Yard. Doç. Dr. Tarkan Erdik
Probability Distributions
Uniform and Normal Distributions- Week 7
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Probability Distributions
It has been observed that certain functions F(x) and f(x) can
successfully express the distributions of many random
variables.
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Several continuous distributions play useful roles in
engineering as in numerous other disciplines. The more
important ones are:
Uniform,
Normal,
Exponential,
Gamma,
Beta,
Weibull,
Lognormal distributions.
3
What is the difficulty is in choosing the
distribution function?
There are no general rules for this.
So how should/can we choose it?
The engineer has to make a choice based on his experience
and knowledge as regards the properties of the commonly
used distribution.
4
The comparison of the histogram of the observed data with
the chosen probability density function helps in the decision
making.
5
Uniform Distribution
The simplest type of continuous distribution is the uniform.
As implied by the name, the pdf is constant over a given
interval (for example, from a to b, where a < b).
f(x)=constant, F(x)=cx; c is constant.
6
The density function of the continuous uniform random
variable X on the interval [A, B] is
1/(3-1)
?
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The mean and variance of the uniform distribution are
Example: Assume that the length X of a conference has a uniform
distribution on the interval [0, 4]. P[X≥3]=?
?,
P[X≥3]=
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Normal Distributions
The normal distribution arose originally in the study of
experimental errors.
Such errors pertain to unavoidable differences between
observations when an experiment is repeated under similar
conditions.
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An alternative term is noise, which is used in
telecommunication engineering and elsewhere when
referring to the difference between the true state of nature
and the signal received.
The uncertainties which are manifest in the errors may arise
from different causes that are not easily identifiable.
THE NORMAL DISTRIBUTION IS AN IDEAL CANDIDATE TO
REPRESENT SUCH ERRORS WHEN THEY ARE OF AN
ADDITIVE NATURE.
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A large number of random variables encountered in practical
applications fit to the Normal (Gaussian) distribution with the
following probability density function:
f ( x) 
1

X
2

exp  ( x   X ) / 2
2
2
X
   x  
This distribution is shown briefly as N(, 2).
It has two parameters: X and X .
Normal distribution is symmetrical (Cs=0), with a kurtosis coefficient equal to 3
(K=3).
The mode and median are equal to the mean because of the symmetry.
The sample values of x and sX can be taken as the estimates of the parameters X and X
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(a) Probability density function, f X (x), and (b) probability distribution function, FX (x), of X
for m=0 and σ=1
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[-3:.1:3];
[-3:.1:3];
Once μ and σ are specified, the normal curve is completely
determined. For example,if μ = 50 and σ = 5, then the
ordinates n(x; 50, 5) can be computed for various values of
x and the curve drawn.
Example: Lets draw to two normal distribution functions
having the same mean but different standart deviations
such as n(x; 0, 1) and n(x; 0.5, 1) on the axis x=[-3 3].
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x = [-3:.1:3];
norm = normpdf(x,0,1);
norm1 = normpdf(x,0.5,1);
figure;
plot(x,norm,'r')
hold on
plot(x,norm1,'b')
𝜇1 = 0
𝜇2 = 0.5
𝜎1 = 𝜎2 = 1
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x = [-3:.1:3];
norm = normpdf(x,0,1);
norm1 = normpdf(x,0,2);
figure;
plot(x,norm,'r')
hold on
plot(x,norm1,'b')
0.4
𝜎1 = 1
0.35
0.3
0.25
0.2
0.15
𝜎2 = 2
0.1
0.05
0
-3
-2
-1
0
𝜇1 = 𝜇2 = 0
1
2
3
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The analytical form of the probability distribution function F(x)
of the normal distribution cannot be obtained by
integration, but is tabulated numerically.
A single table for the normal distribution can be prepared by
standardizing the random variable as follows
Z  (X  X ) / 
X
where the standard normal variable Z has the mean 0 and
standard deviation 1. The distribution N(0,1) of the variable Z is
called the STANDARD NORMAL DISTRIBUTION.
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Probability
distribution
function of
standard
normal
distribution
• Since the normal distribution is
symmetrical, this table is
prepared for the positive values
of Z only.
• The probabilities of Z exceeding a
certain positive value z, F1(z)=A
are given.
• For positive z we can compute
the probability of nonexceedance
as F(z)=1-F1(z), and for negative z
we have F(z)=F1(z) because of
symmetry around the mean 0.
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The probability density
function is bell-shaped around
the mean X .
The mode and median are
equal to the mean because of
the symmetry.
The probabilities of the normal variable to remain in the intervals around the
mean of width one, two and three standard deviations are equal to 0.683,
0.955 and 0.9975 (nearly 1), respectively.
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The ordinate axis of this
probability paper is scaled
such that the cumulative
distribution function of the
normal distribution appears
as a straight line
The probability paper of the normal distribution
What is σ?
The standard deviation can be computed as X=X0.84-X or X=X -X0.16
because the probability of the normal variable to remain in an interval of
two standard deviations around the mean is about 0.68.
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Central Limit Theorem states that, the distribution of the variable
n
X= 
ci X i
i 1
where Xi are independent random variables approaches the
normal distribution with the increase of n, whatever the
distributions of the variables Xi are.
The approach is rather fast such that the normal distribution can
be assumed for n10.
Thus, if a random variable is affected by a large number of
independent variables such that the effects are additive,
then it can be assumed to be distributed normally.
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21
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Example: Given a standard normal distribution, find the area under the curve that lies
(a) to the right of z = 1.84 and
(b) between z = −1.97 and z = 0.86.
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Solution (a)
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Solution (b)
1-0.0244=0.9756
.9756-0.1949=0.7807
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Example: Given a standard normal distribution, find the value of k such that
(a) P(Z > k) = 0.3015 and
(b) P(k < Z < −0.18) = 0.4197
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Solution (a)
k = 0.52
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Solution (b)
0.4286-P(k)=0.4197
P(k)=0.0089
k=-2.37
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Example:Given a random variable X having a normal
distribution with μ = 50 and σ = 10,find the probability that
X assumes a value between 45 and 62.
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Solution : The z values corresponding to x1 = 45 and x2 = 62 are
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Solution (b)
1-0.1151=0.8849
0.8849-0.3085=0.57
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Example:Given a normal distribution with μ = 40 and σ = 6, find
the value of x that has
(a) 45% of the area to the left and
(b) 14% of the area to the right.
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Solution
(a) x = (6)(−0.13) + 40 = 39.22
(b) x = (6)(1.08) + 40 = 46.48
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Example:A certain type of storage battery lasts, on average,
3.0 years with a standard deviation of 0.5 year. Assuming
that battery life is normally distributed, find the probability
that a given battery will last less than 2.3 years.
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Example:An electrical firm manufactures light bulbs that have a
life, before burn-out, that is normally distributed with mean
equal to 800 hours and a standard deviation of 40 hours. Find
the probability that a bulb burns between 778 and 834 hours.
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-0.55
1-0.2912= 0.7088
0.85
0.7088-0.1977=0.5111
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Example: In an industrial process, the diameter of a ball bearing is an important measurement. The
buyer sets specifications for the diameter to be 3.0 ± 0.01 cm. The implication is that no part falling
outside these specifications will be accepted. It is known that in the process the diameter of a ball
bearing has a normal distribution with mean μ = 3.0 and standard deviation σ = 0.005. On average,
how many manufactured ball bearings will be scrapped?
-2
2
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P(Z < −2.0) + P(Z > 2.0) = 2(0.0228) = 0.0456
As a result, it is anticipated that, on average, 4.56% of manufactured ball bearings
will be scrapped.
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Example:Gauges are used to reject all components for which a certain dimension is
not within the specification 1.50 ± d. It is known that this measurement is normally
distributed with mean 1.50 and standard deviation 0.2. Determine the value d such
that the specifications “cover” 95% of the measurements.
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P(−1.96 < Z < 1.96) = 0.95
(X-1.5)/0.2=1.96; x=1.89 and d =1.89-1.5=0.39
0.95
-1.96
1.96
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If X and Y are two independent normal variables,
the distribution of X+Y is
N(X + Y , σx2 + σy2 ) and
The distribution of X-Y is N (X - Y , σx2 + σy2 )
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How can we tell if a variable is normally distributed?
1. The first check is by sketching the cumulative frequency
distribution of the data on the normal probability paper. We
can resume normal distribution if the plot is nearly a
straight line.
2. The closeness of the skewness coefficient Cs of the
sample to 0 (its absolute value below 0.10 or 0.05) and that
of the kurtosis coefficient K to 3 (between 2.5 and 3.5) are
further checks.
3. If the data pass these tests, the assumption of the normal
distribution can be tested by statistical tests to be discussed
in Chapter 6.
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The normal distribution is certainly not valid in many cases
because the variable is skewed.
Most hydrologic variables (such as the discharge in a
stream, the precipitation depth at a location) are NOT
symmetrically distributed.
For such variables, distributions other than normal must be
used.
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Exercises
Given a standard normal distribution, find the value of k
such that
 (a) P(Z > k) = 0.2946;
 (b) P(Z < k) = 0.0427;
 (c) P(−0.93 < Z < k) = 0.7235.
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(a) 0.54; (b) −1.72; (c) 1-0.1762=0.8238; 0.8238-0.7235=0.1003
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EXAMPLE 4.3. (M. Bayazıt, page 83)
The load of a footing consists of the sum of the dead load and moving load. These loads are assumed to be random variables.
The dead load X has the mean 100 t, standard deviation 10 t. The moving load Y has the mean 40 t, standard deviation 10 t.
The design load corresponds to the risk of exceedance of 5%. Let us determine the design load assuming that loads follow the
normal distribution.
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μu=100+40=140t
σu= 102 + 102 =14.1t
for the excedence probability of 0.05, Z=1.65
Ud= μu+1.65× σu=140+1.65×14.1=163t
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EXAMPLE 4.5 (M. Bayazıt, page 85)
Water is transmitted from A to B by two parallel pipelines.
The capacities of these pipelines are assumed to be normal
variables with parameters:
X = 5 m3/s
CvX = 0.10
Y = 8 m3/s
CvY = 0.15
Find the probability that the total discharge is below 12
m3/s.
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μu=5+8=13t
σu= (Cvxμx)2 + (Cvyμy)2 =1.3m3/s
12−13
1.3 =−0.77
F(-0.77)=F1(0.77)=0.2206
The probability of discharge less than 12m3/s is 0.2206
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