1. Continuous Probability Distributions
Chapter 8
Continuous Probability Distributions

2. 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?
The probability of each value
1/ / / /4 = 1
1/ / /3 = 1
1/ /2 = 1
1/2 1
1/3
2/3

3. 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.
The probability of each value
1/ / / /4 = 1
1/ / /3 = 1
1/ /2 = 1
1/2 1
1/3
2/3

4. Probability Density Function
To calculate probabilities we define a probability density function f(x).
The density function satisfies the following conditions
f(x) is non-negative,
The total area under the curve representing f(x) equals 1.
Area = 1 x1 x2
P(x1<=X<=x2)
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.

5. Uniform Distribution
A random variable X is said to be uniformly distributed if its density function is
The expected value and the variance are

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/( ) = 1/3000 for x: [2000,5000]
P(2500£X£3000) = ( )(1/3000) = .1667
1/3000 x
2000
2500
3000
5000

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/( ) = 1/3000 for x: [2000,5000]
P(X³4000) = ( )(1/3000) = .333
1/3000 x
2000
4000
5000

8. 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/( ) = 1/3000 for x: [2000,5000]
P(X=2500) = ( )(1/3000) = 0
1/3000 x
2000
2500
5000

9. 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.

10. Normal Distribution
A random variable X with mean m and variance s2 is normally distributed if its probability density function is given by

11. The Shape of the Normal Distribution
The normal distribution is bell shaped, and symmetrical around m. 90 m
110
Why symmetrical? Let m = Suppose x = 110.
Now suppose x = 90

12. The effects of m and s 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

13. 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?

14. 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.
Therefore, once probabilities for Z
are calculated, probabilities of any
normal variable can be found.
E(Z) = m = 0
V(Z) = s2 = 1

15. Finding Normal Probabilities
Example - continued 45
- 50 X
- m 60
- 50
P(45<X<60) = P( < < ) 10 s 10
= P(-0.5 < Z < 1)
To complete the calculation we need to compute
the probability under the standard normal distribution

16. Using the Standard Normal 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
Z = 0
Z = z0

17. Finding Normal Probabilities
Example - continued
P(45<X<60) = P( < < ) 45 X 60
- m
- 50 s 10
= P(-.5 < Z < 1)
We need to find the shaded area
z0 = 1
z0 = -.5

18. Finding Normal Probabilities
Example - continued
P(45<X<60) = P( < < ) 45 X 60
- m
- 50 s 10
= P(-.5<Z<1) =
P(-.5<Z<0)+ P(0<Z<1)
P(0<Z<1
.3413
z=0
z0 = 1
z0 =-.5

19. 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
+z0
P(-z0<Z<0) = P(0<Z<z0)

20. Finding Normal Probabilities
Example - continued
.1915
.3413
-.5 .5

21. Finding Normal Probabilities
Example - continued
.1915
.1915
.1915
.1915
.3413
-.5 .5
1.0
P(-.5<Z<1) = P(-.5<Z<0)+ P(0<Z<1) = = .5328

22. 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
10% 0%
0 - 10 5
(i) P(X< 0 ) = P(Z< ) = P(Z< - 2)
.4772 Z -2 2
=P(Z>2) =
0.5 - P(0<Z<2) = = .0228

23. Finding Normal Probabilities
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
10% 0% 1
0 - 10 10
(ii) P(X< 0 ) = P(Z< )
.3413 Z -1
= P(Z< - 1) = P(Z>1) =
0.5 - P(0<Z<1) = = .1587

24. 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

25. Finding Values of Z Example 8.3 & 8.4 Solution
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
Z0.05
1.645

26. 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.

27. 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

28. Exponential distribution for l = .5, 1, 2
f(x) = 2e-2x
f(x) = 1e-1x
f(x) = .5e-.5x
P(a<x<b) = e-la - e-lb a b

29. 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)

30. 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?

31. 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

32. 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.

33. Exponential Distribution
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) =

34. 8.5 Other Continuous Distribution
Three new continuous distributions:
Student t distribution
Chi-squared distribution
F distribution

35. The Student t Distribution
The Student t density function
n is the parameter of the student t distribution
E(t) = V(t) = n/(n – 2)
(for n > 2)

36. The Student t Distribution

37. 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

38. Using the t Table
The table provides the t values (tA) for which P(tn > tA) = A t t t t A
= .05
-tA A
= .05
The t distribution is
symmetrical around 0 tA
=-1.812
=1.812
t.100
t.05
t.025
t.01
t.005

39. The Chi – Squared Distribution
The Chi – Squared density function:
The parameter n is the number of degrees of freedom.

40. The Chi – Squared Distribution

41. Determining Chi-Squared Values
Chi squared values can be found from the chi squared table, from Excel, or from Minitab.
The c2-table entries are the c2 values of the right hand tail probability (A), for which P(c2n > c2A) = A. A
c2A

42. Using the Chi-Squared Table
To find c2 for which P(c2n<c2)=.01, lookup the column labeled c or c2.99 A
c2A
=.05
A =.99
c2.05
c c c c c2.005

43. The F Distribution
The density function of the F distribution: n1 and n2 are the numerator and denominator degrees of freedom. !

44. The F Distribution
This density function generates a rich family of distributions, depending on the values of n1 and n2
n1 = 5, n2 = 10
n1 = 50, n2 = 10
n1 = 5, n2 = 10
n1 = 5, n2 = 1

45. 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.