Lecture 5
Dan Piett
STAT 211-019
West Virginia University
Test 1 Recap
Grade Distribution for Test 1
40
36
35
Median Score – 85%
30
30
30
Frequency
25
20
18
16
15
11
10
7
5
0
0
25
1
30
0
0
35
40
1
45
2
2
2
50
55
60
65
Grade
8
70
75
80
85
90
95
100
Last Week
 Probability Distributions
 Expected Value of a Probability Distribution

 x  p(x)
Overview
 Binomial Distributions and Probabilities
Binomial Distribution
 Suppose an experiment possesses the following properties:
1.
2.
3.
4.
5.
There are a fixed number of trials, n
Each trial results in one of two possible outcomes
(success/failure)
The probability of a success (p) is the same for each trial
The trials are independent of one another
X = Number of Successes
 This is a binomial experiment
 Note that Binomial Distributions are Discrete (You cannot have
1.9976 successes)
Example: Flipping a Coin 50 Times and
Recording the Number of Heads
Requirements
This Experiment
 There are a fixed number of
 There are n = 50 trials in this




trials, n.
Each trial results in a success
or a failure
Same probability of success
over each trial
The trials are independent of
one another
X = The number of successes




experiment
Heads = Success
Tails = Failure
The probability of getting a
heads remains constant
Tosses are independent of
one another
X = number of heads
General Binomial Distribution
 Suppose X counts the number of successes in a binomial
experiment consisting of n trials. Then X follows a Binomial
Distribution
 Notation: X~B( n, p )
 B stands for binomial distribution
 p stands for the probability of success on a single trial
 For the previous example X~B(50,.5)
Formula for a Binomial Distribution
Problem on Board
 Assume that the probability of a child developing a particular
respiratory illness is as an infant is 15%. A family has two
children. Assume that the illness is not contagious.
 Does this constitute a binomial experiment?
 Find:
 The probability that none of the children develop the illness
 The probability that exactly 1 child develops the illness
 Using the rule that all probabilities must add to 1. Find:
 The probability that exactly 2 children develop the illness
Cumulative Binomial Probabilities
 The previous formula can be used to find the probability that
X equal to exactly some value
 What about other probabilities of interest?
 X equal to less than some value?
 X equal to more than some value?
 X is between two values?
 How do we do this?
Back to the Previous Example
 What is the probability that at most 1 child gets the illness?
 At most = less than or equal to
 At most 1 child = {0, 1, 2}
 P(At most 1 child) = P(X=0)+P(X=1)
 Note: The probability of this event is defined as the sums of
the probabilities.
 Remember that this only works because Binomial Distributions
are discrete
 This is great, but what if n and x are large?
Same Example, New Problem
 Suppose that a small town has 20 infants. What is the
probability that 18 or less develop the respiratory disease?
 18 or less = {0, 1, 2, … , 17, 18, 19, 20}
 P(18 or less) = P(X=0) + P(X=1) +…+ P(X=18)
 We would need to compute 19 probabilities to solve this.
 Is there a better way?
 Actually, there are two
 Using cumulative probability tables
 Using our knowledge of Complementary Probabilities
Cumulative Probability Tables
 Because of the difficulty of calculating these probabilities
(and how common the binomial distribution is). Cumulative
probabilities for specific values of n and x have been
tabulated.
 Note: These tables will be provided on quizzes and exams.
 How to read the table:
 Find the appropriate n and p value, look for x
 This is the probability that X is less than or equal to that
value
Example: Less Than Probabilities
 We have our town of 20 infants. Find the following




probabilities:
At most 5 develop the disease
Less than 8 develop the disease
At most 2 develop the disease
Less than 3 develop the disease
Greater than Probabilities
 So we now know how to calculate the probability that X is
equal to exactly some value or the probability that X is less
than/less than or equal to some value.
 What about the probability that X is greater than/greater
than or equal to some value?
 Think back to complementary probabilities
Headed back to our Example, n=20
 What is the probability that 19 or more children develop the
disease?
 19 or more = {0, 1, 2, … , 18, 19, 20}
 P(19 or more) = P(19) + P(20)
 Remember back to the previous example: P(At most 18)
 P(At most 18) and P(19 or more) are complementary
events
 What does this mean?
 P(19 or more) = 1 – P(At most 18)
 This can be very effective for probabilities such as:
 P(At least 1) = 1 – P(At most 0) = 1 – P(X=0)
Greater than Probabilities
 Remember back to our use of the tables for calculating less
than or equal to probabilities
 We can likewise calculate greater than/greater than or
equal to probabilities using the table.
 Watch the =
 We want to get our greater than probabilities in terms of less
than or equal to
 For n = 6
 P(X>3) = 1 – P(X<=3)
 {1, 2, 3, 4, 5, 6)
 P(X>=3) = 1 – P(X<3) = P(X<=2)
 {1,2 ,3, 4, 5, 6}
Example: Greater Than Probabilities
 We have our town of 20 infants. Find the following




probabilities:
6 or more develop the disease
At least 8 develop the disease
3 or more develop the disease
At least 4 develop the disease
In-between Probabilities
 So far we’ve done
 P(X=x), P(X<x), P(X>x)
 One more to go (The probability the X is between 2 values)




P(a < =x <= b)
Example with the disease: P(X is between 2 and 6)
Between 2 and 6 = {0, 1, 2, 3, 4, 5, 6, 7, … , 19, 20)
P(X is between 2 and 6) = P(X<=6) – P(X<=1)
 Why?
 P(X<=6) = P(0) + P(1) + P(2)+P(3)+P(4)+P(5)+P(6)
 P(X<=1) = P(0) + P(1)
 Subtract these and the 0 and 1 cancel leaving:
 P(2)+P(3)+P(4)+P(5)+P(6)
 This is what we want
Example: In-between Probabilities
 We have our town of 20 infants. Find the following
probabilities:
 Between 3 and 7 develop the disease
 At least 1 develops the disease, but less than 14
Coming back to Exact Probabilities
 We can use the cumulative table to find exact probability as
well
 P(X=2) = P(X<=2) – P(X<=1)
 Same logic as the previous examples
 P(X<=2) = P(X=0) + P(X=1) + P(X=2)
 P(X<=1) = P(X=0) + P(X=1)
 Subtract and you are left with P(X=2)
Mean and Standard Deviation of a
Binomial Distribution
 Expected Value (Mean) of a Binomial Distribution
 n*p
 Standard Deviation of a Binomial Distribution
 Sqrt(n*p*(1-p))