Download Chapter 6.2 The Binomial Probability Distribution

Chapter 6 Discrete Probability Distributions
Ch 6.1 Discrete Random Variables
Objective A: Discrete Probability Distribution
A1. Distinguish between Discrete and Continuous Random Variables
(counting numbers only)
(can take on decimal values)
Example 1: Determine whether the random variable is discrete or continuous.State the possible values
of the random variable.
(a) The number of fish caught during the fishing tournament.
Discrete: 1, 2, 3, ….
(b) The distance of a baseball travels in the air after being hit.
Continuous: can be 40.3 feet for example
A2. Discrete Probability Distributions (for counting numbers)
∑=1
(odds)
X: 0, 1, 2, 3 are counting numbers representing # of successful free throws
P(x): odds of getting 0, 1, 2, 3 free throws on three tries
#successes
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1. The sum of all the individual probabilities is 1
2. Each individual probability is between 0 and 1 (0% and 100%).
Example 1: Determine whether the distribution is a discrete probability distribution. If not, state why.
no
∑ = 0.73
yes ∑ = 1.00
and each is between 0 and 1
Example 2: (a) Determine the required value of the missing probability to make the distribution
a discrete probability distribution.
(b) Draw a probability histogram.
Where x is the # of times absent a student was for a semester in a history class
a) 0.30 + 0.15 + 0.20 + 0.15 + 0.05 = 0.85 Thus 1 – 0.85 = 0.15 (The probability that a student was absent two
times is 0.15 or 15%.)
b) probability histogram:
probability distribution:
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Objective B: The Mean and Standard Deviation of a Discrete Random Variable
Example 1: Find the mean, variance, and standard deviation of the discrete random variable x which represents
the number of times a person was late to work (in one year) for a small company.
(a) Mean x  [ x  P( x)]
x
0
1
2
3
4
P( x)
0.073
0.117
0.258
0.322
0.230
x  P( x)
0(0.073) = 0
1(0.117) = 0.117
2(0.258) = 0.516
3(0.322)=0.966
4(0.230) = 0.920
Mean 𝜇𝑥 = ∑ x  P ( x ) =2.519
The average number of times a
person was late to work was
about 2.5 times per year.
(b) Variance ---> Use the definition formula
 x 2  [( x  x )2  P( x)]
Formula (2a) in the textbook
x
0
P( x)
0.073
𝑥 − 𝜇𝑥
0 – 2.519 = - 0.2519
(𝑥 − 𝜇𝑥 )2
(− 0.2519)2
(𝑥 − 𝜇𝑥 )2 ∙ 𝑃(𝑥)
(− 0.2519)2 ∙ 0.073 = 0.00463211353
1
0.117
1 – 2.519 = - 1.519
(− 1.519)2
(− 1.519)2 ∙ 0.117 = 0.269961237
2
0.258
2 – 2.519 = - 0.519
(− 0.519)2
(− 0.519)2 ∙ 0.258
3
4
0.322
0.230
3 – 2.519 = 0.481
4 – 2.519 = 1.481
(0.481)2
(1.481)2
(0.481)2 ∙ 0.322
(1.481)2 ∙ 0.230
𝜎𝑥 = 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣. = √𝑠𝑢𝑚𝑚𝑎𝑡𝑖𝑜𝑛
= √1.381639 ≅ 1.18 ≅ 1.2
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*A typical employee was late 2.5 ± 1.2 times a year. Thus, on average, an employee is late between 1.3 and 3.7
times per year.
Objective C : Expected Value (long run probability)
The mean of a random variable is the expected value, E( x)   x  P( x) , of the probability
experimentin the long run. In game theory x is positive for money gained and x is negative for
money lost.
Example 1: A life insurance company sells a $250,000 1-year term life insurance policy to a 20year-old malefor $350. According to the National Vital Statistics Report, 56(9),
the probability that the male survives the year is 0.998734. Compute and interpret
the expected value of this policy to the insurance company.
𝑥 ∙ 𝑃(𝑥)
Gain/Loss
X ($)
P(x)
Company gain (if person
$350
.998734
$349.5569
survives
Company loss (if person
(250000-350) =
(1 - 0.998734) = 0.001266 $ - 316.0569
does not survive
$ - 249650
E(x) = ∑ = 33.5
*In the long run, the company gains $33.50 per each 20 year male customer.
Chapter 6.2 The Binomial Probability Distribution
Objective A : Criteria for a Binomial Probability Experiment
The binomial probability distribution is a discrete probability distribution that obtained from a
binomial experiment. For experiments that have only two outcomes.
Examples: toss a coin (Head/Tail); gender of a baby (Male/Female); color of a card from a deck (Black/Red)
Example: What is the probability that today is a weekday? weekday/ not weekday
Success: 5/7 (Failure: 2/7)
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Example 1: Determine which of the following probability experiments represents a binomial
experiment.If the probability experiment is not a binomial experiment, state why.
(a) A random sample of 30 cars in a used car lot is obtained, and their mileages
recorded. No, more than two outcomes (30 outcomes)
(b) A poll of 1,200 registered voters is conducted in which the repondents are asked
whether they believe Congress should reform Social Security.
Yes, only two outcomes (yes/no)
Fixed number of trials (1200)
Independent trials (one person’s response does not depend on another person’s response)
Assume probability of success is same for each trial
Objective B : Binomial Formula
Let the random variable x be the number of successes in n trials of a binomial experiment.
Where
𝑛!
𝑛 𝐶𝑥 =𝑥!(𝑛−𝑥)!
Example of n!
6! = 6(5)(4)(3)(2)(1) = 720
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Example 1: A binomial probability experiment is conducted with the given parameters.
Compute the probability of x successes in the n independent trials of the
experiment.
n  15, p  0.85, x  12 (Round to four decimal places as needed)
The chance of rain in one city for the next 15 days is 85%. What is the probability that it will rain 12 days out of the
15 days?
The probability of 12 successes out of 15 trials is: P(x=12)= 15 𝐶12 (.85)12 (.15)3 ≅ 0.2184
Where
15 𝐶12
15!
= 12!3! = 455
There is a 21.8% chance there will be 12 successes from the 15 trials. In context: There is a
21.8% chance that it will rain 12 days out of the 15 days.
Example 2: (a) Let’s say the chance of rain for the next 4 days in another city is 65%. Use StatCrunch to
compute a Binomial table of n  4 and p  0.65 .
First enter the possible values of the random variable x in first column then select
Stat  Calculators  Binomial and input n  4 and p  0.65 then each x value: x = 0
compute, then x = 1 compute, etc.
x
0
1
2
3
4
P( x)
0.01500625
0.111475
0.3105375
0.384475
0.17850625
(b) Use the Binomial table from part (a), find P ( x  2) , the probability that it will rain more
than two days.
= P(3) + P(4) = 0.384475+ 0.17850625 = 0.56298125 (statcrunch can do this as well)
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(c) Use the Binomial table from part (a), find P(0  x  3) , the probability that it will rain
beween 0 and less than 3 days.
= 𝑃(0 ≤ 𝑥 < 3) = P(0) + P(1) + P(2) = 0.43701875
(statcrunch only does ≤ in compounds)
Objective C : Binomial Table by StatCunch
Example 1: Use StatCrunch with Binomial Distribution to find P ( x  6) with n  12 and p=0.5. For example, if you
flip a coin 12 times what is the probability of getting 6 or less tails. Before you compute it, would you expect a
high or low probability? P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6)
Example 2: According to the American Lung Association, 90% of adult smokers started smoking
before turning 21 years old. Ten smokers 21 years old or older are randomly selected,
and the number of smokers who started smoking before 21 is recorded.
(a) Explain why this is a binomial experiment.
* two outcomes (yes/no)
* fixed number of trials (10)
* independent trials (one person’s response has no affect on another’s response)
*probability of a success is same each time (0.90)
(b) Use StatCrunch to find the probability that exactly 8 of them started smoking
before 21 years of age.
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P(x = 8) = 0.19371024
(c) Use StatCrunch to find the probability that at least 8 of them started smoking
before 21 years of age.
P(x ≥ 8) = 0.92980917
(d) Use StatCrunch to find the probability that between 7 and 9 of them, inclusive,
started smoking before 21 years of age.
P(7≤X≤9) = 0.63852636
Objective D : Mean and Standard Deviation of a Binomial Random Variable
Example 1: A binomial probability experiment is conducted with the given parameters.
Compute the mean and standard deviation of the random variable x .
n = 15 p = 0.9
The probability of a smoker starting to smoke before the age of 21 is 0.90. Find the mean and standard deviation
of a sample of 15 people.
a. mean
μ = np = 15(.90) = 13.5
*You can expect the mean number of smokers who began smoking before age 21 to be 13.5 smokers in a random
sample of 15 smokers.
Thus, you can expect about 13.5 smokers out of 15 smokers to have started before an age of 21.
b) standard deviation
𝜎 = √15(0.9)(1 − 0.9) = √1.35 ≈ 1.2
*The expected number out of 15 smokers who started smoking before 21 is 13.5 ±1.2. Thus, you can expect out of
15 smokers that between 12.3 and 14.7 started smoking before an age of 21.
Example 2: According to the 2005 American Community Survey, 43% of women aged 18 to 24
were enrolled in college in 2005.
(a) For 500 randomly selected women ages 18 to 24 in 2005, compute the mean
and standard deviation of the random variable x , the number of women who
were enrolled in college.
n = # trials = 500
p = 0.43 (success)
q=1- p = 1 - 0.43 = 0.57 (failure)
mean: μ = np =500 (0.43) = 215
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standard deviation: 𝜎 = √500(0.43)(0.57) ≈ 11.1
(b) Interpret the mean and standard deviation.
For a group of 500 randomly selected women one can expect on average 215 ± 11.1 females to be enrolled in
college. Thus, you can expect between 203.9 and 226.1 women to be enrolled in college out of 500 females.
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