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Continuous Probability Distributions
Chapter Contents
Uniform Continuous Distribution
Exponential Distribution
Describing a Continuous Distribution
Normal Distribution
Standard Normal Distribution
Normal Approximations
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Continuous Probability Distributions
Chapter Learning Objectives
Define a continuous random variable.
Calculate uniform probabilities.
Know the form and parameters of the normal distribution.
Find the normal probability for given z or x using tables or Excel.
Solve for z or x for a given normal probability using tables or Excel.
Use the normal approximation to a binomial or a Poisson
distribution.
Find the exponential probability for a given x.
Solve for x for given exponential probability.
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Uniform Continuous Distribution
LO7-2
Characteristics of the Uniform
Distribution
If X is a random variable that is
uniformly distributed between
a and b, its PDF has
constant height.
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Denoted U(a, b)
Area =
base x height =
(b-a) x 1/(b-a) = 1
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Characteristics of the Uniform Distribution
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Example: Anesthesia Effectiveness
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An oral surgeon injects a painkiller prior to extracting a tooth. Given the
varying characteristics of patients, the dentist views the time for
anesthesia effectiveness as a uniform random variable that takes
between 15 minutes and 30 minutes.
X is U(15, 30)
a = 15, b = 30, find the mean and standard deviation.
Find the probability that the effectiveness anesthetic takes between
20 and 25 minutes.
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Example: Anesthesia Effectiveness
P(20 < X < 25) = (25 – 20)/(30 – 15) = 5/15 = 0.3333 = 33.33%
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1. The time that the customers at the “self serve” check out stations
at the Mejers store spend checking out follows a uniform distribution
between 0 and 3 minutes.
a.
Determine the height and draw this uniform distribution.
b. How long does the typical customer wait to check out?
c. Determine the standard deviation of the wait time.
d. What is the probability a particular customer will wait less than one
minute?
e. What is the probability a particular customer will wait between 1.5 and
2 minutes?
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2. A statistics instructor collected data on the time it takes the students
to complete a test. The test taking time is uniformly distributed within a
range of 35 minutes to 55 minutes.
a. Determine the height.
b. How long does the typical test taking time?
c. Determine the standard deviation of the test taking time.
d. What is the probability a particular student will take
less than 45 minutes?
e. What is the probability a particular student will take
between 45 and 50 minutes?
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3. A tube of Listerine Tartar Control toothpaste contains 4.2 ounces. As people
use the toothpaste, the amount remaining in any tube is random. Assume the
amount of toothpaste left in the tube follows a uniform distribution. From this
information, we can determine the following information about the amount
remaining in a toothpaste tube without invading anyone’s privacy.
a. How much toothpaste would you expect to be remaining in the tube?
b. What is the standard deviation of the amount remaining in the tube?
c. What is the likelihood there is less than 3.0 ounces remaining in the
tube?
d. What is the probability there is more than 1.5 ounces remaining in
the tube?
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Characteristics of the Exponential Distribution
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If events per unit of time follow a Poisson distribution, the time until the
next event follows the Exponential distribution.
The time until the next event is a continuous variable.
NOTE: Here
we will find
probabilities
> x or ≤ x.
7-10
Characteristics of the Exponential Distribution
Probability of waiting less than or
equal to x
Probability of waiting more than x
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Example Customer Waiting Time
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Between 2P.M. and 4P.M. on Wednesday, patient insurance
inquiries arrive at Blue Choice insurance at a mean rate of 2.2 calls
per minute.
What is the probability of waiting more than 30 seconds (i.e., 0.50
minutes) for the next call?
Set  = 2.2 events/min and x = 0.50 min
P(X > 0.50) = e–x = e–(2.2)(0.5) = .3329
or 33.29% chance of waiting more than 30 seconds for the next
call.
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Example Customer Waiting Time
P(X > 0.50)
P(X ≤ 0.50)
7-13
Inverse Exponential
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If the mean arrival rate is 2.2 calls per minute, we want the 90th
percentile for waiting time (the top 10% of waiting time).
Find the x-value
that defines the
upper 10%.
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Inverse Exponential
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Mean Time Between Events
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5. If arrivals occur at a mean rate of 3.6 events per hour, the exponential
probability of:
waiting more than 0.5 hour for the next arrival is:
P(X > .50) = exp(-3.6 × 0.50) = .1653.
6. waiting less than 0.5 hour for the next arrival is:
P(X < .50) = 1 - exp(-3.6 × 0.50) = 1 - .1653 = .8347
7. If arrivals occur at a mean rate of 2.6 events per minute, the
exponential probability of waiting more than 1.5 minutes for the next
arrival is:
P(X > 1.5) = exp(-2.6 × 1.50) = .0202
7-17
8. If arrivals occur at a mean rate of 1.6 events per minute, the
exponential probability of waiting less than 1 minute for the next
arrival is:
(X < 1) = 1 - exp(-1.6 × 1) = 1 - .2019 = .7981.
9. On average, a major earthquake (Richter scale 6.0 or above)
occurs 3 times a decade in a certain California county. What is the
probability that less than six months will pass before the next
earthquake?
Set λ = 3/120 = 0.025 earthquake/month so
P(X < 6) = 1 - exp(-0.025 × 6) = 1 - .8607 = .1393
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Events as Intervals
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Discrete Variable – each value of X has its own probability P(X).
Continuous Variable – events are intervals and probabilities are
areas under continuous curves. A single point has no probability.
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PDF – Probability Density Function
Continuous PDF’s:
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Denoted f(x)
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Must be nonnegative
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Total area under
curve = 1
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Mean, variance and
shape depend on
the PDF parameters
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Reveals the shape
of the distribution
7-20
CDF – Cumulative Distribution Function
Continuous CDF’s:
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Denoted F(x)
Shows P(X ≤ x), the
cumulative proportion
of scores
Useful for finding
probabilities
7-21
Probabilities as Areas
Continuous probability functions:
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Unlike discrete
distributions, the
probability at any
single point = 0.
The entire area under
any PDF, by definition,
is set to 1.
Mean is the balance
point of the distribution.
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Characteristics of the Normal Distribution
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Defined by two parameters, µ and .
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Domain is –  < X < +  (continuous scale).
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Almost all (99.7%) of the area under the normal curve is included in the
range µ – 3 < X < µ + 3.
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Symmetric and unimodal about the mean.
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Characteristics of the Normal Distribution
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Characteristics of the Normal Distribution
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Normal PDF f(x) reaches a maximum at µ and has points of inflection at
µ ±
Bell-shaped curve
NOTE: All normal
distributions
have the same
shape but differ
in the axis scales.
7-25
Characteristics of the Normal Distribution
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Normal CDF
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Characteristics of the Standard Normal Distribution
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Since for every value of µ and , there is a different normal distribution, we transform a normal
random variable to a standard normal distribution with µ = 0 and  = 1 using the formula.
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Characteristics of the Standard Normal
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Standard normal PDF f(x) reaches a maximum at z = 0 and has
points of inflection at +1.
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Shape is unaffected by
the transformation.
It is still a bell-shaped
curve.
Figure 7.11
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Characteristics of the Standard Normal
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Standard normal CDF
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A common scale
from -3 to +3 is used.
Entire area under the
curve is unity.
The probability of an
event P(z1 < Z < z2)
is a definite integral
of f(z).
However, standard
normal tables or
Excel functions can
be used to find the
desired probabilities.
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Normal Areas from Table 1
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The table allows you to find the area under the curve from 0 to z.
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For example, find P(0 < Z < 1.96):
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Normal Areas from Table 1
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Now find P(-1.96 < Z < 1.96).
Due to symmetry, P(-1.96 < Z) is the same as P(Z < 1.96).
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So, P(-1.96 < Z < 1.96) = .4750 + .4750 = .9500 or 95% of the
area under the curve.
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Basis for the Empirical Rule
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Approximately 68% of the area under the curve is between + 1
Approximately 95% of the area under the curve is between + 2
Approximately 99.7% of the area under the curve is between + 3
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Normal Areas from Cumulative Table 2
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Cumulative table allows you to find the area under the curve from the left
of z (similar to Excel).
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For example,
P(Z < 1.96)
P(Z < -1.96)
P(-1.96 < Z < 1.96)
7-33
Normal Areas from Tables
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Both tables yield identical results.
Use whichever table is easiest.
Finding z for a Given Area
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Both tables can be used to find the
z-value corresponding to a given probability.
For example, what z-value defines the top 1% of a normal
distribution?
This implies that 49% of the area lies between 0 and z which
gives z = 2.33 by looking for an area of 0.4900 in Table1.
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Finding Areas by using Standardized Variables
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Suppose John took an economics exam and scored 86 points. The class
mean was 75 with a standard deviation of 7. What percentile is John in?
That is, what is P(X < 86) where X represents the exam scores?
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So John’s score is 1.57 standard deviations about the mean.
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P(X < 86) = P(Z < 1.57) = .9418 (from Appendix C-2)
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So, John is approximately in the 94th percentile.
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The net sales and the number of employees for aluminum fabricators with
similar characteristics are organized into frequency distributions. Both are
normally distributed. For the net sales, the mean is $180 million and the
standard deviation is $25 million. For the number of employees, the mean is
1,500 and the standard deviation is 120. Clarion Fabricators had sales of $170
million and 1,850 employees.
a. Convert Clarion’s sales and number of employees to z values.
b. Locate the two z values.
c. Compare Clarion’s sales and number of employees with those of the other
fabricators.
7-36
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Finding Areas by using Standardized Variables
NOTE: You can use Excel, Minitab, TI83/84 etc. to compute these
probabilities directly.
7-37
LO7-5: Solve for z or x for a normal probability using tables or Excel.
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Inverse Normal
• How can we find the various normal percentiles (5th, 10th, 25th, 75th,
90th, 95th, etc.) known as the inverse normal? That is, how can we
find X for a given area? We simply turn the standardizing
transformation around:
Solving for x in z = (x − μ)/ gives x = μ + zσ
7-38
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Inverse Normal
• For example, suppose that John’s economics professor has decided
that any student who scores below the 10th percentile must retake the
exam.
• The exam scores are normal with μ = 75 and σ = 7.
• What is the score that would require a student to retake the exam?
• We need to find the value of x that satisfies P(X < x) = .10.
• The z-score for with the 10th percentile is z = −1.28.
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Inverse Normal
• The steps to solve the problem are:
• Use Appendix C or Excel to find z = −1.28 to satisfy P(Z < −1.28) = .10.
• Substitute the given information into z = (x − μ)/σ to get
−1.28 = (x − 75)/7
• Solve for x to get x = 75 − (1.28)(7) = 66.03 (or 66 after rounding)
• Students who score below 66 points on the economics exam will be
required to retake the exam.
7-40
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Inverse Normal
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Normal Approximation to the Binomial
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Binomial probabilities are difficult to calculate when n is large.
Use a normal approximation to the binomial distribution.
As n becomes large, the binomial bars become smaller and continuity is
approached.
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Normal Approximation to the Binomial
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Rule of thumb: when n ≥ 10 and n(1- ) ≥ 10, then it is appropriate to use
the normal approximation to the binomial distribution.
In this case, the mean and standard deviation for the binomial distribution
will be equal to the normal µ and , respectively.
Example Coin Flips
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If we were to flip a coin n = 32 times and  = .50, are the
requirements for a normal approximation to the binomial distribution
met?
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Example Coin Flips
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n = 32 x .50 = 16
n(1- ) = 32 x (1 - .50) = 16
So, a normal approximation can be used.
When translating a discrete scale into a continuous scale,
care must be taken about individual points.
For example, find the probability of more than 17 heads in
32 flips of a fair coin.
This can be written as P(X  18).
However, “more than 17” actually falls between 17 and 18
on a discrete scale.
7-44
Example Coin Flips
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Since the cutoff point for “more than 17” is halfway between 17 and 18, we
add 0.5 to the lower limit and find P(X > 17.5).
This addition to X is called the Continuity Correction.
At this point, the problem can be completed as any normal distribution
problem.
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Example Coin Flips
P(X > 17) = P(X ≥ 18)  P(X ≥ 17.5)
= P(Z > 0.53) = 0.2981
7-46