1. Continuous Probability Distributions
Chapter Contents
Uniform Continuous Distribution
Exponential Distribution
Describing a Continuous Distribution
Normal Distribution
Standard Normal Distribution
Normal Approximations

2. Continuous Probability Distributions
Chapter Learning Objectives
Define a continuous random variable.
Calculate uniform probabilities.
Know the form and parameters of the normal distribution.
Find the normal probability for given z or x using tables or Excel.
Solve for z or x for a given normal probability using tables or Excel.
Use the normal approximation to a binomial or a Poisson
distribution.
Find the exponential probability for a given x.
Solve for x for given exponential probability.

3. Characteristics of the Uniform Distribution
Uniform Continuous Distribution
LO7-2
If X is a random variable that is
uniformly distributed between
a and b, its PDF has
constant height.
Characteristics of the Uniform Distribution
Denoted U(a, b)
Area = base x height = (b-a) x 1/(b-a) = 1

4. Characteristics of the Uniform Distribution

5. Example: Anesthesia Effectiveness
An oral surgeon injects a painkiller prior to extracting a tooth. Given the varying characteristics of patients, the dentist views the time for anesthesia effectiveness as a uniform random variable that takes between 15 minutes and 30 minutes.
X is U(15, 30)
a = 15, b = 30, find the mean and standard deviation.
Find the probability that the effectiveness anesthetic takes between 20 and 25 minutes.

6. Example: Anesthesia Effectiveness
P(20 < X < 25) = (25 – 20)/(30 – 15) = 5/15 = = 33.33%

7. 1. The time that the customers at the “self serve” check out stations at the Mejers store spend checking out follows a uniform distribution between 0 and 3 minutes.
Determine the height and draw this uniform distribution.
b. How long does the typical customer wait to check out?
c. Determine the standard deviation of the wait time.
d. What is the probability a particular customer will wait less than one minute?
e. What is the probability a particular customer will wait between 1.5 and 2 minutes?

8. A statistics instructor collected data on the time it takes the students
to complete a test. The test taking time is uniformly distributed within a
range of 35 minutes to 55 minutes.
a. Determine the height.
b. How long does the typical test taking time?
c. Determine the standard deviation of the test taking time.
d. What is the probability a particular student will take
less than 45 minutes?
e. What is the probability a particular student will take
between 45 and 50 minutes?

9. 3. A tube of Listerine Tartar Control toothpaste contains 4. 2 ounces
3. A tube of Listerine Tartar Control toothpaste contains 4.2 ounces. As people use the toothpaste, the amount remaining in any tube is random. Assume the amount of toothpaste left in the tube follows a uniform distribution. From this information, we can determine the following information about the amount remaining in a toothpaste tube without invading anyone’s privacy.
How much toothpaste would you expect to be remaining in the tube?
b. What is the standard deviation of the amount remaining in the tube?
c. What is the likelihood there is less than 3.0 ounces remaining in the tube?
d. What is the probability there is more than 1.5 ounces remaining in the tube?

10. Characteristics of the Exponential Distribution
If events per unit of time follow a Poisson distribution, the time until the next event follows the Exponential distribution.
The time until the next event is a continuous variable.
NOTE: Here
we will find
probabilities
> x or ≤ x.

11. Characteristics of the Exponential Distribution
Probability of waiting less than or
equal to x
Probability of waiting more than x

12. Example Customer Waiting Time
Between 2P.M. and 4P.M. on Wednesday, patient insurance inquiries arrive at Blue Choice insurance at a mean rate of 2.2 calls per minute.
What is the probability of waiting more than 30 seconds (i.e., 0.50 minutes) for the next call?
Set  = 2.2 events/min and x = 0.50 min
P(X > 0.50) = e–x = e–(2.2)(0.5) = or 33.29% chance of waiting more than 30 seconds for the next call.

13. Example Customer Waiting Time
P(X > 0.50)
P(X ≤ 0.50)

14. Inverse Exponential
If the mean arrival rate is 2.2 calls per minute, we want the 90th percentile for waiting time (the top 10% of waiting time).
Find the x-value that defines the upper 10%.

15. Inverse Exponential

16. Mean Time Between Events

17. 5. If arrivals occur at a mean rate of 3
5. If arrivals occur at a mean rate of 3.6 events per hour, the exponential probability of:
waiting more than 0.5 hour for the next arrival is: 
P(X > .50) = exp(-3.6 × 0.50) =
6. waiting less than 0.5 hour for the next arrival is:
P(X < .50) = 1 - exp(-3.6 × 0.50) = = .8347
7. If arrivals occur at a mean rate of 2.6 events per minute, the exponential probability of waiting more than 1.5 minutes for the next arrival is: 
P(X > 1.5) = exp(-2.6 × 1.50) = .0202

18. 8. If arrivals occur at a mean rate of 1
8. If arrivals occur at a mean rate of 1.6 events per minute, the exponential probability of waiting less than 1 minute for the next arrival is: 
(X < 1) = 1 - exp(-1.6 × 1) = =
9. On average, a major earthquake (Richter scale 6.0 or above) occurs 3 times a decade in a certain California county. What is the probability that less than six months will pass before the next earthquake? 
Set λ = 3/120 = earthquake/month so
P(X < 6) = 1 - exp( × 6) = = .1393

19. Events as Intervals
Discrete Variable – each value of X has its own probability P(X).
Continuous Variable – events are intervals and probabilities are areas under continuous curves. A single point has no probability.
7-19

20. PDF – Probability Density Function
Continuous PDF’s:
Denoted f(x)
Must be nonnegative
Total area under curve = 1
Mean, variance and shape depend on the PDF parameters
Reveals the shape of the distribution
7-20

21. Continuous CDF’s: CDF – Cumulative Distribution Function Denoted F(x)
Shows P(X ≤ x), the cumulative proportion of scores
Useful for finding probabilities
Describing a Continuous Distribution:
For a continuous random variable, the function which is used to determine the cumulative probability for the variable is called the cumulative distribution function, denoted by F(x).
F(x) = P(X  x). This function is useful for finding probabilities.
The slide shows an example of the cumulative distribution function for a normal distribution.
7-21

22. Probabilities as Areas
Continuous probability functions:
Unlike discrete distributions, the probability at any single point = 0.
The entire area under any PDF, by definition,
is set to 1.
Mean is the balance point of the distribution.
7-22

23. Characteristics of the Normal Distribution
Defined by two parameters, µ and .
Domain is –  < X < +  (continuous scale).
Almost all (99.7%) of the area under the normal curve is included in the range µ – 3 < X < µ + 3.
Symmetric and unimodal about the mean.

24. Characteristics of the Normal Distribution

25. Characteristics of the Normal Distribution
Normal PDF f(x) reaches a maximum at µ and has points of inflection at µ ± 
Bell-shaped curve
NOTE: All normal
distributions
have the same
shape but differ
in the axis scales.

26. Characteristics of the Normal Distribution
Normal CDF

27. Characteristics of the Standard Normal Distribution
Since for every value of µ and , there is a different normal distribution, we transform a normal random variable to a standard normal distribution with µ = 0 and  = 1 using the formula.

28. Characteristics of the Standard Normal
Standard normal PDF f(x) reaches a maximum at z = 0 and has points of inflection at +1.
Shape is unaffected by the transformation. It is still a bell-shaped curve.
Figure 7.11

29. Characteristics of the Standard Normal
Standard normal CDF
A common scale from -3 to +3 is used.
Entire area under the curve is unity.
The probability of an event P(z1 < Z < z2) is a definite integral of f(z).
However, standard normal tables or Excel functions can be used to find the desired probabilities.

30. Normal Areas from Table 1
The table allows you to find the area under the curve from 0 to z.
For example, find P(0 < Z < 1.96):

31. Normal Areas from Table 1
Now find P(-1.96 < Z < 1.96).
Due to symmetry, P(-1.96 < Z) is the same as P(Z < 1.96).
So, P(-1.96 < Z < 1.96) = = or 95% of the area under the curve.

32. Basis for the Empirical Rule
Approximately 68% of the area under the curve is between + 1
Approximately 95% of the area under the curve is between + 2
Approximately 99.7% of the area under the curve is between + 3

33. Normal Areas from Cumulative Table 2
Cumulative table allows you to find the area under the curve from the left of z (similar to Excel).
For example,
P(Z < 1.96)
P(Z < -1.96)
P(-1.96 < Z < 1.96)

34. Normal Areas from Tables
Both tables yield identical results.
Use whichever table is easiest.
Finding z for a Given Area
Both tables can be used to find the z-value corresponding to a given probability.
For example, what z-value defines the top 1% of a normal distribution?
This implies that 49% of the area lies between 0 and z which gives z = 2.33 by looking for an area of in Table1.

35. Finding Areas by using Standardized Variables
Suppose John took an economics exam and scored 86 points. The class mean was 75 with a standard deviation of 7. What percentile is John in? That is, what is P(X < 86) where X represents the exam scores?
So John’s score is 1.57 standard deviations about the mean.
P(X < 86) = P(Z < 1.57) = (from Appendix C-2)
So, John is approximately in the 94th percentile.

36. The net sales and the number of employees for aluminum fabricators with similar characteristics are organized into frequency distributions. Both are normally distributed. For the net sales, the mean is $180 million and the standard deviation is $25 million. For the number of employees, the mean is 1,500 and the standard deviation is 120. Clarion Fabricators had sales of $170 million and 1,850 employees.
a. Convert Clarion’s sales and number of employees to z values.
b. Locate the two z values.
c. Compare Clarion’s sales and number of employees with those of the other fabricators.

37. Finding Areas by using Standardized Variables
NOTE: You can use Excel, Minitab, TI83/84 etc. to compute these probabilities directly.

38. LO7-5: Solve for z or x for a normal probability using tables or Excel.
Inverse Normal
How can we find the various normal percentiles (5th, 10th, 25th, 75th,
90th, 95th, etc.) known as the inverse normal? That is, how can we
find X for a given area? We simply turn the standardizing
transformation around:
Solving for x in z = (x − μ)/ gives x = μ + zσ

39. Inverse Normal
For example, suppose that John’s economics professor has decided
that any student who scores below the 10th percentile must retake the
exam.
The exam scores are normal with μ = 75 and σ = 7.
What is the score that would require a student to retake the exam?
We need to find the value of x that satisfies P(X < x) = .10.
The z-score for with the 10th percentile is z = −1.28.

40. Inverse Normal
The steps to solve the problem are:
Use Appendix C or Excel to find z = −1.28 to satisfy P(Z < −1.28) = .10.
Substitute the given information into z = (x − μ)/σ to get
−1.28 = (x − 75)/7
Solve for x to get x = 75 − (1.28)(7) = (or 66 after rounding)
Students who score below 66 points on the economics exam will be
required to retake the exam.

41. Inverse Normal

42. Normal Approximation to the Binomial
Binomial probabilities are difficult to calculate when n is large.
Use a normal approximation to the binomial distribution.
As n becomes large, the binomial bars become smaller and continuity is approached.

43. Example Coin Flips Normal Approximation to the Binomial
Rule of thumb: when n ≥ 10 and n(1- ) ≥ 10, then it is appropriate to use the normal approximation to the binomial distribution.
In this case, the mean and standard deviation for the binomial distribution will be equal to the normal µ and , respectively.
Example Coin Flips
If we were to flip a coin n = 32 times and  = .50, are the requirements for a normal approximation to the binomial distribution met?

44. Example Coin Flips n = 32 x .50 = 16 n(1- ) = 32 x (1 - .50) = 16
So, a normal approximation can be used.
When translating a discrete scale into a continuous scale, care must be taken about individual points.
For example, find the probability of more than 17 heads in 32 flips of a fair coin.
This can be written as P(X  18).
However, “more than 17” actually falls between 17 and 18 on a discrete scale.

45. Example Coin Flips
Since the cutoff point for “more than 17” is halfway between 17 and 18, we add 0.5 to the lower limit and find P(X > 17.5).
This addition to X is called the Continuity Correction.
At this point, the problem can be completed as any normal distribution problem.

46. Example Coin Flips
P(X > 17) = P(X ≥ 18)  P(X ≥ 17.5) = P(Z > 0.53) =