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 previously-mentioned (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 (%) 0-9.9 10-19.9 20-29.9 30-39.9 40-49.9Frequency (No. of students) 1 4 8 17 47

Mark Scored (%) 50-59.9 60-69.9 70-79.9 80-89.9 90-100Frequency (No. of students) 53 39 25 11 3

0 10 20 30 40 50 60 70 80 90 1000

10

20

30

40

50

Mark (per cent)

Frequency (No. of students)

1

4

8

17

47

53

39

25

11

3

Frequency polygonsThe 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.

0 10 20 30 40 50 60 70 80 90 1000

10

20

30

40

50

60

Mark (per cent)

Frequency (No. of students)

1

4

8

17

47

53

39

25

11

3

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):

2

2

1

22

1)(

x

xx

x

exP

The resulting plot of P(x) as x varies is a ‘bell-shaped’ curve, as

shown in the next slide.

0

-4 -3 -2 -1 0 1 2 3 4

0.1

0.2

0.3

0.4P(x)

z = no. of standard deviations of x from its mean valuez = 0 for mean of x

x

Notes1. The “x axis” is in STANDARD DEVIATIONS2. 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.

ExampleSay 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 -3 -2 -1 0 1 2 3 4

0.4P(x)

z for x = 55 per cent

x

z for x = 70 per centz for x = 40 per cent

area = probabilitythat student fails

area = probability thatstudent gets a ’first’

SD from Area SD from Areamean mean2.00 0.0227 0.95 0.17101.95 0.0256 0.90 0.18401.90 0.0287 0.85 0.19761.85 0.0321 0.80 0.21181.80 0.0359 0.75 0.22661.75 0.0400 0.70 0.24191.70 0.0445 0.65 0.25781.65 0.0495 0.60 0.27421.60 0.0548 0.55 0.29111.55 0.0606 0.50 0.30851.50 0.0668 0.45 0.32631.45 0.0735 0.40 0.34461.40 0.0807 0.35 0.36321.35 0.0885 0.30 0.38211.30 0.0968 0.25 0.40131.25 0.1056 0.20 0.42071.20 0.1150 0.15 0.44041.15 0.1250 0.10 0.46021.10 0.1356 0.05 0.48011.05 0.1468 0.00 0.50001.00 0.1586

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 real-world situations, more

‘standardised’ probability distributions are required.

The other one we shall briefly look at is the negative-

exponential 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):

N

t

tN

period time

failures ofnumber

period time

components ofnumber in the change

This is called a differential equation and would normally be

written

Ndt

dN

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:

tNen

We can plot this negative exponential function as the following curve relating the

remaining number of components n to time:

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

Time units

Multiple of initial numberof components (N)

n = Ne -t

The End