Transcript 8.1 The Binomial Distribution

8.1 The Binomial Distribution
AP Statistics
Two of the most important and useful random
variable distributions are the Binomial
distribution and the Geometric distribution.
In this chapter, we will examine both.
The binomial distributions are an important
class of discrete probability distributions.
Four conditions must be present for a random
variable distribution to be considered “binomial”
The Binomial Distribution
1. Each observation can be categorized as
“success” or “failure.”
2. The probability of success is the same for
each observation.
3. The observations are independent.
4. There are a fixed number of observations.
A binomial setting will be noted as B(n, p)
where B indicates the conditions for a
binomial distribution are met with “n” as
the fixed number of observations and “p”
as the probability for success. Because
the distribution is a random variable (X),
the values of X will be whole numbers
from 0 to n.
Example 1: Suppose you receive a shipment of five monkeyscooters (nothing is funnier than a monkey on a scooter). Each
scooter has a 15% chance of not working. What is the
probability 3 or more scooters in your shipment will be
defective?
Why does this situation satisfy the binomial setting?
1.
2.
3.
4.
Defective or not defective
15% are defective
Independent
5 observations
X = # of defective scooters out of a shipment of 5
X is B(5, 0.15)
Use methods from chapter 7 to complete the
probability distribution for X = # defective
scooters out of 5.
X
0
1
2
P(X) 0.4437 0.3915 0.13818
3
4
0.024 0.00215
5
7.6e-5
Find:
• P(3 or more scooters are defective) =
P(X > 3) = 0.02661
• P(no more than 1 scooter is defective) =
P(X < 1) = 0.83521
• P(odd number of defective scooters) =
0.415576
TI NOTE: (Woo Hoo!) Many binomial
calculations can be done on your calculator…
look under your DISTR menu:
P(X = k) = binompdf(n, p, k)
Binompdf (# trials, probability, point of interest)
P(X < k) = binomcdf(n, p, k)
Binomcdf (#trials, probability, point of interest)
NOTE: cdf calculates to the left
Practice Examples
1. Suppose I am successful on 75% of my penalty shot attempts.
What is the probability I will make 5 or fewer of my next 9
attempts? Define a binomial random variable and calculate the
probability.
X = # successful penalty shots
X ~ B(9, 0.75)
P(X < 5) = 0.1657
binomcdf(9, 0.75, 5)
2.
The Los Angeles Times reported that 80% of airline
passengers prefer to sleep on long flights rather than watch
movies, read, etc. Consider randomly selecting 25 passengers
from a particular long flight. Define a random variable X and
answer the following questions:
X = # of passengers who prefer to sleep
X ~ B(25, 0.80)
• Calculate and interpret P(X = 12) 0.000293
binompdf(25, 0.80, 12)
• Calculate and interpret P(X = 25) 0.003778
binompdf(25, 0.80, 25)
• Calculate and interpret P(X ≥ 20) 0.616689
1 – binomcdf(25, 0.80, 19)
Mean and Standard Deviation of a
Binomial Distribution
If a Random Variable X is B(n, p), what is the
expected number of successes? How much
variability will there be from trial to trial?
Mean and Standard Deviation of a Binomial
Random Variable
If a Random Variable X is B(n, p), the mean and
standard deviation of X are:
 x  np
 x  np1  p   npq
Normal Approximation to Binomial
Distribution
As the number of trials n gets larger, the
binomial distribution of X gets closer to a
normal distribution.
B(n, p) =

N np, np1  p 

When n is large, we can use normal probability
calculations to approximate binomial
probabilities.
Rule of C: Use the normal approximation when
np > 10 and n(1-p) > 10.
NOTE: The normal approximation is most
accurate when p is close to ½ and least
accurate when p is close to 0 or 1.
• Example 2: Sample surveys show that fewer
people enjoy shopping than in the past. A
recent survey asked a nationwide random
sample of 2500 adults if they agreed or
disagreed that “I like buying new clothes, but
shopping is often frustrating and timeconsuming”. The population that the poll
wants to draw conclusions about is all U. S.
Residents ages 18 and over. Suppose that in
fact 60% of all adult U. S. residents would say
“agree” if asked the same question. What is
the probability that 1520 or more of the
sample agree?
Justify using Binomial Distribution:
1. Agree or Disagree
2. 60% success
3. Observations are independent
4. Fixed number (2500) of observations
B( 2500, .60)
1 – Binomcdf (2500, .6, 1519) = .2131
Use the Normal distribution:
np = 2500 (.6) = 1500 > 10
2500(.4) = 1000 > 10
N( np, np1  p  ) = N ( 1500, 24.49)
Normalcdf( 1520, 1E99, 1500, 24.49)
= .2071
The difference in the two calculations is only .006
• Example 3: Suppose 55% of adults have credit card
debt. If we survey 3200 adults, what is the
probability more than 1800 would have credit card
debt?
X = # of adults who have credit card debt out of 3200
X is B(3200, 0.55)
We want P(X > 1800). Can we use a normal
approximation? What would the mean and standard
deviation of the normal approximation be? Use this
to calculate the probability.
np = (3200)(.55) = 1760 > 10
nq = (3200)(.45) = 1440 > 10
N(1760, 28.14)
P(X > 1800) = .0776
Normalcdf(1800, 1E99, 1760, 28.14)
• Youtube video from educator.com on binomial
example:
http://www.youtube.com/watch?v=XzS3lUh0
VoA