Download Continuous Probability Distribution

Continuous Probability Distributions
Types:
1. Uniform Distribution
2. Exponential Distribution
3. Normal Distribution
4. Gamma Distribution
5. Lognormal Distribution
6. Beta Distribution
7. Weibull Distribution
Many random variables seen in practice have more than a countable
collection of possible values. For example:
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Diameter of machine rods from a certain industrial process may be
anywhere from 1.2 to 1.5 centimeters
Proportions of impurities in ore samples may run from 0.10 to 0.80
The height of students in the College of Engineering, 4.8 to 6.5 feet
Water depth along the coast of Manila Bay, 0.5 feet to 20.5 feet
Etc
A random variable X is said to be continuous if it can take on the infinite
number of possible values associated with intervals of real numbers.
A function f(x) of a random variable X is called the probability density
function if
1. f(x)  0 for all x
2.
3.
Note that for a continuous random variable X
for any specific value of a.
If random variable X is continuous with the probability density function
f(x), then the distribution function X is
Expected Values of Continuous Random Variables
For a continuous random variable X having probability density function
f(x),
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The expected value, also known as the mean, of X is given by
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The variance of X is given by
For a continuous random variable and constants a and b,
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E(aX + b) = aE(X) + b
V(aX + b) = a2V(X)
Example:
For a lathe in a machine shop, let X denote the percentage of time out of
a 40-hour workweek that the lathe is actually in use. Suppose X has a
probability density function given by
Find the mean and variance of X.
Solution:
From the definition of expected value we have,
Thus, on the average, the lathe is in use 75% of the time.
To compute V(X), we first find E(X2) as follows
Then,
.
The Uniform Distribution
Example:
The failure of a circuit board causes a computing system to shut down
until a new board is delivered. The delivery time X is uniformly distributed
over the interval of 1 to 5 days. The cost C of this failure and shut down
consists of a fixed cost, c0, for the new part plus lost time and a cost (c1)
that increases proportional to X2, so that
a. Find the probability that the delivery time is two or more days
b. Find the expected cost of a single component failure.
Solution:
The delivery time X is uniformly distributed form 1 to 5 days, which gives
a. The probability that the delivery time is two or more days (See
Figure 6.7) is
Figure 6.7
0.25
0
b. We know that
get
Thus,
Because
1
2
3
4
5
, we
The Exponential Distribution
E(X) = 
and V(X) = 2
F(x) =
where  is a constant that determines the rate at which the curve
decreases.
Example: A sugar refinery has three processing plants, all receiving raw
sugar in bulk. The amount of raw sugar (in tons) that one plant can
process in one day can be modelled using an exponential distribution with
mean of 4 tons for each of three plants. If each plant operates
independently,
a. Find the probability that any given plant processes more than 5 tons
of raw sugar on a given day.
b. Find the probability that exactly two of the three plants process
more than 5 tons of raw sugar on a given day.
c. How much raw sugar should be stocked for the plant each day so
that the chance of running out of the raw sugar is only 0.05?
Solution:
Let X = amount of sugar processed in a day.
a.
Knowledge of the distribution function could allow us to evaluate this
probability easily as
b. Assuming that three of the plants operate independently, the
problem is to find the probability of two successes out of three tries,
where 0.37 denotes the probability of success. This is a binomial
problem, and the solution is
P(exactly 2 use more than 5 tons) =
c. Let a = the amount of raw sugar to be stocked. The chance of
running out of the raw sugar means there is a probability that more
than the stocked amount is needed. We want to choose a so that
Solving this equation yields a = 11.98 tons.
The Normal Distribution
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The most widely used of all continuous probability distributions is
the normal distribution.
The normal probability density function is bell-shaped and centered
at the mean value .
Its spread is measured by the standard deviation .
The functional form is given by
Some useful properties of the normal distribution
 The mean  determines the location of the distribution
 The standard deviation  determines the spread of the distribution.
The normal distribution with larger  is shorter and more spread
(Figure 6.4)
 Any linear function of the normally distributed random variable is
also normally distributed. That is, if X has a normal distribution with
mean  and variance 2 and Y = aX + b for constants a and b, then
Y is also normally distributed with mean a and b and variance a22.
Standard Normal Distribution
A normal distribution with  = 0 and  = 1 is known as a standard normal
distribution. The letter Z is used to indicate the standard normal variable.
For any normally distributed random variable X, with parameters  and
2,
will have a standard normal distribution.
Because any normally distributed random variable can be transformed to
the standard normal, probabilities can be evaluated for any normal
distribution by having a table of standard normal integrals available and
provided in Table 16 of the Appendix.
Table 16 gives numerical values for the cumulative distribution function
using
Example:
Find P(Z  1.53) for a standard normal variable Z.
Solution:
The P(Z  1.53) is equal to the shaded area in Figure 6.15.
From the table, this is found in the cross-section of the row corresponding
to 1.5 and the column corresponding to 0.03. Hence,
P(Z  1.53) = 0.9370
Example:
If Z denotes a standard normal variable, find
a. P(Z  1)
b. P(Z < -1.5)
c. P(Z > 1)
d. P(-1.5  Z  0.5)
e. Find the value of Z, say z0, such that P(Z  z0) = 0.99.
Solution:
a. P(Z  1) = 0.8413 as shown in Figure 6.17
b. P(Z < -1.5) = 0.0668, as shown in Figure 6.18
Figure 6.17 P(Z  1)
Figure 6.18 P(Z < -1.5)
c. P(Z > 1) = 1 – P(Z  1) = 1 – 0.8413 = 0.1587, shown in Figure
6.19
d. P(-1.5  Z  0.5) = P(Z  0.5) – P(Z < -1.5) = 0.6915 – 0.0668 =
0.6247 as shown in Figure 6.20.
Figure 6.19 P(Z > 1)
Figure 6.20 P(-1.5  Z  0.5)
e. Find the value of Z, say z0, such that P(Z  z0) = 0.99.
We must look for the given probability 0.99 on the area side of
Appendix Table 4. The closest we can come is 0.9901, which
corresponds to the z-value of 2.33. Hence, z0 = 2.33.
Figure 6.21 (Z  z0) = 0.99
Example: A firm that manufactures and bottles apple juice has a machine
that automatically fills 16-ounce bottles. There is, however, some
variation in the amount of liquid dispensed (in ounces) into each bottle by
the machine. Over a long period of time, the average amount dispensed
into the bottles was 16 ounces, but there is a standard deviation of 1
ounce in these measurements. If the amount filled per bottle can be
assumed to be normally distributed, find the probability that the machine
will dispense more than 17 ounces of liquid into any one bottle.
Solution: Let Z denote the ounces of liquid dispensed into one bottle by
the filling machine. Then X is assumed to be normally distributed with
mean 16 and standard deviation 1 (Figure 6.23). Hence,
See Figure 6.23.
Figure 6.23
Example: Suppose that another machine, similar to the one in the
previous example that operate so that ounces of fill have a mean equal to
the dial setting for “amount of liquid” but have a standard deviation of 1.2
ounces. Find the proper setting for the dial so that 17-ounce bottles will
overflow only 5% of the time. Assume that the amounts dispensed have a
normal distribution.
Solution: Letting X denote the amount of liquid dispensed, we are now
looking for a value of  such that P(X > 17) = 0.05, as depicted in Figure
6.24. Hence
From Appendix Table 4, we know that if P(Z > z0) = 0.05, then z0 =
1.645. Thus, it must be that
and
See Figure 6.24
Figure 6.24 P(X > 17) = 0.05
The Relationship between the Poisson and the Exponential
Distribution
The Gamma Distribution
Some useful results:
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The Lognormal Distribution
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If Y is a normal random variable, then X = eY follows a lognormal
distribution.
The lognormal distribution is very commonly used to describe lifetime distribution, such as lifetime of products, components, and
systems.
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It is a right-skewed distribution and takes values in the range (0, )
The lognormal distribution has the following form, and the graph of
the density function is shown in Figure 6.38.
Because the probabilities of a lognormal random variable are
obtained by transforming a normal random variable, we can obtain
the cumulative distribution function of lognormal using the
cumulative distribution function of a standard normal variable, Z, as
follows:
Figure 6.38 The lognormal density function
The Lognormal Distribution
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If Y = ln (X) has a normal distribution with mean  and variance 2,
then then X has a lognormal distribution with probability density
function
Example: The Department of Energy (DOE) has a Chronic Beryllium
Disease Prevention Program. The focus of this program is to monitor
equipment and buildings for beryllium contamination and exposure of
workers to beryllium in the workplace. Data given in Table 6.3 were
collected from a smelter facility used to recycle metal, where beryllium
contaminants tend to deposit on all surface. Wipe samples were taken
randomly from the surface of equipment in the workplace and tested for
possible beryllium contaminations (measured in gm/100 cm2).
The distribution of data (Figure 6.39) shows the lognormal distribution
(given by the curve) is a reasonable choice to describe beryllium
contamination at this smelter. From the samples, the estimated  is 2.291 and  to be 1.276.
Find the following:
a. If a wipe sample is taken randomly form this smelter, what is the
probability that it will have beryllium contamination exceeding 0.50
gm/100 cm2?
b. The DOE has established a safe limit for beryllium contamination to
be L0.95 = 0.20 gm/100 cm2, where L0.95 indicate the 95th
percentile. Use this data to determine the safety of workers’ health
at this smelter
c. The expected amount of beryllium and the variance.
Solution:
a. Let X the beryllium contamination then ln X follows a normal
distribution with mean – 2.291 and standard deviation 1.276.
Therefore, using the above stated result and standard normal
cumulative distribution function, we get
b. The DOE has established a safe limit for beryllium contamination to
be L0.95 = 0.20 gm/100cm2, where L0.95 indicates 95th percentile.
Use this data to determine the safety of workers’ health at this
smelter.
We need to determine L0.95 value for this smelter such that P(X  L0.95) =
0.95. Therefore, from the above result relating normal and lognormal
distributions we get
Using the standard normal cumulative distribution function,
Solving this equation, we get L0.95 = exp[-2.291] +
1.645(1.276)]=0.8253 > 0.20. Because this estimate of 95 th
percentile exceeds the DOE safety limit of 0.20, we can conclude
that the beryllium contamination at this smelter is at an unhealthy
level for workers.
c. The expected amount of beryllium and the variance can be
calculated as follows:
and
Even the average amount of beryllium exceeds the safety limit.
The Beta Distribution
For ,  > 0
The following relation is used by the beta distribution. For ,  > 0,
Example: A gasoline wholesale distributor has bulk storage tanks holding
a fixed supply. The tanks are filled every Monday. Of interest to the
wholesaler is the proportion of this supply that is sold during the week.
Over many weeks, this proportion has been observed to be modelled
fairly well by a beta distribution with  = 4 and  = 2.
a. Find the expected value of this proportion.
b. Is it highly likely that the wholesaler will sell at least 90% of the
stock in a given week?
Solution:
Let x denote the proportion of the total supply sold in a given week.
a. The expected proportion of supply sold in a given week is
b. We are interested in
It is not very likely that 90% of the stock will be sold in a given week.
The Weibull Distribution
Example: The length of service time during which a certain type of
thermistor produces resistance within its specifications has been observed
to follow a Weibull distribution with  = 50 and  = 2. The measurements
are in thousands of hours.
a. Find the probability that one of these randomly selected
thermistors, to be installed in a system today, will function properly
for over 10,000 hours.
b. Find the expected lifetimes for thermistors of this type.
Solution:
Let X represent the lifetime of the thermistor in question.
a. As seen earlier, the Weibull distribution has a closed-form
expression for F(x). Thus,
P(X > 10) = 1 – F(10) =
b. The expected lifetime for thermistors of this type is
Thus, the average service time for these thermistors is 6,270 hours.