QBM117
Business Statistics
Probability and Probability Distributions
Continuous Probability Distributions
1
Objectives
•
To differentiate between a discrete probability
distribution and a continuous probability distribution
•
To introduce the probability density function and its
relationship to a continuous random variable
•
To introduce the uniform distribution
•
To introduce the normal distribution
•
To introduce the standard normal random variable
•
Use the standard normal tables to find probabilities
2
Continuous Random Variables
• A continuous random variable has an infinite number
of possible values.
• It can assume any value in an interval.
• We cannot list all the possible values of a continuous
random variable.
3
Continuous Probability Distribution
• A continuous random variable has an infinite number
of possible values.
• It is not possible to calculate the probability that a
continuous random variable will take on a specific
value.
• Instead we calculate the probability that a continuous
random variable will lie in a specific interval.
4
Probability Density Function
• A smooth curve is used to represent the probability
distribution of a continuous random variable.
• The curve is called a probability density function and
is denoted by f (x) .
5
•
1.
2.
A probability density function f ( x) must satisfy two
conditions:
f ( x) is non-negative, f ( x)  0
The total area under the curve f ( x) is equal to 1
6
• If X is a continuous random variable with probability
density function f ( x) , the probability that X will take a
value between a and b, P(a  X  b), is given by the
area under the curve f ( x) between a and b.
f (x)
P (a  X  b)
a
b
x
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The Uniform Distribution
• A continuous random variable X, defined over an
interval a  x  b , is uniformly distributed if its
probability density function is given by
 1

f ( x)   b  a
0

for a  x  b
elsewhere
8
f ( x)
1
ba
x
a
b
9
• To calculate the probability that X falls between
x1 and x2 we need to find the area of the rectangle
whose base is x1  x2 and whose height is 1
ba
f ( x)
P( x1  X  x2 )  ( x1  x2 )
1
ba
1
ba
x
a
x1
x2 b
10
Expected Value and Variance of a
Uniform Random Variable
• The expected value and variance of a uniform
random variable are given by
E( X ) 
ab
2
V (X ) 
( a  b)
2
12
11
Example 1
The total time to process a loan application is
uniformly distributed between 3 and 7 days.
1. Define the probability density function for loan
processing time.
2. What is the probability that a loan will be
processed in less than 3 days?
3. What is the probability that a loan will be
processed in 5 days or less?
4. Find the expected processing time and its
standard deviation.
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1. The probability density function for loan processing
time is given by
1
 1
  0.25

f ( x)   7  3 4
0

3 x7
elsewhere
f ( x)
0.25
x
3
7
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2. The probability that a loan will be processed in less
than 3 days:
P( X  3)  0
3. The probability that a loan will be processed in 5
days or less:
P( X  5)  (5  3)  0.25
 2  0.25
 0.5
14
f ( x)
P ( X  5)  0.5
0.25
x
3
5
7
15
4. The expected processing time:
E( X ) 
3 7

2
10
5
2
Variance:
V (X ) 
(7  3)
2

12
4
2
 1.33
(2d.p.)
12
Standard deviation:
  1.33  1.15
(2d.p.)
16
The Normal Distribution
• A continuous random variable X with mean 
and standard deviation  is normally distributed if its
probability density function is given by
f ( x) 
where
1
 2
e
1  x 
 

2  
2
  x  
e = 2.71828… and  = 3.14159…
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f (x)
x

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• A normal random variable is normally distributed with
mean  and standard deviation 
X ~ N ( , )
• The normal distribution is described by two
parameters,  and  .

•
is the mean and determines the location of the
curve.
•
 is the standard deviation and determines the
spread about the mean, that is the width of the curve.
• There is a different normal distribution for each
combination of a mean and a standard deviation.
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 = 10  = 11  = 12
= 2
 =3
 =4
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For the area under the normal curve
• 68% lies between    and   
• 95% lies between   2 and   2
• 99.7% lies between   2 and   2
21
Calculating Normal Probabilities
• If X is a normal random variable with probability
density function f ( x) , the probability that X will take a
value between a and b, P(a  X  b), is given by the
area under the normal curve f ( x) between a and b.
f (x)
P (a  X  b)
x
a
b
22
Standard Normal Distribution
• There are an infinite number of normal distributions
because  and  can take an infinite number of
possible values.
• All normal distributions are related to the standard
normal distribution.
• A standard normal random variable Z is normally
distributed with a mean   0 and a standard
deviation   1.
Z ~ N (0,1)
• We can convert any normal random variable X to a
standard normal random variable Z.
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Standard Normal Probability Table
• Areas under the standard normal probability density
function have been calculated and are available in
Table 3 of Appendix 3 of the text.
• This table gives the probability that the standard
normal random variable Z lies between 0 and z.
24
P (0  Z  z )
f (z)
z
0
z
25
Using Table 3
• Always draw a diagram.
• Shade in the area representing the probability you
are trying to find.
26
Example 2
What is the probability of finding a z value
1. between 0 and 1.43
2. between –0.87 and 0
3. greater than 2.35
4. less than –1.98
5. less than 1.06
6. greater than –1.30
7. between –1.83 and 2.01
8. between 0.89 and 2.12
9. between –2.15 and –1.68
10. less than –1.48 or greater than 1.13
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1. The probability of finding a z value between 0 and
1.43:
P (0  Z  1.43)
3.00
2.50
2.00
1.50
1.00
0.50
0.00
-0.50
-1.00
-1.50
-2.00
-2.50
-3
P (0  Z  1.43)  0.4236
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2. The probability of finding a z value between –0.87
and 0:
P (  0.87  Z  0)
3.00
2.50
2.00
1.50
1.00
0.50
0.00
-0.50
-1.00
-1.50
-2.00
-2.50
-3
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The normal distribution is symmetric, therefore
P (  0.87  Z  0)  P (0  Z  0.87 )
3.00
2.50
2.00
1.50
1.00
0.50
0.00
-0.50
-1.00
-1.50
-2.00
-2.50
-3
P (  0.87  Z  0)  P (0  Z  0.87 )
 0.3078
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3. The probability of finding a z value greater than 2.35:
P ( Z  2.35)
3.00
2.50
2.00
1.50
1.00
0.50
0.00
-0.50
-1.00
-1.50
-2.00
-2.50
-3
31
Since the total area under the normal curve equals
one, and since the curve is symmetric about 0, the
area to the right of 0 is 0.5. Therefore
P ( Z  2.35)  0.5  P (0  Z  2.35)
3.00
2.50
2.00
1.50
1.00
0.50
0.00
-0.50
-1.00
-1.50
-2.00
-2.50
-3
P ( Z  2.35)  0.5  P (0  Z  2.35)
 0.5  0.4906
 0.0094
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4. The probability of finding a z value less than –1.98:
P ( Z   1.98)
3.00
2.50
2.00
1.50
1.00
0.50
0.00
-0.50
-1.00
-1.50
-2.00
-2.50
-3
P ( Z   1.98)  P ( Z  1.98)
 0.5  P (0  Z  1.98)
 0.5  0.4761
 0.0239
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5. The probability of finding a z value less than 1.06:
3.00
2.50
2.00
1.50
1.00
0.50
0.00
-0.50
-1.00
-1.50
-2.00
-2.50
-3
P ( Z  1.06)  0.5  P (0  Z  1.06)
 0.5  0.3554
 0.8554
34
6. The probability of finding a z value greater than
–1.30:
3.00
2.50
2.00
1.50
1.00
0.50
0.00
-0.50
-1.00
-1.50
-2.00
-2.50
-3
P ( Z   1.30)  P ( Z  1.30)
 0.5  P (0  Z  1.30)
 0.5  0.4032
 0.9032
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7. The probability of finding a z value between –1.83
and 2.01:
3.00
2.50
2.00
1.50
1.00
0.50
0.00
-0.50
-1.00
-1.50
-2.00
-2.50
-3
P (  1.83  Z  2.01)  P (  1.83  Z  0)  P (0  Z  2.01)
 P (0  Z  1.83)  P (0  Z  2.01)
 0.4664  0.4778
 0.9442
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8. The probability of finding a z value between 0.89
and 2.12:
3.00
2.50
2.00
1.50
1.00
0.50
0.00
-0.50
-1.00
-1.50
-2.00
-2.50
-3
P (0.89  Z  2.12)  P (0  Z  2.12)  P (0  Z  0.89)
 0.4830  0.31 33
 0.1697
37
9. The probability of finding a z value between –2.15
and –1.68:
3.00
2.50
2.00
1.50
1.00
0.50
0.00
-0.50
-1.00
-1.50
-2.00
-2.50
-3
P (  2.15  Z   1.68)  P (1.68  Z  2.15)
 P (0  Z  2.15)  P (0  Z  1.68)
 0.4842  0.4535
 0.0307
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10. The probability of finding a z value less than –1.48
or greater than 1.13:
P (  1 .4 8  Z )  P ( Z  1 .1 3)
 P ( Z  1 .4 8)  P ( Z  1 .1 3)
  0 .5  P (0  Z  1 .4 8)    0 .5  P (0  Z  1 .1 3) 
 (0 .5  0 .4 3 0 6 )  (0 .5  0 .3 7 0 8)
 0 .0 6 9 4  0 .1 2 9 2
 0 .1 9 8 6
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Reading for next lecture
• Chapter 5, reread section 5.7
Exercises
• 5.50
• 5.51
• 5.52
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