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Chapter 6
Continuous Probability Distributions
1
Chapter Outline
 Uniform Probability Distribution
 Normal Probability Distribution
f (x)
Uniform
f (x)
x
Normal
x
2
Continuous Probability Distributions
 A continuous random variable assume values that
have no gap or jump between them.
 Since between any two values, a continuous
random variable assumes infinite number of
values, the probability that any particular value
occurs is zero.
 Therefore, we study the probability of the random
variable assuming a value within a given interval.
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Continuous Probability Distributions
 The probability of the random variable assuming a
value within some given interval from x1 to x2 is
defined to be the area under the line of the
probability density function between x1 and x2.
f (x)
Uniform
x1 x2
f (x)
x
Normal
x 1 x2
x
4
Uniform Probability Distribution
 A random variable is uniformly distributed whenever the
probability is proportional to the interval’s length.
 The uniform probability density function is:
f (x) = 1/(b – a) for a < x < b
=0
elsewhere
where: a = smallest value the variable can assume
b = largest value the variable can assume
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Uniform Probability Distribution
Expected Value of x
E(x) = (a + b)/2
Variance of x
Var(x) = (b - a)2/12
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Uniform Probability Distribution
 Example: Slater's Buffet
Slater customers are charged for the amount of
salad they take. Sampling suggests that the amount
of salad taken is uniformly distributed between 5
ounces and 15 ounces.
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Uniform Probability Distribution
 Example: Slater's Buffet
 Uniform Probability Density Function
f(x) = 1/10 for 5 < x < 15
=0
elsewhere
where:
x = salad plate filling weight
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Uniform Probability Distribution
 Expected Value of x
E(x) = (a + b)/2
= (5 + 15)/2
= 10
 Variance of x
Var(x) = (b - a)2/12
= (15 – 5)2/12
= 8.33
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Uniform Probability Distribution
 Uniform Probability Distribution for Salad Plate Filling
Weight
f(x)
1/10
0
5
10
Salad Weight (oz.)
x
15
10
Uniform Probability Distribution
What is the probability that a customer will take
between 12 and 15 ounces of salad?
f(x)
P(12 < x < 15) = (1/10)(3) = .3
1/10
0
5
10 12
Salad Weight (oz.)
x
15
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Area as A Measure of Probability
 The area under the graph of f(x) is simply the
measure of probability.
 This is valid for all continuous random variables.
 The probability that x takes on a value between some
lower value x1 and some higher value x2 can be found
by computing the area under the graph of f(x) over
the interval from x1 to x2.
 The overall area under the graph of f(x) over the
value range of x is 1 or 100%.
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Normal Probability Distribution
 The normal probability distribution is the most important
distribution for describing a continuous random variable.
 It is widely used in statistical inference.
 It has been used in describing a wide variety of real-life
applications including:
• Heights of people
• Rainfall amounts
• Test scores
• Scientific measurements
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Normal Probability Distribution
 Normal Probability Density Function
1
 ( x   )2 /2 2
f (x) 
e
 2
where:
 = mean
 = standard deviation
 = 3.14159
e = 2.71828
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Normal Probability Distribution
 Characteristics
The distribution is symmetric; its skewness
measure is zero.
x
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Normal Probability Distribution
 Characteristics
The entire family of normal probability
distributions is defined by its mean  and its
standard deviation .
Standard Deviation 
Mean 
x
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Normal Probability Distribution
 Characteristics
The highest point on the normal curve is at the
mean, which is also the median and mode.
Mean 
x
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Normal Probability Distribution
 Characteristics
The mean can be any numerical value. When only
the mean (the central location) changes, the whole
normal curve simply shifts horizontally.
x
-8
0
24
18
Normal Probability Distribution
 Characteristics
The standard deviation determines the width of the
curve. A smaller  results in a narrower, taller curve.
 =5
 = 12
x
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Normal Probability Distribution
 Characteristics
Probabilities for the normal random variable are
given by areas under the curve. The total area under
curve is 1 (.5 to the left of the mean and
.5 to the right.
.5
.5
Mean 
x
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Normal Probability Distribution
 Characteristics (basis for the empirical rule)
68.26% of values of a normal random variable
are between  -  and  + .
Expected
95.44% of values of a normal random variablenumber of
correct
are between  - 2 and  + 2.
answers
99.72% of values of a normal random variable
are between  - 3 and  + 3.
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Normal Probability Distribution
 Characteristics (basis for the empirical rule)
99.72%
95.44%
68.26%
Expected
number of
correct
answers
 – 3
 – 1
 – 2

 + 3
 + 1
 + 2
x
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Standard Normal Probability Distribution
 Characteristics (basis for the empirical rule)
A normally distributed random variable
with a mean of 0 and a standard deviation of 1 is
said to have a standard normal probability
Expected
distribution. Any other normal distribution can
be
converted to the standard normal distribution.number of
correct
answers
23
Standard Normal Probability Distribution
 Characteristics
The letter z is used to designate the stand normal
random variable.
Expected
number of
correct
answers
1
z
0
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Standard Normal Probability Distribution
 Converting to the Standard Normal Distribution
z
x

We can think of z as a measure of the number of
standard deviations x is from .
0
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Standard Normal Probability Distribution
 Example: What is the probability that z  1.58?
Area of z  1.58 is the probability.
z
0 1.58
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Standard Normal Probability Distribution
 Example: What is the probability that z  1.58?

z
.
Cumulative Probability Table for the Standard Normal
Distribution: it provides the areas (probabilities) to the LEFT of
the z values.
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.
.
.
.
.
.
.
.
.
.
1.5 .9332 .9345 .9357 .9370 .9382
1.6 .9452 .9463 .9474 .9484 .9495
1.7 .9554 .9564 .9573 .9582 .9591
1.8 .9641 .9649 .9656 .9664 .9671
.9394 .9406 .9418 .9429 .9441
.9505 .9515 .9525 .9535 .9545
.9599 .9608 .9616 .9625 .9633
.9678 .9686 .9693 .9699 .9706
1.9 .9713 .9719 .9726 .9732 P(z
.9738 1.58)
.9744 .9750 .9756 .9761 .9767
.
.
.
.
.=.9429
.
.
.
.
.
.
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Standard Normal Probability Distribution
 Example: What is the probability that z  1.58?
Area = 1-.9429
= .0571
Area = .9429
z
0 1.58
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Standard Normal Probability Distribution
 Example: Computer Prices
The average price of personal computers manufactured by
Company APO is $1,000 with a standard deviation of $150.
Furthermore, it is known that the computer prices
manufactured by APO are normally distributed.
a. What is the probability that a randomly selected
computer will have a price of at least $1125?
P(x  1125) = ?
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Standard Normal Probability Distribution
 Example: Computer Prices
Step 1: Convert x to the standard normal distribution.
z = (x - )/
= (1125 – 1000)/150
= .83
Step 2: Find the area under the standard normal
curve to the left of z = .83.
see next slide
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Standard Normal Probability Distribution
 Example: Computer Prices

Cumulative Probability Table for the Standard Normal
Distribution:
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.
.
.
.
.
.
.
.
.
.
.
.5
.6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224
.6
.7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549
.7
.7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
.8
.7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133
.9
.8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389
.
.
.
.
.
.
.
.
.
.
.
P(z < .83)
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Standard Normal Probability Distribution
 Example: Computer Prices
Step 3: Compute the area under the standard normal
curve to the right of z = .83.
P(z  .83) = 1 – P(z < .83)
= 1- .7967
= .2033
P(x  1125)
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Standard Normal Probability Distribution
 Example: Computer Prices
Area = 1 - .7967
Area = .7967
= .2033
0
.83
z
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Standard Normal Probability Distribution
 Example: Computer Prices
b. To attract buyers like college students, Company APO
decides to provide promotions on the cheaper
computers. What is the maximum price such that only
15% of all the computers are eligible for a promotion
sale?
---------------------------------------------------------------------(Hint: Given a probability (area), we can use the
standard normal table in an inverse fashion to
find the corresponding z value.)
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Standard Normal Probability Distribution
 Example: Computer Prices
Area = .15
Area = .15
z.15
0
z.85
z
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Standard Normal Probability Distribution
 Example: Computer Prices
Step 1: Determine whether the z value is negative or
positive. The question is asking for the
maximum computer price such that the
cheapest computers (15% of all the computers)
can enjoy a promotion sale. It is translated as:
P(x <pricemax) = .15
Therefore, the z value should be negative and
it cuts off an area of .15 in the left tail of the
standard normal distribution.
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Standard Normal Probability Distribution
 Example: Computer Prices
Step 2: Find the z-value. The following table shows
the positive z-values. To find the negative
z-values, we can utilize the symmetric
feature of the Normal Distributions.
z
1
1.1
1.2
1.3
1.4
.00
.01
.02
.03
.04
.05
0.8413 0.8438 0.8461 0.8485 0.8508 0.8531
0.8643 0.8665 0.8686 0.8708 0.8729 0.8749
0.8849 0.8869 0.8888 0.8907 0.8925 0.8944
0.9032 0.9049 0.9066 0.9082 0.9099 0.9115
0.9192
0.9207
0.9222
The
closest
z-value
that0.9236
cuts off0.9251
the left 0.9265
area of .85 is 1.04. Therefore, the z-value
we look for is –1.04.
.06
0.8554
0.8770
0.8962
0.9131
0.9279
.07
.08
.09
0.8577
0.8790
0.8980
0.9147
0.9292
0.8599
0.8810
0.8997
0.9162
0.9306
0.8621
0.8830
0.9015
0.9177
0.9319
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Standard Normal Probability Distribution
 Example: Computer Prices
Step 3: Convert z.15 to the corresponding value of x.
x =  + z.15
= 1000 + (-1.04)(150)
= 844
So, the maximum price at which a computer can enjoy
a promotion sale is $844, and there are about 15% of all
the computers manufactured by APO are eligible for the
promotion sale.
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Standard Normal Probability Distribution
 Example: Computer Prices
 = 150
Area = .15
844
z
1000
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