1. Analysis of Economic Data
Dr. Ka-fu Wong
ECON1003
Analysis of Economic Data

2. The Continuous Probability Distribution
Chapter Five
The Continuous Probability Distribution
GOALS
Uniform probability distribution
Normal probability distribution.
Define and calculate z values.
Determine the probability an observation will lie between two points using the standard normal distribution.
Determine the probability an observation will be above or below a given value using the standard normal distribution.
Compare two or more observations that are on different probability distributions.
Use the normal distribution to approximate the binomial probability distribution. l

3. Continuous Probability Distributions
The curve f(x) is the continuous probability distribution (or probability curve or probability density function) of the random variable x if the probability that x will be in a specified interval of numbers is the area under the curve f(x) corresponding to the interval.
Properties of f(x)
f(x)  0 for all x
The total area under the curve of f(x) is equal to 1

4. Mean and Variance for Continuous Probability Distributions
Mean and Variance for Discrete Probability Distributions

5. The Uniform Distribution
If c and d are numbers on the real line, the probability curve describing the uniform distribution is
The mean and standard deviation of a uniform random variable x are

6. The Uniform Probability Curve

7. The Normal Probability Distribution
The normal probability distribution is defined by the equation
and  are the mean and standard deviation,  = … and e = is the base of natural or Naperian logarithms.

8. Importance of Normal Distribution
Describes many random processes or continuous phenomena
Basis for Statistical Inference

9. Characteristics of a Normal Probability Distribution
bell-shaped and single-peaked (unimodal) at the exact center of the distribution.

10. Characteristics of a Normal Probability Distribution
Symmetrical about its mean. The arithmetic mean, median, and mode of the distribution are equal and located at the peak. Thus half the area under the curve is above the mean and half is below it.

11. Characteristics of a Normal Probability Distribution
The normal probability distribution is asymptotic. That is the curve gets closer and closer to the X-axis but never actually touches it.

12. The density functions of normal distributions with zero mean value and different standard deviations
Bell-shaped
Symmetric
Mean=median = mode
Unimodal
Asymptotic

13. Difference Between Normal Distributionsx
(a) x
(b) x
(c)

14. Normal Distribution Probability
Probability is the area under the curve!
A table may be constructed to help us find the probability
f(X) X c d

15. Infinite Number of Normal Distribution Tables
Normal distributions differ by mean & standard deviation.
f(X)
Each distribution would require its own table. X

16. The Standard Normal Probability Distribution
The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1.
It is also called the z distribution.
A z-value is the distance between a selected value, designated X, and the population mean , divided by the population standard deviation, . The formula is:

17. Working with mean and Standard Deviation
Set
Data
Mean
St Dev
(1) 19 20 21
20.00
0.82
(2) -1 1
0.00
(3)
0.71
(4) 38 40 42
40.00
1.63
(5) 57 60 63
60.00
2.45
(2) = (1) – mean(1):
Mean(2)=0; Stdev(2)=Stdev(1)
(3) = (1) + mean(1)
Mean(3)=Mean(1); Stdev(3)<Stdev(1).
(4) = (1)*2; (5) = (1)*3
Mean(4)=mean(1)*2; mean(5)=mean(1)*3
Stdev(4)=stdev(1)*2; stdev(5)=stdev(1)*3

18. Transform to Standard Normal Distribution A numerical example
Any normal random variable can be transformed to a standard normal random variable x
x-
(x-)/σ
x/σ -2 1 -1
0.7071 2
1.4142 3
2.1213 4
2.8284
Mean
std

19. The Standard Normal Probability Distribution
Any normal random variable can be transformed to a standard normal random variable
Suppose X ~ N(µ,  2)
Z=(X-µ)/  ~ N(0,1)
P(X<k) = P [(X-µ)/  < (k-µ)/  ]

20. Standardize the Normal Distribution
Standardized Normal Distribution  
= 1 z  X 
= 0 Z Z
Because we can transform any normal random variable into standard normal random variable, we need only one table!

21. Standardizing Example
Normal Distribution
Standardized Normal Distribution 
= 10 
= 1 Z 
= 5 X 
= 0
.12 Z
6.2 Z

22. Obtaining the Probability
Standardized Normal Probability Table (Portion)
.02 Z
.00
.01 
= 1 Z
0.0
.0000
.0040
.0080
0.0478
0.1
.0398
.0438
.0478
0.2
.0793
.0832
.0871 
= 0
0.12 Z Z
0.3
.1179
.1217
.1255
Shaded Area Exaggerated
Probabilities

23. Example P(3.8  X  5) Normal Distribution
Standardized Normal Distribution 
= 10 
= 1 Z
0.0478
3.8 
= 5 X
-0.12 
= 0 Z Z
Shaded Area Exaggerated

24. Example (2.9  X  7.1) X   2 . 9  5 Z     . 21  10 X   7 .
Normal Distribution Z    . 21
Standardized Normal Distribution  10 
= 10 
= 1 Z
.1664
.0832
.0832
2.9 5
7.1 X
-.21
.21 Z
Shaded Area Exaggerated

25. Example (2.9  X  7.1) X   2 . 9  5 Z     . 21  10 X   7 .
Normal Distribution Z    . 21
Standardized Normal Distribution  10 
= 10 
= 1 Z
.1664
.0832
.0832
2.9 5
7.1 X
-.21
.21 Z
Shaded Area Exaggerated

26. Example P(X  8) X   8  5 Z    . 30 Normal Distribution  10
Standardized Normal Distribution 
= 10 
= 1 Z
.5000
.3821
.1179 
= 5 8 X 
= 0
.30 Z Z
Shaded Area Exaggerated

27. Example P(7.1  X  8) X   7 . 1  5 Z    . 21  10 X   8  5
Normal Distribution Z    . 30
Standardized Normal Distribution  10 
= 10 
= 1 Z
.1179
.0347
.0832 
= 5
7.1 8 X 
= 0 Z
.21
.30 z
Shaded Area Exaggerated

28. Normal Distribution Thinking Challenge
You work in Quality Control for GE. Light bulb life has a normal distribution with µ= 2000 hours &  = 200 hours. What’s the probability that a bulb will last
between 2000 & hours?
less than 1470 hours?

29. Solution P(2000  X  2400) Normal Distribution
= P[(X-µ)/  < (2400-µ)/  ] – P [(X-µ)/  < (2000-µ)/  ]
= P[(X-µ)/  < (2400-µ)/  ] – 0.5
Normal Distribution
Standardized Normal Distribution X  
2400 
2000 Z    2 . 
200 
= 200 
= 1 Z
.4772 
= 2000
2400 X 
= 0
2.0 Z Z

30. Solution P(X  1470) X   1470  2000 Z     2 . 65  200
P(X<1470) = P [(X-µ)/  < (1470-µ)/  ] X  
1470 
2000 Z     2 . 65 
200
Normal Distribution
Standardized Normal Distribution 
= 200 
= 1 Z
.5000
.0040
.4960 X Z
1470 
= 2000
-2.65 
= 0 Z

31. Finding Z Values for Known Probabilities
What Is Z Given P(Z) = ?
Standardized Normal Probability Table (Portion) 
= 1
.01 Z
.00
.02 Z
.1217
0.0
.0000
.0040
.0080
0.1
.0398
.0438
.0478 
= 0
.31 Z
0.2
.0793
.0832
.0871 Z
0.3
.1179
.1217
.1255
Shaded Area Exaggerated

32. Finding X Values for Known Probabilities
Normal Distribution
Standardized Normal Distribution 
= 10 
= 1 Z
.1217
.1217 ? 
= 5 X 
= 0
.31 Z Z
Shaded Area Exaggerated

33. EXAMPLE 1
The bi-monthly starting salaries of recent MBA graduates follows the normal distribution with a mean of $2,000 and a standard deviation of $200. What is the z-value for a salary of $2,200?

34. EXAMPLE 1 continued What is the z-value of $1,700 ?
A z-value of 1 indicates that the value of $2,200 is one standard deviation above the mean of $2,000.
A z-value of –1.50 indicates that $1,700 is 1.5 standard deviation below the mean of $2000.

35. Areas Under the Normal Curve
About 68 percent of the area under the normal curve is within one standard deviation of the mean.  ± 
P( -  < X <  + ) =
About 95 percent is within two standard deviations of the mean.  ± 2 
P( - 2  < X <  + 2 ) =
Practically all is within three standard deviations of the mean.  ± 3 
P( - 3  < X <  + 3 ) =

36. Areas Under the Normal Curve
Between:
± 1  %
± 2  %
± 3  %
µ-2σ
µ+2σ
µ-3σ
µ-1σ
µ+1σ
µ+3σ

37. EXAMPLE 2
The daily water usage per person in New Providence, New Jersey is normally distributed with a mean of 20 gallons and a standard deviation of 5 gallons. About 68 percent of those living in New Providence will use how many gallons of water?
About 68% of the daily water usage will lie between 15 and 25 gallons.

38. EXAMPLE 3
What is the probability that a person from New Providence selected at random will use between 20 and 24 gallons per day?
P(20<X<24)
=P[(20-20)/5 < (X-20)/5 < (24-20)/5 ]
=P[ 0<Z<0.8 ]

39. Example 3 continued
The area under a normal curve between a z-value of 0 and a z-value of 0.80 is
We conclude that percent of the residents use between 20 and 24 gallons of water per day.
See the following diagram.

40. x
EXAMPLE 3
P(0<z<.8)=.2881
0<x<.8

41. How do we know P(0<z<0.8)=0.2881
P(z<c)
Computer software (such as MegaStat in Excel), based on an numerical integration of the standard normal density function.
Numerical table for “Standard normal” c
P(0<z<c) c

42. How do we find P(0<z<0.8)
P(0<z<c)
P(z<c) c c
P(0<z<0.8) =
P(0<z<0.8)
= P(z<0.8) – P(z<0)
= – 0.5
=0.2881

43. EXAMPLE 3 continued
What percent of the population use between 18 and 26 gallons of water per day?
Suppose X ~ N(µ,  2)
Z=(X-µ)/  ~ N(0,1)
P(X<k) = P [(X-µ)/  < (k-µ)/  ]

44. How do we find P(-0.4<z<1.2)
P(0<z<c)
P(z<c) c c
P(-0.4<z<1.2)
= P(z<1.2) - P(z<-0.4)
= P(z<1.2) - P(z>0.4)
= P(z<1.2) – [1- P(z<0.4)]
= – [ ]
=0.5403
P(-0.4<z<1.2)
= P(-0.4<z<0) + P(0<z<1.2)
=P(0<z<0.4) + P(0<z<1.2) =
=0.5403
P(-0.4<z<0) =P(0<z<0.4) because of symmetry of the z distribution.

45. EXAMPLE 4
Professor Mann has determined that the scores in his statistics course are approximately normally distributed with a mean of 72 and a standard deviation of 5. He announces to the class that the top 15 percent of the scores will earn an A. What is the lowest score a student can earn and still receive an A?

46. Example 4 continued Those with a score of 77.2 or more earn an A.
To begin let k be the score that separates an A from a B.
15 percent of the students score more than k, then 35 percent must score between the mean of 72 and k.
Write down the relation between k and the probability:
P(X>k) = and P(X<k) =1-P(X>k) = 0.85
Transform X into z:
P[(X-72)/5) < (k-72)/5 ] = P[z < (k-72)/5]
P[0<z < s] = = 0.35
Find s from table:
P[0<z<1.04]=0.35
Compute k:
(k-72)/5=1.04 implies K=77.2
Those with a score of 77.2 or more earn an A.

47. The Normal Approximation to the Binomial
The normal distribution (a continuous distribution) yields a good approximation of the binomial distribution (a discrete distribution) for large values of n.
The normal probability distribution is generally a good approximation to the binomial probability distribution when n  and n(1-  ) are both greater than 5.
Why can we approximate binomial by normal?
Because of the Central Limit Theorem.

48. The Normal Approximation continued
Recall for the binomial experiment:
There are only two mutually exclusive outcomes (success or failure) on each trial.
A binomial distribution results from counting the number of successes.
Each trial is independent.
The probability is fixed from trial to trial, and the number of trials n is also fixed.

49. The Normal Approximation
binomial

50. Continuity Correction Factor
Because the normal distribution can take all real numbers (is continuous) but the binomial distribution can only take integer values (is discrete), a normal approximation to the binomial should identify the binomial event "8" with the normal interval "(7.5, 8.5)" (and similarly for other integer values). The figure below shows that for P(X > 7) we want the magenta region which starts at 7.5.

51. Continuity Correction Factor
Example: If n=20 and p=.25, what is the probability that X is greater than or equal to 8?
The normal approximation without the continuity correction factor yields
z=(8-20 × .25)/(20 × .25 × .75)0.5 = 1.55,
P(X ≥ 8) is approximately (from the table).
The continuity correction factor requires us to use 7.5 in order to include 8 since the inequality is weak and we want the region to the right.
z = ( × .25)/(20 × .25 × .75)0.5 = 1.29,
P(X ≥ 7.5) is
The exact solution from binomial distribution function is
The continuity correct factor is important for the accuracy of the normal approximation of binomial.
The approximation is quite good.

52. EXAMPLE 5
A recent study by a marketing research firm showed that 15% of American households owned a video camera. For a sample of 200 homes, how many of the homes would you expect to have video cameras?
What is the mean?
What is the variance?
What is the standard deviation?

53. EXAMPLE 5 continued
What is the probability that less than 40 homes in the sample have video cameras?
“Less than 40” means “less or equal to 39”. We use the correction factor, so X is 39.5.
The value of z is 1.88.

54. Example 5 continued
From Standard Normal Table the area between 0 and 1.88 on the z scale is
So the area to the left of 1.88 is =
The likelihood that less than 40 of the 200 homes have a video camera is about 97%.

55. EXAMPLE 5
P(z<1.88) =
=.9699
z=1.88 z

56. The Exponential Distribution
If  is positive, then the exponential distribution is described by the probability density function
mean mx=1/l
standard deviation sx=1/l

57. Example: Computing Exponential Probabilities
Given mx=3.0 or l=1/3=.333,
0.05
0.1
0.15
0.2
0.25
0.3
0.35 5 9
l=0.333 x mx

58. Chapter Five
The Continuous Probability Distribution
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