Yard. Do. Dr. Tarkan Erdik Probability Distributions Uniform
and Normal Distributions- Week 7 1
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Probability Distributions 2 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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3 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.
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What is the difficulty is in choosing the distribution
function? There are no general rules for this. 4 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.
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5 The comparison of the histogram of the observed data with the
chosen probability density function helps in the decision
making.
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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
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7 The density function of the continuous uniform random
variable X on the interval [A, B] is ? 1/(3-1)
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8 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[X3]=? ?, P[X3]=
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Normal Distributions The normal distribution arose originally
in the study of experimental errors. 9 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. 10
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: 11 This distribution is
shown briefly as N( , 2 ). It has two parameters: X and X. Normal
distribution is symmetrical (C s =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 s X can be taken as the
estimates of the parameters X and X
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12 (a) Probability density function, f X (x), and (b)
probability distribution function, F X (x), of X for m=0 and
=1
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Once and are specified, the normal curve is completely
determined. For example,if = 50 and = 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]. 13 [-3:.1:3];
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14 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')
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15 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')
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16 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 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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17 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, F 1 (z)=A
are given. For positive z we can compute the probability of
nonexceedance as F(z)=1-F 1 (z), and for negative z we have F(z)=F
1 ( z ) because of symmetry around the mean 0.
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18 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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19 The probability paper of the normal distribution 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 standard deviation can be computed as X =X
0.84 - X or X = X -X 0.16 because the probability of the normal
variable to remain in an interval of two standard deviations around
the mean is about 0.68. What is ?
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Central Limit Theorem states that, the distribution of the
variable X= where X i are independent random variables approaches
the normal distribution with the increase of n, whatever the
distributions of the variables X i 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. 20
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23 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.
32 Example:Given a normal distribution with = 40 and = 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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33 Solution (a) x = (6)(0.13) + 40 = 39.22 (b) x = (6)(1.08) +
40 = 46.48
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34 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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35 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.
38 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 = 3.0 and standard deviation = 0.005. On
average, how many manufactured ball bearings will be scrapped? -2
2
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39 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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40 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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41 P(1.96 < Z < 1.96) = 0.95 -1.96 1.96 0.95
(X-1.5)/0.2=1.96; x=1.89 and d =1.89-1.5=0.39
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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. 43 2. The
closeness of the skewness coefficient C s 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. 44
Slide 45
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. 45
EXAMPLE 4.3. (M. Bayazt, 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. 47
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EXAMPLE 4.5 (M. Bayazt, 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 m 3 /sC vX =
0.10 Y = 8 m 3 /sC vY = 0.15 Find the probability that the total
discharge is below 12 m 3 /s. 49
