Chapter 8
Continuous Probability
Distributions
1
8.2 Continuous Probability Distributions
• A continuous random variable has an
uncountably infinite number of values in the
interval (a,b).
• The probability that a continuous variable X will
assume any particular value is zero. Why?
1/4
1/3
1/2
0
The probability of each value
+
1/4
+
1/4
+
+
1/3
+
+
1/3
1/2
2/3
1/4 = 1
1/3 = 1
1/2 = 1
1
2
8.2 Continuous Probability Distributions
As the number of values increases the probability of each
value decreases. This is so because the sum of all the
probabilities remains 1.
When the number of values approaches infinity (because X
is continuous) the probability of each value approaches 0.
1/4
1/3
1/2
0
The probability of each value
+
1/4
+
1/4
+
+
1/3
+
+
1/3
1/2
2/3
1/4 = 1
1/3 = 1
1/2 = 1
1
3
Probability Density Function
• To calculate probabilities we define a probability
density function f(x).
• The density function satisfies the following
Area = 1
conditions
P(x1<=X<=x2)
– f(x) is non-negative,
x1 x2
– The total area under the curve representing f(x) equals 1.
• The probability that X falls between x1 and x2 is
found by calculating the area under the graph of f(x)
between x1 and x2.
4
Uniform Distribution
– A random variable X is said to be uniformly
distributed if its density function is
1
f ( x) 
a  x  b.
ba
– The expected value and the variance are
2
ab
(b  a )
E(X) 
V( X ) 
2
12
5
Uniform Distribution
• Example 8.1
– The daily sale of gasoline is uniformly distributed between
2,000 and 5,000 gallons. Find the probability that sales are:
– Between 2,500 and 3,500 gallons
– More than 4,000 gallons
– Exactly 2,500 gallons
f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000]
P(2500X3000) = (3000-2500)(1/3000) = .1667
1/3000
2000 2500 3000
5000
x
6
Uniform Distribution
• Example 8.1
– The daily sale of gasoline is uniformly distributed between
2,000 and 5,000 gallons. Find the probability that sales are:
– Between 2,500 and 3,500 gallons
– More than 4,000 gallons
– Exactly 2,500 gallons
f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000]
P(X4000) = (5000-4000)(1/3000) = .333
1/3000
2000
4000
5000
x
7
Uniform Distribution
• Example 8.1
– The daily sale of gasoline is uniformly distributed between
2,000 and 5,000 gallons. Find the probability that sales are:
– Between 2,500 and 3,500 gallons
– More than 4,000 gallons
– Exactly 2,500 gallons
f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000]
P(X=2500) = (2500-2500)(1/3000) = 0
1/3000
2000 2500
5000
x
8
8.3 Normal Distribution
• This is the most important continuous distribution.
– Many distributions can be approximated by a normal
distribution.
– The normal distribution is the cornerstone distribution
of statistical inference.
9
Normal Distribution
• A random variable X with mean m and variance
s2 is normally distributed if its probability density
function is given by
 x m 
(1/ 2)

s


e
2
1
f ( x) 
  x  
s 2
w here   3.14159... and e  2.71828...
10
The Shape of the Normal Distribution
The normal distribution is bell shaped, and
symmetrical around m.
90
m
Why symmetrical? Let m = 100. Suppose x = 110.
f (110) 
1
s 2
 110100 
(1/ 2)

s 

e
2

1
s 2
 10 
(1/ 2) 
s
e
110
Now suppose x = 90
2
f (90) 
1
s 2
 90100 
(1/ 2)

s 

e
2

1
s 2
 10 
(1/ 2)

s 

e
11
2
The effects of m and s
How does the standard deviation affect the shape of f(x)?
s= 2
s =3
s =4
How does the expected value affect the location of f(x)?
m = 10 m = 11 m = 12
12
Finding Normal Probabilities
• Two facts help calculate normal probabilities:
– The normal distribution is symmetrical.
– Any normal distribution can be transformed into a
specific normal distribution called…
• “STANDARD NORMAL DISTRIBUTION”
Example
The amount of time it takes to assemble a computer is normally
distributed, with a mean of 50 minutes and a standard
deviation of 10 minutes. What is the probability that a
computer is assembled in a time between 45 and 60 minutes?
13
Finding Normal Probabilities
• Solution
– If X denotes the assembly time of a computer, we
seek the probability P(45<X<60).
– This probability can be calculated by creating a new
normal variable the standard normal variable.
Every normal variable
with some m and s, can
be transformed into this Z.
X  mx
Z
sx
E(Z) = m = 0
Therefore, once probabilities for Z
are calculated, probabilities of any
normal variable can be found.
V(Z) = s2 = 1
14
Finding Normal Probabilities
• Example - continued
45 - 50
X m
60 - 50
P(45<X<60) = P(
<
<
)
s
10
10
= P(-0.5 < Z < 1)
To complete the calculation we need to compute
the probability under the standard normal distribution
15
Using the Standard Normal Table
Standard normal probabilities have been
calculated and are provided in a table .
The tabulated probabilities correspond
to the area between Z=0 and some Z = z0 >0
z
0.0
0.1
.
.
1.0
.
.
1.2
.
.
0
0.0000
0.0398
.
.
0.3413
.
.
0.3849
.
.
0.01
0.0040
0.0438
.
.
0.3438
.
.
0.3869
.
.
…….
…….
.
.
0.05
0.0199
0.0596
.
.
0.3531
.
.
0.3944
.
.
P(0<Z<z0)
Z=0
0.06
0.0239
0.0636
.
.
0.3554
.
.
0.3962
.
.
Z = z0
16
Finding Normal Probabilities
• Example - continued
45 - 50
X m
60 - 50
P(45<X<60) = P(
<
<
)
s
10
10
= P(-.5 < Z < 1)
We need to find the shaded area
z0 = -.5
z0 = 1
17
Finding Normal Probabilities
• Example - continued
45 - 50
X m
60 - 50
P(45<X<60) = P(
<
<
)
s
10
10
= P(-.5<Z<1) = P(-.5<Z<0)+ P(0<Z<1
P(0<Z<1)
z
0.0
0.1
.
.
1.0
.
0
0.0000
0.0398
.
.
0.3413
.
0.1
0.0040
0.0438
.
.
0.3438
.
…….
0.05
0.0199
0.0596
.
.3413
.
0.3531
.
z0 =-.5z=0 z0 = 1
0.06
0.0239
0.636
.
.
0.3554
.
18
Finding Normal Probabilities
• The symmetry of the normal distribution makes it
possible to calculate probabilities for negative
values of Z using the table as follows:
-z0
0
+z0
P(-z0<Z<0) = P(0<Z<z0)
19
Finding Normal Probabilities
• Example - continued
0
0.0000
0.0398
.
.
0.1915
.
0.1
0.0040
0.0438
.
.
….
.
…….
.3413
.1915
z
0.0
0.1
.
.
0.5
.
-.5
0.05
0.0199
0.0596
.
.
….
.
0.06
0.0239
0.636
.
.
….
.
.5
20
Finding Normal Probabilities
• Example - continued
0
0.0000
0.0398
.
.
0.1915
.
0.1
0.0040
0.0438
.
.
….
.
…….
.3413
.1915
.1915
.1915
z
0.0
0.1
.
.
0.5
.
-.5
0.05
0.0199
0.0596
.
.
….
.
0.06
0.0239
0.636
.
.
….
.
.5 1.0
P(-.5<Z<1) = P(-.5<Z<0)+ P(0<Z<1) = .1915 + .3413 = .5328
21
Finding Normal Probabilities
• Example 8.2
– The rate of return (X) on an investment is normally distributed
with mean of 10% and standard deviation of (i) 5%, (ii) 10%.
– What is the probability of losing money?
0%
10%
0 - 10
(i) P(X< 0 ) = P(Z<
) = P(Z< - 2)
5
X
.4772
-2
=P(Z>2) = 0.5 - P(0<Z<2) = 0.5 - .4772 = .0228
0
Z
2
22
Find Normal
Probabilities
Finding Normal Probabilities
• Example 8.2
– The rate of return (X) on an investment is normally distributed
with mean of 10% and standard deviation of (i) 5%, (ii) 10%.
– What is the probability of losing money?
X
0%
10%
0 - 10
(ii) P(X< 0 ) = P(Z<
)
10
.3413
-1
= P(Z< - 1) = P(Z>1) = 0.5 - P(0<Z<1) = 0.5 - .3413 = .1587
Z
1
23
Finding Values of Z
• Sometimes we need to find the value of Z for a
given probability
• We use the notation zA to express a Z value for
which P(Z > zA) = A
A
zA
24
Finding Values of Z
• Example 8.3 & 8.4
– Determine z exceeded by 5% of the population
– Determine z such that 5% of the population is below
• Solution
z.05 is defined as the z value for which the area on its right under the
standard normal curve is .05.
0.45
0.05
0.05
-Z0.05
0
Z0.05
1.645
25
Exponential Distribution
• The exponential distribution can be used to model
– the length of time between telephone calls
– the length of time between arrivals at a service station
– the life-time of electronic components.
• When the number of occurrences of an event
follows the Poisson distribution, the time between
occurrences follows the exponential distribution.
26
Exponential Distribution
A random variable is exponentially distributed
if its probability density function is given by
f(x) = le-lx, x>=0.
l is the distribution parameter (l>0).
E(X) = 1/l
V(X) = (1/l)2
27
Exponential distribution for l = .5, 1, 2
2.5
f(x) = 2e-2x
2
f(x) = 1e-1x
1.5
f(x) = .5e-.5x
1
0.5
0
0
1
2
3
4
5
2.5
2
1.5
1
P(a<x<b) = e-la - e-lb
0.5
0
a
b
28
Exponential Distribution
• Finding exponential probabilities is relatively
easy:
– P(X > a) = e–la.
– P(X < a) = 1 – e –la
– P(a1 < X < a2) = e – l(a1) – e – l(a2)
29
Exponential Distribution
• Example 8.5
– The lifetime of an alkaline battery is exponentially
distributed with l = .05 per hour.
– What is the mean and standard deviation of the
battery’s lifetime?
– Find the following probabilities:
• The battery will last between 10 and 15 hours.
• The battery will last for more than 20 hours?
30
Exponential Distribution
• Solution
– The mean = standard deviation =
1/l  1/.05 = 20 hours.
– Let X denote the lifetime.
• P(10<X<15) = e-.05(10) – e-.05(15) = .1341
• P(X > 20) = e-.05(20) = .3679
31
Exponential Distribution
• Example 8.6
– The service rate at a supermarket checkout is 6
customers per hour.
– If the service time is exponential, find the following
probabilities:
• A service is completed in 5 minutes,
• A customer leaves the counter more than 10 minutes after
arriving
• A service is completed between 5 and 8 minutes.
32
Compute Exponential
probabilities
Exponential Distribution
• Solution
– A service rate of 6 per hour =
A service rate of .1 per minute (l = .1/minute).
– P(X < 5) = 1-e-lx = 1 – e-.1(5) = .3935
– P(X >10) = e-lx = e-.1(10) = .3679
– P(5 < X < 8) = e-.1(5) – e-.1(8) = .1572
33
8.5 Other Continuous Distribution
• Three new continuous distributions:
– Student t distribution
– Chi-squared distribution
– F distribution
34
The Student t Distribution
• The Student t density function
[(n  1)]!  t 
f (t ) 
1  
n[(n  2)]!  n 
2
 ( n 1) / 2
n is the parameter of the student t distribution
E(t) = 0
V(t) = n/(n – 2)
(for n > 2)
35
The Student t Distribution
0.2
0.15
n=3
0.1
0.05
0
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
0.2
0.15
n = 10
0.1
0.05
0
-6
-4
-2
0
2
4
6
36
Determining Student t Values
• The student t distribution is used extensively in
statistical inference.
• Thus, it is important to determine values of tA
associated with a given number of degrees of freedom.
• We can do this using
– t tables
– Excel
– Minitab
37
Using the t Table
• The table provides the t values (tA) for which P(tn > tA) = A
A = .05
A = .05
The t distribution is
symmetrical around 0
tA =1.812
-tA=-1.812
t.100
t.05
t.025
t.01
3.078
1.886
.
.
1.372
6.314
2.92
.
.
1.812
12.706
4.303
.
.
2.228
31.821
6.965
.
.
2.764
.
.
.
.
.
.
.
.
.
.
200
1.286
1.282
1.653
1.645
1.972
1.96
2.345
2.326
Degrees of Freedom
1
2
.
.
10

t.005
63.657
9.925
.
.
3.169
.
.
2.601
38
2.576
The Chi – Squared Distribution
• The Chi – Squared density function:
f ( ) 
2
1
[(n / 2)  1]!2n / 2
2 (n / 2) 1   2 2
( )
e
2  0
• The parameter n is the number of degrees of
freedom.
39
The Chi – Squared Distribution
0.0018
0.0016
0.0014
0.0012
0.001
0.0008
0.0006
0.0004
0.0002
0
n =5
n = 10
0
5
10
15
20
25
30
35
40
Determining Chi-Squared Values
• Chi squared values can be found from the chi squared
table, from Excel, or from Minitab.
• The 2-table entries are the 2 values of the right hand
tail probability (A), for which P(2n  2A) = A.
A
0
5
10
152A
20
25
30
35
41
Using the Chi-Squared Table
To find 2 for which
P(2n<2)=.01, lookup
the column labeled
21-.01 or 2.99
A =.05
A =.99
0
5
Degrees of 2
freedom
.995
1
0.0000393
.
.
10
2.15585
.
.
.
.

10
20
22.05
A
15
2.990
0.0001571
.
2.55821
.
.
.
.
.
25
30
35
2.05 2.010 2.005
3.84146
6.6349
7.87944
18.307
23.2093
.
.
25.1882
.
.
.
42
The F Distribution
• The density function of the F distribution:
 n1  n 2  2 
n1  2
n1

 !  n  2
F 2
2
1
 
f (F) 
n1  n 2
 n1  2   n 2  2   n 2 
 n1F  2
 2  !  2  !
 1 

n2 

F0
n1 and n2 are the numerator and denominator
degrees of freedom.
43
The F Distribution
• This density function generates a rich family of
distributions, depending on the values of n1 and n2
n1 = 5, n2 = 10
0.01
0.008
n1 = 50, n2 = 10
0.006
0.004
0.002
0
0
1
2
3
0.008
0.007
0.006
0.005
0.004
0.003
0.002
40.001
0
n1 = 5, n2 = 10
n1 = 5, n2 = 1
5
0
1
2
3
4
5
44
Determining Values of F
• The values of the F variable can be found in the
F table, Excel, or from Minitab.
• The entries in the table are the values of the F
variable of the right hand tail probability (A), for
which P(Fn1,n2>FA) = A.
45