BSc/HND IETM Week 9/10 Some Probability Distributions
When we looked at the histogram
a few weeks ago, we were looking
at frequency distributions.
It is possible to convert such
frequency distributions into
probability distributions, such that
the probability of encountering
some particular value (or range of
values) of x is plotted on the
vertical axis, rather than the
number of occurrences of that
value of x.
There are a few standard forms of
such distributions, which make
analysis rather easy - so long as
the data really do fit the chosen
form.
We shall look at two of these
standard forms, the normal and
the negative exponential
distributions.
Probability distributions from
frequency distributions
Suppose that our previouslymentioned (and, sadly,
hypothetical) optional unit for
your course, ‘Flower Arranging
for Engineers’, becomes
extremely popular.
In fact, it becomes so popular that
it is studied by 208 students,
from all the various BSc courses
in the School.
In an effort to analyse the
performance of the students, so as
to determine if any improvements
to the unit are required, we might
decide to plot a histogram of the
final marks obtained.
As we know, this is a frequency
distribution, and might be
obtained from the following
summary of the students’ scores,
as shown:
Mark Scored (%)
Frequency (No. of students)
Mark Scored (%)
Frequency (No. of students)
0-9.9
1
50-59.9
53
10-19.9
4
60-69.9
39
20-29.9
8
70-79.9
25
30-39.9
17
80-89.9
11
40-49.9
47
90-100
3
53
F req u en cy (N o . o f stu d en ts)
50
47
39
40
30
25
20
17
11
10
8
4
3
1
0
0
10
20
30
40
50
60
70
80
90
100
M ark (p er cen t)
Frequency polygons
The first step in the conversion is
to change from the histogram to
what is called a frequency
polygon. This is simply a line
graph, joining the centres of each
of the chosen data intervals.
At the ends, our frequency
polygon reaches the zero axis as
shown, since no student can
obtain less than zero or more
than 100 per cent. In situations
when this doesn’t apply, it is
conventional to terminate the
polygon on the zero axis, half
way through the next interval.
60
Frequency (No. of students)
53
50
47
39
40
30
25
20
17
11
10
8
4
3
1
0
0
10
20
30
40
50
60
70
80
90
100
Mark (per cent)
It is very easy to obtain
probability distributions from
diagrams such as those above. All
that is necessary is to divide each
frequency by the total number of
(in this case) students, to obtain
the probability of any individual
student, selected at random,
obtaining a mark in a particular
range.
For example, to convert the
histogram on page 1, or the
frequency polygon on page 2, into
probability distributions, simply
divide every number on the
vertical axis (and therefore also
the numbers written on the plots)
by 208.
Thus, the vertical axes would
now be calibrated in
probabilities from zero to
53/208 = 0.255.
The probability of any given
student obtaining a mark in the
range 40 to 49.9 per cent will be
47/208 = 0.226. The probability
of a student scoring 90 per cent or
more will be 3/208 = 0.0144, etc.
The normal distribution
It is not very surprising that the
marks distribution (frequency or
probability) looks like the
diagrams above.
In a fair examination, taken by a
large number of students, we
would expect that only a few
students would obtain either
abysmally low marks or
astronomically high marks.
We would expect the majority of
marks to be ‘somewhere in the
middle’, with a ‘tail’ at both the
low and the high ends of the
range.
We would expect the majority of
marks to be ‘somewhere in the
middle’, with a ‘tail’ at both the
low and the high ends of the
range.
This is what we see above.
Several real-life situations fit this
general form of distribution,
where it is most likely that results
will be clustered around the
centre of some range, with
outlying values tailing off
towards the ends of the range.
Wisniewski, in his ‘Foundation’
text, uses an example based on
the distributions of the weights
of breakfast cereal packed by
machines into boxes.
There should always ideally be
the stated amount in a box but,
inevitably, some boxes will be
lighter, and some heavier. There
will be the odd ‘rogue’ boxes a
long way from the mean.
To make it easier to cope with
such situations, they are often
assumed to fit a standardised
probability distribution, called the
normal distribution.
By doing this, it is possible to use
standard printed tables to make
predictions such as (for example),
how many students would be
expected to score less than 40 per
cent
To allow standard tables to be
used, we need to assume a certain
fixed shape of probability
distribution, and we also need to
define it in terms of mean and
standard deviation.
We cannot define it in terms of
actual data values (e.g.
examination marks, or weight of
cereal in a box), otherwise we
would need a different set of
tables for every new problem.
The normal distribution curve is
actually defined by a rather
unpleasant formula (but we don’t
need to use it, as we are going to
use tables which have been
derived from it by someone else).
If the variable in which we are
interested is x (e.g. a mark in per
cent, or the weight of cereal in a
box in kg), the mean value of x
is and the standard deviation of
the data set is x,
then the normal distribution curve
is defined by the probability that x
will take a particular value (P(x))
obeying the following relationship
(I believe there is an error in
Wisniewski’s version):
P( x ) 
1
2
2
x
e
1  x x 

 
2  x 
2
The resulting plot of P(x) as x
varies is a ‘bell-shaped’ curve, as
shown in the next slide.
0.4
P(x)
0.3
0.2
0.1
x
0
-4
-3
-2
-1
z = 0 for mean of x
0
1
2
3
4
z = no. of standard deviations
of x from its mean value
Notes
1. The “x axis” is in STANDARD
DEVIATIONS
2. The total area under the graph is
1 unit.
3. The area under the graph
between two values of x gives the
probability that the quantity will
be between those values.
Example
Say that a large set of
examination results has a mean
of 55 per cent, and a standard
deviation of 15 per cent.
How many students would we
expect to fail the examination (if
we define a failure as obtaining
less than 40 per cent), and how
many students would we expect
to get a first-class result (defined
as obtaining 70 per cent or more)?
0.4
P(x)
area = probability that
student gets a ’first’
area = probability
that student fails
x
0
-4
-3
-2
z for x = 40 per cent
-1
0
z for x = 55 per cent
1
2
3
z for x = 70 per cent
4
SD from
mean
2.00
1.95
1.90
1.85
1.80
1.75
1.70
1.65
1.60
1.55
1.50
1.45
1.40
1.35
1.30
1.25
1.20
1.15
1.10
1.05
1.00
Area
0.0227
0.0256
0.0287
0.0321
0.0359
0.0400
0.0445
0.0495
0.0548
0.0606
0.0668
0.0735
0.0807
0.0885
0.0968
0.1056
0.1150
0.1250
0.1356
0.1468
0.1586
SD from
mean
0.95
0.90
0.85
0.80
0.75
0.70
0.65
0.60
0.55
0.50
0.45
0.40
0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.00
Area
0.1710
0.1840
0.1976
0.2118
0.2266
0.2419
0.2578
0.2742
0.2911
0.3085
0.3263
0.3446
0.3632
0.3821
0.4013
0.4207
0.4404
0.4602
0.4801
0.5000
X = 1.0 (1 SD from mean)
First : Probability 0.1587
Fail: Also 0.1587 !
The negative exponential
distribution
To cover a wider range of realworld situations, more
‘standardised’ probability
distributions are required.
The other one we shall briefly
look at is the negativeexponential distribution.
This is also sometimes called a
‘failure-rate’ curve, because it
tends to describe how
components fail with time.
If a certain number of components
is manufactured and put into
service, it is reasonable to assume
that they will all eventually fail.
If a certain number of components
is manufactured and put into
service, it is reasonable to assume
that they will all eventually fail.
The probability of any one of the
components failing during a given
time period might well depend on
how many components are left in
service.
Choose to measure time t in the
best units for the problem
(seconds, months, years, etc.).
Technically, the unit chosen
should be short compared with the
expected lifetime of a component,
so that any given component is
expected to last for many time
units.
Let  be the failure rate, that is,
the proportion of components
expected to fail in one time unit.
This means that  must have
‘dimensions’ of (1/time). In the
example above, we said that 1 per
cent of components might fail in
three years so, in that case, the
failure rate   0.01/3 (proportion
per year).
This can also be viewed as a
probability - there is a probability
of 0.01/3 that any given
component will fail in a given
period of one year.
Therefore, to find the proportion
of components expected to fail
over a time t (measured in our
chosen units), we need the
quantity t. This is now
dimensionless - it is actually the
probability that any given
component will fail over the
stated time period.
We can now state the rate of
change of the number of
components as follows (it is
negative, because the number
decreases as time passes):
changein thenumber of components
timeperiod
number of failures

timeperiod
N t

  N
t
This is called a differential
equation and would normally be
written
dN
  N
dt
in which the quantity dN / dt is
to be interpreted as the rate of
change of N as the time
progresses.
It turns out that:
n  Ne
 t
We can plot this negative
exponential function as the
following curve relating the
remaining number of
components n to time:
Multiple of initial number
of components (N)
1
0.8
0.6
n = Ne -t
0.4
0.2
0
0
1
2
3
4
5
6
Time units
The End