Continuous Probability Distributions

Chapter 7

Continuous Probability
Distributions

1


7.1 Probability Density Function
A continuous random variable has an
uncountable infinite number of values in the
interval (a,b).
To calculate probabilities of continuous random
variables we define a probability density function
f(x), which satisfies the following conditions
Area = 1
– f(x) is non-negative,
– The total area under the curve
representing f(x) is equal to 1.

2


The probability that a continuous variable X will assume
any particular value is zero.
The probability that x falls between ‘a’ and ‘b’ is the area
under the graph of f(x) between ‘a’ and ‘b’.

P(a≤ x≤ b)
a

b

3


Uniform Distribution
A random variable X is said to be uniformly
distributed if its density function
is
1
f ( x) =

with

b−a

a+b
E(X) =
2

f(X)

a

b

a ≤ x ≤ b.

(b − a) 2
V(X ) =

12

X
4


Example 7.1: 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 3,000 gallons

P(2500≤ X≤ 3000) = (3000-2500)(1/3000) = .1667
1/3000

2000 2500 3000

5000

x
5


7.2 Normal Distribution
A random variable X with mean µ and variance σ 2 is normally distributed if its probability density function is

We denote a normal distribution by N(µ, σ 2 ) .
Normal distributions range from minus infinity to plus infinity

2
x
−

µ


x
−
µ


−−(1(1/ /22)) σ 
 σ 
e

2

1
1
∞≤≤ xx≤≤∞
∞
ff((xx))==
e
−−∞
σσ 22ππ
where ππ==33..14159
14159...... and
and ee ==22..71828
71828......
where

6

The Shape of the Normal Distribution

µ

bell shaped, and symmetrical around µ.
7


Increasing the mean shifts the curve to
the right…

8.8


Increasing the standard deviation “flattens” the
curve…

8.9


Two facts help calculate normal probabilities:
-

The normal distribution is symmetrical.
2
Any random variable (rv) X having N(µ , σ ) can be transformed into a rv Z having
N(0 , 1 ) , called the
“STANDARD NORMAL DISTRIBUTION”
or the Z-distribution by

X −− µµ
X
ZZ ==
σσ
10


We can use the following function to convert any normal random variable to a standard
normal random variable:

0

Some advice: always
draw a picture!

8.11

The “standard normal distribution” or the Zdistribution N(0,1)


This shifts the
mean of X to zero…

0

8.12


0

This changes the
shape of the curve…

8.13


Example 1: 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 between 45 and 60
minutes?
•Solution
Let X denote the assembly time of a computer.
We seek P(4514


45 - 50
P(45<
10

X− µ
60 - 50
<
)
σ
10

(45-50)/10 = -.5

(60 – 50)/10 = 1

= P(-0.5
To complete the
calculation we need to
compute
the probability under the
standard normal
distribution

z0 = -.5

z0 = 1


The
probability
provided by
the Z-Table
covers the
area
between
‘-infinity’
and some
‘z0’.

z0

16


Using the Normal Table, or the Z table

P(-.5
.3413

0
17


Example 2: The rate of return (X) on an investment is normally
distributed with mean of 10% and standard deviation of 5%.
What is the probability of losing money?
Solution

Money is lost if the
return is negative
0%

10%

X

P(X< 0 ) = P(Z< 0 - 10 ) = P(Z< - 2) = .0228
5

18


Example 2: The standard deviation of the rate of return (X) is
now 10%. What is the probability of losing money?
The curve for σ =5%
The curve for σ = 10%
X
0%

P(X< 0 ) = P(Z<

10%

0 - 10
)=
10

P(Z< - 1) = .1587

Comment: When the standard deviation is 10% rather than
5%, more values fall away from the mean, so the probability
of finding values at the distribution tail increases from .0228
to .1587.

Z
19

Using Excel to Find Normal Probabilities
For P(X=normdist(k,m,s,True).
Example: Let m = 50 and s = 10.
P(X < 30): =normdist(30,50,10,True)
P(X > 45): =1 - normdist(45,50,10,True)
P(30=normdist(60,50,10,True) – normdist(30,50,10,True).

Using “normsdist” if the “Z” value is known
P(Z<1.2234): =normsdist(1.2234)
20


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
21


Example 3:
What percentage of N(0,1) the standard normal population is located to the right of z .10?
Answer: 10%

A = .10

z.10

What percentage of the standard normal population
is located to the left of z.30?
Answer: 70%

1 – A = .70

A = .30
z.30

What percentage of the standard normal population is located between z .95 and
z.40:
55%
Comment: z.95 has a negative value

A1 = .95

A2 = .40
z.95

z.40


Example 4: Determine z not exceeded by 5% of
the population; that is, z is exceeded by 95% of
the population.
Solution: Because of the symmetry of the normal
distribution it is the negative value of z.05.

0.05
-1.645

0.95
-Z0.05

0

Z0.05

1.645

23


7.5 Other Continuous Distribution
Student t-distribution
Chi-squared distribution
F distribution

24


The Student t density function
− ( ν +1) / 2
22 − ( ν +1) / 2

[(νν−−11)]!
)]!  tt 
[(

1
+
ff((tt))==
1
+




νπ[([(νν−−22)]!
)]! νν
νπ

ν (nu) is called the degrees of freedom
E(t) = 0

V(t) = n/(n – 2)
(for n > 2)
25