Probability Theory and the use of the Normal Curve.
By: Pat McGillion, Examiner: Business Mathematics and Quantitative Methods
Introduction. Question 3 sometimes relates to probability theory and the use of the
Normal Curve. This area is underpinned by Section 4 of the Syllabus. When asked
the question is often inadequately attempted. When a probability is derived it may not
be supported by an understanding of the theory and principles of probability which is
usually required as a subsection of the question. Appropriate diagrams are often
requested to support the calculations. The general principles, theory and process is
set out in the following and supported by a simple example.
Probability and the Normal Distribution.
Many quantitative techniques treat situations with certainty, that is, they use
deterministic models to find the solution. This gives a single point estimate. It is
often possible, however, for data to take a range of values. Since there is variability
in the data used decisions are made with some degree of uncertainty. In some
cases it is often prudent to examine the range of values. This can be done by using
a numerical measure of likelihood to account for variability and probability gives a
clearer understanding of the effects of variability by measuring the likelihood of
occurrence. Probability distributions are used to help with the calculation or
probability. Such distributions are ways of describing how likely the various
outcomes from a process will be and are often presented by means of a graph. In
estimating probabilities a normal distribution of data naturally occurs in many events
and practical business situations. Thus, many business phenomena generate
random variables with probability distributions that are well approximated to a normal
distribution.
However, the adequacy of the normal approximation to an existing population can be
determined by comparing the relative frequency of a large sample of the data to the
normal probability distribution. The characteristics of the normal distribution are: it is
perfectly symmetric about its mean, μ and its spread is determined by the value of its
standard deviation, σ. To graph the normal curve it is necessary to know the
numerical values of μ and σ. Although there are an infinitely large number of normal
curves, one for each pair of values of μ and σ, a simple table called the ‘standard
normal distribution’ with μ = 0 and σ = 1, is used. A random variable with a standard
normal distribution is denoted by z. All normal random variables are converted to
standard normal to find probabilities. Statistical tables are developed to permit the
value of probability to be read after the normalization process is completed. Since all
probabilities associated with standard normal random variables can be depicted as
areas under the standard normal curve, it is advisable to draw the curve and equate
the desired probability to an area.
Property of Normal distributions: If x is a normal random variable with mean μ and
standard deviation σ, then the random variable z, defined by z = z – μ
σ
2
has a standard normal distribution denoted by Z = N(0,1 ). The value of z describes
the number of standard deviations between x and μ. In order to find a probability
corresponding to a normal random variable the following steps should be followed:
sketch the normal distribution and indicate the mean of the random variable.
Highlight the area, that is, the probability that is required.
Convert the x values to standard normal random variable z values by using z
= z – μ. Show the z values on the sketch.
σ
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Using the Normal tables, find the area corresponding to the z values. The
symmetry of the normal distribution can be used to find areas corresponding
to negative z values. It should be noted that different Normal tables may be
used – the most common either reading from the centre of the distribution or
the left side tail; the tables used in this example are provided by the Institute
at examinations and are available upon request.
To demonstrate the use of probability distributions, the following example is used:
Example:
CPA Distributors has a policy of cash payment. It has recently revised this policy to
cater for larger customers with a good credit rating who purchase bulk volumes of
goods. The auditors report that the mean time between sending an invoice and
receiving payment is 20 days with a standard deviation of 5 days. Assuming that the
distribution is normally distributed
(i)
Find the probability that a bill will not be paid until after 25 days;
(ii)
If 2000 bills were issued, what percentage would be paid in less than
10 days?
(iii)
From 2000 bills how many would you expect to be paid between 15
and 21 days?
(i)
Transform the 25 days from X = (20, 52) to Z = N(0, 12)
z = x – μ = 25 – 20 = 1
σ
5
From the tables at z = 1, probability = 0.8413
The prob (x > 25) = prob (z > 1) = 1 – 0.8413 = 0.1587, represented by the shaded
area in the diagram.
Therefore, the probability that a bill will not be paid until after 25 days is 16%.
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(ii)
Transforming the 10 days from X to Z; z = x – μ = 10 – 20 = -2
σ
5
Prob (x, 10) = Prob (z, -2); from tables at z = -2, prob = 1 – 0.9773 = 0.0227,
represented by the shaded area in the diagram.
That is, the expected number of bills paid within 10 days is 2000 x 0.0227 = 45.
(iii)
Transforming the 15 and 21 days from X to Z:
(x = 15) z = x – μ = 15 – 20 = -1;
σ
5
(x = 21) z = x – μ = 21 – 20 = 0.2
σ
5
Prob (15 < x < 21) = Prob (-1 < z < 1.2), represented by the shaded area in the
diagram.
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This area, and the probabilities, are found by:
(a)
(b)
(c)
1 – area in tail at z = 1, that is, 1 – 0.8413 = 0.1587
Area at z = 0.2, that is, 0.5793
Required area is 0.5793 – 0.1587 = 0.4206
The expected number of bills paid between 15 and 21 days is 2000 x 0.4206 = 841,
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