Transcript Binomial Review

Lesson 8 – 1R
Binomial Review
Knowledge Objectives
• Describe the conditions that need to be present to
have a binomial setting.
• Define a binomial distribution.
• Explain when it might be all right to assume a
binomial setting even though the independence
condition is not satisfied.
• Explain what is meant by the sampling distribution
of a count.
• State the mathematical expression that gives the
value of a binomial coefficient. Explain how to find
the value of that expression.
• State the mathematical expression used to calculate
the value of binomial probability.
Construction Objectives
• Evaluate a binomial probability by using the
mathematical formula for P(X = k).
• Explain the difference between binompdf(n, p, X) and
binomcdf(n, p, X).
• Use your calculator to help evaluate a binomial
probability.
• If X is B(n, p), find µx and x (that is, calculate the
mean and variance of a binomial distribution).
• Use a Normal approximation for a binomial
distribution to solve questions involving binomial
probability
Criteria for a Binomial Setting
A random variable is said to be a binomial 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
Most important skill for using binomial distributions is
the ability to recognize situations to which they do and
don’t apply
English Phrases
Math
Symbol
≥
>
<
≤
=
≠
English Phrases
At least
More than
Fewer than
No more than
Exactly
Different from
No less than
Greater than
Less than
At most
Equals
P(x ≤ A) = cdf (A)
Greater than or equal to
Less than or equal to
Is
P(x = A) = pdf (A)
P(X)
∑P(x) = 1
Cumulative
probability
or cdf
P(x ≤ A)
Values of Discrete Variable, X
P(x > A) = 1 – P(x ≤ A)
X=A
TI-83 Binomial Support
• For P(X = k) using the calculator: 2nd VARS
binompdf(n,p,k)
• For P(k ≤ X) using the calculator: 2nd VARS
binomcdf(n,p,k)
• For P(X ≥ k) use 1 – P(k < X) = 1 – P(k-1 ≤ X)
Binomial Mean and Std Dev
A binomial experiment with n independent trials and
probability of success p has
Mean μx = np
Standard Deviation σx = √np(1-p)
Using Normal Apx to Binomials
As binominal’s number of trials increases the
formula for a binomial becomes unworkable (a
situation alleviated with statistical software).
So statisticians developed a procedure to use a
continuous distribution, the normal, to estimate
a discrete distribution.
This procedure is used later with proportions.
Example 1a
Suppose that in its lifetime, a female kangaroo gives
birth to exactly 10 young. Suppose further that each
kangaroo baby, independently of all the others, has a
20% chance of surviving to maturity.
p = 0.2 n = 10 x = 4
(a) Find the probability that exactly four of the
kangaroo’s young will survive to maturity
With x = 4, then we must use binompdf,
binompdf(10, 0.2, 4) = 0.0881
Example 1a
Suppose that in its lifetime, a female kangaroo gives
birth to exactly 10 young. Suppose further that each
kangaroo baby, independently of all the others, has a
20% chance of surviving to maturity.
p = 0.2 n = 10 x ≥ 4
(b) Find the probability that at least four of the
kangaroo’s young will survive to maturity.
With x ≥ 4, then we must use binomcdf and the
complement rule,
binomcdf(10, 0.2, 3) = 0.8791
P(x ≥ 4) = 1 – P(x ≤ 3) = 1 – 0.8791 = 0.1209
Example 1c
Suppose that in its lifetime, a female kangaroo gives
birth to exactly 10 young. Suppose further that each
kangaroo baby, independently of all the others, has a
20% chance of surviving to maturity.
p = 0.2 n = 10 x ≥ 4
(c) What is the expected (mean) number of kangaroo
young that will survive to maturity?
E(x) = np = 10 (0.2) = 2
Example 1d
Suppose that in its lifetime, a female kangaroo gives
birth to exactly 10 young. Suppose further that each
kangaroo baby, independently of all the others, has a
20% chance of surviving to maturity.
p = 0.2 n = 10 x ≥ 4
(d) What is the standard deviation of the number of
kangaroo young that will survive to maturity?
V(x) = n p (1 – p) = 10 (0.2) (0.8) = 1.6
σ(x) = 1.265
Example 2
State the conditions that must be satisfied for a
random variable X to have a binomial distribution.
1) Outcomes are mutually exclusive (success or failure)
2) Probability of success is the same for each event
3) Each event is independent
4) Fixed number of trials
Example 3
What conditions must be satisfied for a normal
distribution to provide a reasonable approximation for
a binomial distribution?
The number of events must be large enough and the
probability of success or failure not too small
or
1) np ≥ 10
2) n(1 – p) ≥ 10
Example 4a
In a test for ESP, a subject is told that cards the
experimenter can see (but the subject cannot see) contain
either a star, circle, triangle, or square. As the
experimenter looks at a card the subject names the shape
on the card. The answers of a subject who does not have
ESP should be independent observations, each with
probability 1/4 of success.
p = 0.25, n = 100, and x ≤ 30
(a) Find the exact probability that a subject without ESP
will be successful at most 30 times if he makes 100
attempts. Provide support for your work.
Meets binomial criteria: since x ≤ 30 we use binomcdf
binomcdf (100, 0.25, 30) = 0.8962
Example 4b
In a test for ESP, a subject is told that cards the
experimenter can see (but the subject cannot see) contain
either a star, circle, triangle, or square. As the
experimenter looks at a card the subject names the shape
on the card. The answers of a subject who does not have
ESP should be independent observations, each with
probability 1/4 of success.
p = 0.25, n = 100, and x ≤ 30
μx = 25
σx = 4.33
(b) Use the normal approximation to estimate the
probability that a subject without ESP will be successful
at most 30 times if he makes 100 attempts. Show work!
Meets normal criteria for estimation :
np = 25 > 10 n(1 – p) = 75 > 10
normcdf (-e99, 30, 25, 4.33) = 0.8759
Example 4c
In a test for ESP, a subject is told that cards the
experimenter can see (but the subject cannot see) contain
either a star, circle, triangle, or square. As the
experimenter looks at a card the subject names the shape
on the card. The answers of a subject who does not have
ESP should be independent observations, each with
probability 1/4 of success.
Geometric
p = 0.25, n = 100, and x = 5
(c) What is the probability that the subject’s first correct
identification will occur on the 5th card?
P(x=5) = (1 – p)  (1 – p)  (1 – p)  (1 – p)  (p)
= (0.75)4(0.25)
= 0.0791
Example 4d
In a test for ESP, a subject is told that cards the
experimenter can see (but the subject cannot see) contain
either a star, circle, triangle, or square. As the
experimenter looks at a card the subject names the shape
on the card. The answers of a subject who does not have
ESP should be independent observations, each with
probability 1/4 of success.
Geometric
p = 0.25, n = 100, and x > 5
(d) What is the probability that the subject’s first correct
identification will occur after the 5th card?
geometcdf(0.25, 5) = 0.7627
P(x > 5) = 1 – P(x ≤ 5) =
= 1 – 0.7627 = 0.2373