Transcript Document

Lesson 6 - 2
Binomial Probability Distribution
Objectives
• Determine whether a probability experiment is a
binomial experiment
• Compute probabilities of binomial experiments
• Compute the mean and standard deviation of a
binomial random variable
• Construct binomial probability histograms
Vocabulary
• Trial – each repetition of an experiment
• Success – one assigned result of a binomial
experiment
• Failure – the other result of a binomial experiment
• PDF – probability distribution function
• CDF – cumulative (probability) distribution function,
computes probabilities less than or equal to a
specified value
Criteria for a Binomial Probability
Experiment
An experiment is said to be a binomial experiment
provided:
1. The experiment is performed a fixed number of
times. Each repetition is called a trial.
2. The trials are independent
3. For each trial there are two mutually exclusive
(disjoint) outcomes: success or failure
4. The probability of success is the same for each trial
of the experiment
Binomial Notation
There are n independent trials of the experiment
Let p denote the probability of success and then
1 – p is the probability of failure
Let x denote the number of successes in n
independent trials of the experiment. So 0 ≤ x ≤ n
Determining probabilities:
With your calculator:
2nd VARS 0 yields
binompdf(n,p,x)
Book: Table II in Appendix A
Table III
Binomial PDF
The probability of obtaining x successes in n
independent trials of a binomial experiment, where the
probability of success is p, is given by:
P(x) = nCx px (1 – p)n-x,
x = 0, 1, 2, 3, …, n
A binomial experiment with n independent trials and
probability of success p has
Mean μx = np
Standard Deviation σx = √np(1-p)
English Phrases
Math
Symbol
≥
At least
>
More than
<
Fewer than
≤
No more than
=
Exactly
≠
Different from
English Phrases
No less than
Greater than
Less than
At most
Equals
Greater than or equal to
Less than or equal to
Is
∑P(x) = 1
P(X)
Cumulative
probability
or cdf
P(x ≤ A)
Values of Discrete Variable, X
cdf(x > A) = 1 – P(x ≤ A)
X=A
Example 1
In the “Pepsi Challenge” a random sample of 20 subjects
are asked to try two unmarked cups of pop (Pepsi and
Coke) and choose which one they prefer. If preference is
based solely on chance what is the probability that:
P(d=P) = 0.5
a) 6 will prefer Pepsi?
P(x) = nCx px(1-p)n-x
P(x=6 [p=0.5, n=20]) = 20C6 (0.5)6(1- 0.5)20-6
= 20C6 (0.5)6(0.5)14 = 0.037
b) 12 will prefer Coke?
P(x=12 [p=0.5, n=20]) = 20C12 (0.5)12(1- 0.5)20-12
= 20C12 (0.5)12(0.5)8 = 0.1201
Example 1 cont
P(d=P) = 0.5
P(x) = nCx px(1-p)n-x
c) at least 15 will prefer Pepsi?
P(at least 15) = P(15) + P(16) + P(17) + P(18) + P(19) + P(20)
Use cumulative PDF either Table III in book or calculator
P(X ≥ 15) = 1 – P(X ≤ 14) = 1 – 0.9793 = 0.0207
d) at most 8 will prefer Coke?
P(at most 8) = P(0) + P(1) + P(2) + … + P(6) + P(7) + P(8)
Use cumulative PDF either Table III in book or calculator
P(X ≤ 8) = 0.2517
Example 2
A certain medical test is known to detect 90% of the people
who are afflicted with disease Y. If 15 people with the
disease are administered the test what is the probability
that the test will show that:
P(x) = nCx px(1-p)n-x
P(Y) = 0.9
a) all 15 have the disease?
P(x=15 [p=0.9, n=15]) = 15C15 (0.9)15(1- 0.9)15-15
= 15C15 (0.9)15(0.1)0 = 0.20589
b) at least 13 people have the disease?
P(at least 13) = P(13) + P(14) + P(15)
Use cumulative PDF either Table III in book or calculator
P(X ≥ 13) = 1 – P(X ≤ 12) = 1 – 0.1841 = 0.8159
Example 2 cont
P(Y) = 0.9
P(x) = nCx px(1-p)n-x
c) 8 have the disease?
P(x=8 [p=0.9, n=15]) = 15C8 (0.9)8(1- 0.9)15-8
= 15C8 (0.9)8(0.1)7 = 0.000277
Example 3
A university claims that 80% of its basketball players
get their degree. An investigation examines the fates of
a random sample of 20 players who entered the
program over a period of several years. Of these
players, 10 graduated and 10 are no longer in school. If
the university's claim is true, what is the probability
that exactly 10 out of 20 graduate? Can you conclude
anything about the university's claim?
Example 3 cont
If the university's claim is true, what is the probability
that exactly 10 out of 20 graduate?
P(Y) = 0.8
P(x) = nCx px(1-p)n-x
P(x=10 [p=0.8, n=20]) = 20C10 (0.8)10(1- 0.8)20-10
= 20C10 (0.8)10(0.2)10 = 0.00203
Can you conclude anything about the university's
claim?
It is either a false claim or we got a very unusual sample!
10 – np
10 – 18
-8
-8
Z = ----------- = ----------------- = -------- = --------------- = -4.47
np(1-p)
20(.8)(.2)
3.2
1.7889
Summary and Homework
• Summary
– Binomial experiments have 4 specific
criteria that must be met
– E(X) = np and V(X) = np(1-p)
– Calculator has pdf and cdf functions
– Tables in book for pdf and cdf
• Homework
– pg 340-343: 11, 12, 17 – 20, 30 – 32, 41, 46