Unformatted text preview: Chapter 8 Continuous
Continuous
Probability
Distributions
Distributions
1 8.2
Continuous Probability
8.2
Distributions
Distributions
• A continuous random variable has an
uncountably infinite number of values
in the interval (a,b).
• The probability that a continuous
The
variable X will assume any particular
value is zero. Why? value
value The probability of each
1/4
1/3
1/2 + 1/4 +
0 1/3 +
1/3
+ 1/4 1/2 2/3 + 1/4 = 1
1/3 = 1
1/2 = 1 +
1 2 8.2
Continuous Probability
8.2
Distributions
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 The probability of each value
+
1/4
+
1/4
+
+
1/3
+
+
0 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
To
probability density function f(x).
probability
• The density function satisfies the
The
Area = 1
following conditions <=X<=x )
following
P(x
1 2 x1 x2
– f(x) is nonnegative,
– The total area under the curve representing
The
f(x) equals 1.
f(x)
• The probability that X falls between x1 and x2 is found by calculating the area
and
under the graph of f(x) between x1 and
under
x2. 4 Uniform Distribution
– A random variable X is said to be
random
uniformly distributed if its density function
is
is
1
f(x) =
a ≤ x ≤ b.
b− a a+ b
(b − a)2
– The expected valueV(X) = variance are
and the
E(X)=
2
12
5 Uniform Distribution • Example 8.1 – The daily sale of gasoline is uniformly distributed
The
between 2,000 and 5,000 gallons. Find the
probability that sales are:
probability
– Between 2,500 and 3,500 gallons
– More than 4,000 gallons
– Exactly 2,500 gallons
f(x) = 1/(50002000) = 1/3000 for x: [2000,5000]
P(2500≤ X≤ 3000) = (30002500)(1/3000) = .1667
1/3000
2000 25003000 5000 x
6 Uniform Distribution • Example 8.1 – The daily sale of gasoline is uniformly distributed
The
between 2,000 and 5,000 gallons. Find the
probability that sales are:
probability
– Between 2,500 and 3,500 gallons
– More than 4,000 gallons
– Exactly 2,500 gallons
f(x) = 1/(50002000) = 1/3000 for x: [2000,5000]
P(X≥ 4000) = (50004000)(1/3000) = .333
1/3000
2000 4000 5000 x
7 Uniform Distribution • Example 8.1 – The daily sale of gasoline is uniformly distributed
The
between 2,000 and 5,000 gallons. Find the
probability that sales are:
probability
– Between 2,500 and 3,500 gallons
– More than 4,000 gallons
– Exactly 2,500 gallons
f(x) = 1/(50002000) = 1/3000 for x: [2000,5000]
P(X=2500) = (25002500)(1/3000) = 0
1/3000
2000 2500 5000 x
8 8.3 Normal Distribution
• This is the most important continuous
This
distribution.
distribution.
– Many distributions can be approximated by
Many
a normal distribution.
normal
– The normal distribution is the cornerstone
The
distribution of statistical inference.
distribution 9 Normal Distribution
• A random variable X with mean µ and
random
variance σ 2 is normally distributed if
variance
its probability density function is given
by
by x−µ −(1/ 2) σ e 2 1
f(x) =
− ∞ ≤ x≤ ∞
σ 2π
where = 3.14159 and e = 2.71828
π
...
...
10 The Shape of the Normal
The
Distribution
Distribution
The normal distribution is bell
shaped, and
symmetrical around µ . 90
110
µ
symmetrical? Let µ = 100. Suppose x = 110.Now suppose x = 90
f(110 =
) 1
σ 2π − 110 100
−(1/ 2) σ e 2 = 1
σ 2π 10
−(1/ 2) σ
e 2 f(90 =
) 1
σ 2π − 90 100
−(1/ 2) σ
e 2 = 1
σ 2π −10
−(1/ 2) σ
e 11 2 The effects of µ and σ
The
How does the standard deviation affect the shape of f(x)?
σ= 2
σ =3
σ =4 How does the expected value affect the location of f(x)?
µ = 10 µ = 11 = 12
µ 12 Finding Normal Probabilities
• Two facts help calculate normal
Two
probabilities:
probabilities:
– The normal distribution is symmetrical.
– Any normal distribution can be
Any
transformed into a specific normal
distribution called…
distribution
Example
• “STANDARD NORMAL
STANDARD
The DISTRIBUTION”
amount of time it takes to assemble a
DISTRIBUTION”
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
13
minutes? Finding Normal Probabilities
• Solution – If X denotes the assembly time of a
If
computer, we seek the probability
P(45<X<60).
P(45<X<60).
– This probability can be calculated by
This
creating a new normal variable the
Every normal variable normal variable. once probabilities for Z
Therefore,
standard
standard
X−µ Z= with some µ and σ , can
be transformed into this Z. E(Z) = µ = 0 σx xare calculated, probabilities of an normal variable can be found. V(Z) = σ2 = 1 14 Finding Normal Probabilities
• Example  continued
45 50
X− µ
P(45<X<60) = P(
<σ
10 60  50
<
10 ) = P(0.5 < Z < 1) o complete the calculation we need to compute
e probability under the standard normal distribut
15 Using the Standard Normal
Using
Table
Table
Standard normal probabilities have been
calculated and are provided in a table . P(0<Z<z0) The tabulated probabilities correspond
to the area between Z=0 and some Z = z0 >0= 0 Z = z0
Z
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
.
. 0.06
0.0239
0.0636
.
.
0.3554
.
.
0.3962
.
. 16 Finding Normal Probabilities
• Example  continued
45 50
X− µ
P(45<X<60) = P(
<σ
10 60  50
<
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− µ
60  50
P(45<X<60) = P(
<σ
<
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
. z=0
z0 =.5 z0 = 1 ) 0.06
0.0239
0.636
.
.
0.3554
. 18 Finding Normal Probabilities
• The symmetry of the normal
The
distribution makes it possible to
calculate probabilities for negative
values of Z using the table as follows:
values 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
.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
The
normally distributed with mean of 10% and
standard deviation of (i) 5%, (ii) 10%.
standard
– 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) = P(0<Z<2) = 0.5  .4772 = .0228
0.5  0 2 Z
22 Find Normal
Probabilities Finding Normal Probabilities
• Example 8.2
– The rate of return (X) on an investment is
The
normally distributed with mean of 10% and
standard deviation of (i) 5%, (ii) 10%.
standard
– What is the probability of losing money?
X
0% 0  10
(ii) P(X< 0 ) = P(Z<
10 10%
.3413 )
1 1
= P(Z<  1) = P(Z>1)  P(0<Z<1) = 0.5  .3413 = .1587
0.5 = Z
23 Finding Values of Z
• Sometimes we need to find the value
Sometimes
of Z for a given probability
of
• We use the notation zA to express a Z
value for which P(Z > zA) = A
value 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.
right
0.45
0.05 0.05
Z0.05 0 Z0.05 1.645 25 Exponential Distribution
• The exponential distribution can be used
The
to model
to
– the length of time between telephone calls
– the length of time between arrivals at a
the
service station
service
– the lifetime of electronic components. • When the number of occurrences of an
When
event follows the Poisson distribution,
the time between occurrences follows 26 Exponential Distribution
A random variable is exponentially
random
distributed if its probability density
function is given by
function
f(x) = λ eλx,
f(x) x>=0. λ is the distribution parameter (λ >0). E(X) = 1/λ V(X) = (1/λ) 2 27 Exponential distribution for λ = .5, 1, 2
2.5 f(x) = 2e2x 2 f(x) = 1e1x 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) = eλa  eλb 0.5
0 a b 28 Exponential Distribution
• Finding exponential probabilities is
Finding
relatively easy:
relatively
– P(X > a) = e–λa.
– P(X < a) = 1 – e –λa
P(X
λ
λ
– P(a1 < X < a2) = e – λ((a1) – e – λ((a2) 29 Exponential Distribution
• Example 8.5
– The lifetime of an alkaline battery is
The
exponentially distributed with λ = .05 per
hour.
hour.
– What is the mean and standard deviation
What
of the battery’s lifetime?
of
– 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 =
The
1/λ = 1/.05 = 20 hours.
1/
1/.05
– 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
The
checkout is 6 customers per hour.
– If the service time is exponential, find the
If
following probabilities:
following
• A service is completed in 5 minutes,
service
• A customer leaves the counter more than 10
customer
minutes after arriving
minutes
• A service is completed between 5 and 8
service
minutes.
minutes.
32 Compute Exponential
probabilities Exponential Distribution
Exponential • Solution
– A service rate of 6 per hour =
service
A service rate of .1 per minute (λ = .
service
1/minute).
– P(X < 5) = 1eλx = 1 – e.1(5) = .3935
– P(X >10) = eλx = e.1(10) = .3679
– P(5 < X < 8) = e.1(5) – e.1(8) = .1572
33 8.5 Other Continuous
8.5
Distribution
Distribution
• Three new continuous distributions:
– Student t distribution
– Chisquared distribution
– F distribution 34 The Student t Distribution
• The Student t density function
[(ν − 1
)]! t f(t) =
1+ νπ[(ν − 2)]! ν 2 − (ν +1) / 2 ν is the parameter of the student t
distribution
distribution
(for n > 2) E(t) = 0 V(t) = ν /(ν – 2)
2) 35 The Student t Distribution
0.2
0.15 ν =3 0.1
0.05
0
6 5 4 3 2 1 0 1 2 3 4 5 6 0.2
0.15 ν = 10 0.1
0.05
0
6 4 2 0 2 4 6 36 Determining Student t
Determining
Values
• The student t distribution is used extensively
The
in statistical inference.
in
• Thus, it is important to determine values of
Thus,
tA associated with a given number of
degrees of freedom.
degrees
• We can do this using
–
–
– t tables
tables
Excel
Minitab
37 Using the t Table
• The table provides the t values (tA) for which
for t
P(tν > tA) = A
P(t
A= .05 A= .05
The t distribution is
symmetrical around 0 t=1.812 tA
=1.812
A
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
The
Distribution
• The Chi – Squared density function:
2 f (χ ) = 1
[(ν / 2) − 1]!2ν / 2 2 (ν / 2 ) −1 − χ 2 2 (χ ) e χ2 > 0 • The parameter ν is the number of
The
degrees of freedom.
degrees
39 The Chi – Squared
The
Distribution
0.0018
0.0016
0.0014
0.0012
0.001
0.0008
0.0006
0.0004
0.0002
0 ν =5
ν = 10 0 5 10 15 20 25 30 35 40 Determining ChiSquared
Determining
Values
Values
• Chi squared values can be found from the
Chi
chi squared table, from Excel, or from
Minitab.
Minitab.
• The χ2table entries are the χ2 values of the
The table
right hand tail probability (A), for which
P(χ 2 ν > χ2 Α) = A.
P(
A.
A 0 5 10 15
χ 2A 20 25 30 35 41 Using the ChiSquared Table
o find χ2 for which
(χ2ν<χ2)=.01, lookup
he column labeled
2
2
1.01 or χ .99
0 A=.05 A =.99
5 10 Degrees of
χ 2 .9 9 5
χ 2 .9 9 0
freedom
1
0.0000393 0.0001571
.
.
10
2.15585
2.55821
.
.
.
.
.
. 22
χχ.020
5
Α 15 25 30 35 χ 2 .0 5 χ 2 .0 1 0 . 6.6349 .
. 3.84146
18.307 23.2093 25.1882
.
.
.
. . . 7.87944 χ 2 .0 0 5 42 The F Distribution
• The density function of the F
The
distribution:
distribution: ν1 + ν 2 − 2
ν −2
ν ! ν 2
F2
2 1
1 f(F) = 1 ν1+ ν 2 ν1 − 2 ν 2 − 2 ν 2 ! 2 ! 1+ ν1F 2 2 ν2 ν 1 and ν 2 are the numerator and
denominator degrees of freedom.
denominator F>0 43 The F Distribution
• This density function generates a rich family
This
of distributions, depending on the values of
ν 1 and ν 2
0.01 0.008
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
4
0.001
0 5 0 1 2 3 4 5 44 Determining Values of F
• The values of the F variable can be
The
found in the F table, Excel, or from
Minitab.
Minitab.
• The entries in the table are the values
The
of the F variable of the right hand tail
probability (A), for which P(Fν1,ν2>FA) =
probability
A.
A.
45 ...
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This note was uploaded on 06/06/2011 for the course ADMS 2320 taught by Professor Rochon during the Spring '08 term at York University.
 Spring '08
 ROCHON

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