1. Chapter 6: Probability Distributions
Section 6.1: How Can We Summarize Possible Outcomes and Their Probabilities?

2. Learning Objectives Random variable
Probability distributions for discrete random variables
Mean of a probability distribution
Summarizing the spread of a probability distribution
Probability distribution for continuous random variables

3. Learning Objective 1: Randomness
The numerical values that a variable assumes are the result of some random phenomenon:
Selecting a random sample for a population or
Performing a randomized experiment

4. Learning Objective 1: Random Variable
A random variable is a numerical measurement of the outcome of a random phenomenon.

5. Learning Objective 1: Random Variable
Use letters near the end of the alphabet, such as x, to symbolize
Variables
A particular value of the random variable
Use a capital letter, such as X, to refer to the random variable itself.
Example: Flip a coin three times
X=number of heads in the 3 flips; defines the random variable
x=2; represents a possible value of the random variable

6. Learning Objective 2: Probability Distribution
The probability distribution of a random variable specifies its possible values and their probabilities.
Note: It is the randomness of the variable that allows us to specify probabilities for the outcomes

7. Learning Objective 2: Probability Distribution of a Discrete Random Variable
A discrete random variable X has separate values (such as 0,1,2,…) as its possible outcomes
Its probability distribution assigns a probability P(x) to each possible value x:
For each x, the probability P(x) falls between 0 and 1
The sum of the probabilities for all the possible x values equals 1

8. Learning Objective 2: Example
What is the estimated probability of at least three home runs?
P(3)+P(4)+P(5)= =0.17

9. Learning Objective 3: The Mean of a Discrete Probability Distribution
The mean of a probability distribution for a discrete random variable is
where the sum is taken over all possible values of x.
The mean of a probability distribution is denoted by the parameter, µ.
The mean is a weighted average; values of x that are more likely receive greater weight P(x)

10. Learning Objective 3: Expected Value of X
The mean of a probability distribution of a random variable X is also called the expected value of X.
The expected value reflects not what we’ll observe in a single observation, but rather that we expect for the average in a long run of observations.
It is not unusual for the expected value of a random variable to equal a number that is NOT a possible outcome.

11. Learning Objective 3: Example
Find the mean of this probability distribution.
The mean:
= 0(0.23) + 1(0.38) + 2(0.22) + 3(0.13) + 4(0.03) + 5(0.01) = 1.38

12. Learning Objective 4: The Standard Deviation of a Probability Distribution
The standard deviation of a probability distribution, denoted by the parameter, σ, measures its spread.
Larger values of σ correspond to greater spread.
Roughly, σ describes how far the random variable falls, on the average, from the mean of its distribution

13. Learning Objective 5: Continuous Random Variable
A continuous random variable has an infinite continuum of possible values in an interval.
Examples are: time, age and size measures such as height and weight.
Continuous variables are measured in a discrete manner because of rounding.

14. Learning Objective 5: Probability Distribution of a Continuous Random Variable
A continuous random variable has possible values that form an interval.
Its probability distribution is specified by a curve.
Each interval has probability between 0 and 1.
The interval containing all possible values has probability equal to 1.

15. Chapter 6: Probability Distributions
Section 6.2: How Can We Find Probabilities for Bell-Shaped Distributions?

16. Learning Objectives Normal Distribution
Rule for normal distributions
Z-Scores and the Standard Normal Distribution
The Standard Normal Table: Finding Probabilities
Using the TI-calculator: find probabilities

17. Learning Objectives Using the Standard Normal Table in Reverse
Using the TI-calculator: find z-scores
Probabilities for Normally Distributed Random Variables
Percentiles for Normally Distributed Random Variables
Using Z-scores to Compare Distributions

18. Learning Objective 1: Normal Distribution
The normal distribution is symmetric, bell-shaped and characterized by its mean µ and standard deviation .
The normal distribution is the most important distribution in statistics
Many distributions have an approximate normal distribution
Approximates many discrete distributions well when there are a large number of possible outcomes
Many statistical methods use it even when the data are not bell shaped

19. Learning Objective 1: Normal Distribution
Normal distributions are
Bell shaped
Symmetric around the mean
The mean () and the standard deviation () completely describe the density curve
Increasing/decreasing  moves the curve along the horizontal axis
Increasing/decreasing  controls the spread of the curve

20. Learning Objective 1: Normal Distribution
Within what interval do almost all of the men’s heights fall? Women’s height?

21. Learning Objective 2: 68-95-99.7 Rule for Any Normal Curve
68% of the observations fall within one standard deviation of the mean
95% of the observations fall within two standard deviations of the mean
99.7% of the observations fall within three standard deviations of the mean

22. Learning Objective 2: Example : 68-95-99.7% Rule
Heights of adult women
can be approximated by a normal distribution
= 65 inches; =3.5 inches
Rule for women’s heights
68% are between 61.5 and 68.5 inches
[ µ   = 65  3.5 ]
95% are between 58 and 72 inches
[ µ  2 = 65  2(3.5) = 65  7 ]
99.7% are between 54.5 and 75.5 inches
[ µ  3 = 65  3(3.5) = 65  10.5 ]

23. Learning Objective 2: Example : 68-95-99.7% Rule
What proportion of women are less than 69 inches tall?
? = 84% ?
(height values)
16%
68%
(by Rule) -1 +1

24. Learning Objective 3: Z-Scores and the Standard Normal Distribution
The z-score for a value x of a random variable is the number of standard deviations that x falls from the mean
A negative (positive) z-score indicates that the value is below (above) the mean
z-scores can be used to calculate the probabilities of a normal random variable using the normal tables in the back of the book

25. Learning Objective 3: Z-Scores and the Standard Normal Distribution
A standard normal distribution has mean µ=0 and standard deviation σ=1
When a random variable has a normal distribution and its values are converted to z-scores by subtracting the mean and dividing by the standard deviation, the z-scores have the standard normal distribution.

26. Learning Objective 4: Table A: Standard Normal Probabilities
Table A enables us to find normal probabilities
It tabulates the normal cumulative probabilities falling below the point +z
To use the table:
Find the corresponding z-score
Look up the closest standardized score (z) in the table.
First column gives z to the first decimal place
First row gives the second decimal place of z
The corresponding probability found in the body of the table gives the probability of falling below the z-score

27. Learning Objective 4: Example: Using Table A
Find the probability that a normal random variable takes a value less than 1.43 standard deviations above µ; P(z<1.43)=.9236
TI Calculator = Normcdf(-1e99,1.43,0,1)= .9236

28. Learning Objective 4: Example: Using Table A
Find the probability that a normal random variable takes a value greater than 1.43 standard deviations above µ: P(z>1.43)= =.0764
TI Calculator = Normcdf(1.43,1e99,0,1)=

29. Learning Objective 4: Example:
Find the probability that a normal random variable assumes a value within 1.43 standard deviations of µ
Probability below 1.43σ =
Probability below -1.43σ = ( )
P(-1.43<z<1.43) = =.8472
TI Calculator = Normcdf(-1.43,1.43,0,1)= .8472

30. Learning Objective 5: Using the TI Calculator
To calculate the cumulative probability
2nd DISTR; 2:normalcdf(lower bound, upper bound,mean,sd)
Use –1E99 for negative infinity and 1E99 for positive infinity

31. Learning Objective 5: Find Probabilities Using TI Calculator
Find probability to the left of -1.64
P(z<-1.64)=normcdf(-1e99,-1.64,0,1)=.0505
Find probability to the right of 1.56
P(z>1.56)=normcdf(1.56,1e99,0,1)=.0594
Find probability between -.50 and 2.25
P(-.5<z<2.25)=normcdf(-.5,2.25,0,1)=.6793

32. Learning Objective 6: How Can We Find the Value of z for a Certain Cumulative Probability?
To solve some of our problems, we will need to find the value of z that corresponds to a certain normal cumulative probability
To do so, we use Table A in reverse
Rather than finding z using the first column (value of z up to one decimal) and the first row (second decimal of z)
Find the probability in the body of the table
The z-score is given by the corresponding values in the first column and row

33. Learning Objective 6: How Can We Find the Value of z for a Certain Cumulative Probability?
Example: Find the value of z for a cumulative probability of
Look up the cumulative probability of in the body of Table A.
A cumulative probability of corresponds to z =
Thus, the probability that a normal
random variable falls at least 1.96
standard deviations below the
mean is

34. Learning Objective 6: How Can We Find the Value of z for a Certain Cumulative Probability?
Example: Find the value of z for a cumulative probability of
Look up the cumulative probability of in the body of Table A.
A cumulative probability of corresponds to z = 1.96.
Thus, the probability that a normal
random variable takes a value no more
than 1.96 standard deviations above
the mean is

35. Learning Objective 7: Using the TI Calculator to Find Z-Scores for a Given Probability
2nd DISTR 3:invNorm; Enter
invNorm(percentile,mean,sd)
Percentile is the probability under the curve from negative infinity to the z-score
Enter

36. Learning Objective 7: Examples
The probability that a standard normal random variable assumes a value that is ≤ z is What is z? Invnorm(.975,0,1)=1.96
The probability that a standard normal random variable assumes a value that is > z is
What is z? Invnorm(.975,0,1)=1.96
The probability that a standard normal random variable assumes a value that is ≥ z is
What is z? Invnorm(1-.881,0,1)=-1.18
The probability that a standard normal random variable assumes a value that is < z is
What is z? Invnorm(.119,0,1)= -1.18

37. Learning Objective 7: Example
Find the z-score z such that the probability within z standard deviations of the mean is 0.50.
Invnorm(.75,0,1)= .67
Invnorm(.25,0,1)= -.67
Probability = P(-.67<Z<.67)=.5

38. Learning Objective 8: Finding Probabilities for Normally Distributed Random Variables
State the problem in terms of the observed random variable X, i.e., P(X<x)
Standardize X to restate the problem in terms of a standard normal variable Z
Draw a picture to show the desired probability under the standard normal curve
Find the area under the standard normal curve using Table A

39. Learning Objective 8: P(X<x)
Adult systolic blood pressure is normally distributed with µ = 120 and σ = 20. What percentage of adults have systolic blood pressure less than 100?
P(X<100) =
Normcdf(-1E99,100,120,20)=.1587
15.9% of adults have systolic blood pressure less than 100

40. Learning Objective 8: P(X>x)
Adult systolic blood pressure is normally distributed with µ = 120 and σ = 20. What percentage of adults have systolic blood pressure greater than 100?
P(X>100) = 1 – P(X<100)
P(X>100)= =.8413
Normcdf(100,1e99,120,20)=.8413
84.1% of adults have systolic blood pressure greater than 100

41. Learning Objective 8: P(X>x)
Adult systolic blood pressure is normally distributed with µ = 120 and σ = 20. What percentage of adults have systolic blood pressure greater than 133?
P(X>133) = 1 – P(X<133)
P(X>133)= =.2578
Normcdf(133,1E99,120,20)=.2578
25.8% of adults have systolic blood pressure greater than 133

42. Learning Objective 8: P(a<X<b)
Adult systolic blood pressure is normally distributed with µ = 120 and σ = 20. What percentage of adults have systolic blood pressure between 100 and 133?
P(100<X<133) = P(X<133)-P(X<100)
Normcdf(100,133,120,20)=.5835
58% of adults have systolic blood pressure between 100 and 133

43. Learning Objective 9: Find X Value Given Area to Left
Adult systolic blood pressure is normally distributed with µ = 120 and σ = 20. What is the 1st quartile?
P(X<x)=.25, find x:
Look up .25 in the body of Table A to find z= -0.67
Solve equation to find x:
Check:
P(X<106.6) P(Z<-0.67)=0.25
TI Calculator = Invnorm(.25,120,20)=106.6

44. Learning Objective 9: Find X Value Given Area to Right
Adult systolic blood pressure is normally distributed with µ = 120 and σ = % of adults have systolic blood pressure above what level?
P(X>x)=.10, find x.
P(X>x)=1-P(X<x)
Look up 1-0.1=0.9 in the body of Table A to find z=1.28
Solve equation to find x:
Check:
P(X>145.6) =P(Z>1.28)=0.10
TI Calculator = Invnorm(.9,120,20)=145.6

45. Learning Objective 10: Using Z-scores to Compare Distributions
Z-scores can be used to compare observations from different normal distributions
Example:
You score 650 on the SAT which has =500 and
=100 and 30 on the ACT which has =21.0 and
=4.7. On which test did you perform better?
Compare z-scores
SAT: ACT:
Since your z-score is greater for the ACT, you performed better on this exam

46. Chapter 6: Probability Distributions
Section 6.3: How Can We Find Probabilities When Each Observation Has Two Possible Outcomes?

47. Learning Objectives The Binomial Distribution
Conditions for a Binomial Distribution
Probabilities for a Binomial Distribution
Factorials
Examples using Binomial Distribution
Do the Binomial Conditions Apply?
Mean and Standard Deviation of the Binomial Distribution
Normal Approximation to the Binomial

48. Learning Objective 1: The Binomial Distribution
Each observation is binary: it has one of two possible outcomes.
Examples:
Accept, or decline an offer from a bank for a credit card.
Have, or do not have, health insurance.
Vote yes or no on a referendum.

49. Learning Objective 2: Conditions for the Binomial Distribution
Each of n trials has two possible outcomes: “success” or “failure”.
Each trial has the same probability of success, denoted by p.
The n trials are independent.
The binomial random variable X is the number of successes in the n trials.

50. Learning Objective 3: Probabilities for a Binomial Distribution
Denote the probability of success on a trial by p.
For n independent trials, the probability of x successes equals:

51. Learning Objective 4: Factorials
Rules for factorials:
n!=n*(n-1)*(n-2)…2*1
1!=1
0!=1
For example,
4!=4*3*2*1=24

52. Learning Objective 5: Example: Finding Binomial Probabilities
John Doe claims to possess ESP.
An experiment is conducted:
A person in one room picks one of the integers 1, 2, 3, 4, 5 at random.
In another room, John Doe identifies the number he believes was picked.
Three trials are performed for the experiment.
Doe got the correct answer twice.

53. Learning Objective 5: Example 1
If John Doe does not actually have ESP and is actually guessing the number, what is the probability that he’d make a correct guess on two of the three trials?
The three ways John Doe could make two correct guesses in three trials are: SSF, SFS, and FSS.
Each of these has probability: (0.2)2(0.8)=0.032.
The total probability of two correct guesses is 3(0.032)=0.096.

54. Learning Objective 5: Example 1
The probability of exactly 2 correct guesses is the binomial probability with n = 3 trials, x = 2 correct guesses and p = 0.2 probability of a correct guess.
2nd Vars
0:binampdf(n,p,x)
Binampdf(3,.2,2)=0.096

55. Learning Objective 5: Binomial Example 2
1000 employees, 50% Female
None of the 10 employees chosen for management training were female.
The probability that no females are chosen is:
Binompdf(10,.5,0)= E-4
It is very unlikely (one chance in a thousand) that none of the 10 selected for management training would be female if the employees were chosen randomly

56. Learning Objective 6: Do the Binomial Conditions Apply?
Before using the binomial distribution, check that its three conditions apply:
Binary data (success or failure).
The same probability of success for each trial (denoted by p).
Independent trials.

57. Learning Objective 6: Do the Binomial Conditions Apply to Example 2?
The data are binary (male, female).
If employees are selected randomly, the probability of selecting a female on a given trial is 0.50.
With random sampling of 10 employees from a large population, outcomes for one trial does not depend on the outcome of another trial

58. Learning Objective 7: Binomial Mean and Standard Deviation
The binomial probability distribution for n trials with probability p of success on each trial has mean µ and standard deviation σ given by:

59. Learning Objective 7: Example: Racial Profiling?
Data:
262 police car stops in Philadelphia in 1997.
207 of the drivers stopped were African-American.
In 1997, Philadelphia’s population was 42.2% African-American.
Does the number of African-Americans stopped suggest possible bias, being higher than we would expect (other things being equal, such as the rate of violating traffic laws)?

60. Learning Objective 7: Example: Racial Profiling?
Assume:
262 car stops represent n = 262 trials.
Successive police car stops are independent.
P(driver is African-American) is p =
Calculate the mean and standard deviation of this binomial distribution:

61. Learning Objective 7: Example: Racial Profiling?
Recall: Empirical Rule
When a distribution is bell-shaped, close to 100% of the observations fall within 3 standard deviations of the mean.

62. Learning Objective 7: Example: Racial Profiling?
If there is no racial profiling, we would not be surprised if between about 87 and 135 of the 262 drivers stopped were African-American.
The actual number stopped (207) is well above these values.
The number of African-Americans stopped is too high, even taking into account random variation.
Limitation of the analysis:
Different people do different amounts of driving, so we don’t really know that 42.2% of the potential stops were African-American.

63. Learning Objective 8: Approximating the Binomial Distribution with the Normal Distribution
The binomial distribution can be well approximated by the normal distribution when the expected number of successes, np, and the expected number of failures, n(1-p) are both at least 15.