Transcript Chapter ___ Review: Type the Subject of the Chapter

CHAPTER 8 REVIEW:
THE BINOMIAL AND GEOMETRIC DISTRIBUTIONS
Kate Schwartz & Lexy Ellingwood
THE BIG IDEA
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This chapter focuses on two important classes of discrete random variables, each of
which involves two outcomes or events of interest. Both require independent trails and the
same probability of success on each trial. The binomial random variable requires a fixed
number of trials; the geometric random variable has the property that the number of trials
varies. Both the binomial and the geometric settings occur sufficiently often in applications
that they deserve special attention.
VOCABULARY YOU NEED TO KNOW
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Binomial Setting
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Mean
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Binomial Probability
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Standard Deviation
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Binomial Coefficient
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Normal Approximation
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Factorial n!
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Geometric Distribution
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Probability Distribution Function
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Geometric Probability
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Cumulative Distribution Function
KEY TOPICS COVERED IN THIS CHAPTER
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Binomial
1.
Identify a random variable as a binomial by verifying four conditions: two outcomes (success and failure); fixed
number of trials; independent trials; and the same probability of success for each trial
2.
Use technology or the formula to determine binomial probabilities and to construct probability distribution tables
and histograms
3.
Calculate cumulative distribution functions for binomial random variables, and construct cumulative distribution
tables and histograms
4.
Calculate means (expected values) and standard deviations of binomial random variables
5.
Use Normal approximation to the binomial distribution to compute probabilities
Geometric
1.
Identify a random variable as a geometric by verifying four conditions: two outcomes (success and failure);
independent trials; the same probability of success for each trial; and the count of interest is the number of trials
required to get the first success
2.
Use technology or the formula to determine geometric probabilities and to construct probability distribution tables
and histograms
3.
Calculate cumulative distribution functions for geometric random variables, and construct cumulative distribution
tables and histograms
4.
Calculate means (expected values) and standard deviations of geometric random variables
FORMULAS YOU SHOULD KNOW
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Binomial Probability
𝑃 𝑋=𝑘 =
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𝑘
𝑝 (1 − 𝑝)
𝑛−𝑘
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=
𝑛!
𝑘! 𝑛−𝑘 !
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Factorial n!
Binomial Mean
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Binomial Standard Deviation
𝜎𝑋 =
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𝜇𝑋 = 𝑝
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𝜇𝑋 = 𝑛𝑝
𝑁 𝑛𝑝, 𝑛𝑝 1 − 𝑝
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𝑢𝑠𝑒𝑑 𝑤ℎ𝑒𝑛
𝑛𝑝 ≥ 10 𝑎𝑛𝑑 𝑛(1 − 𝑝) ≥ 10
Geometric Standard Deviation
𝜎𝑋 =
𝑛𝑝(1 − 𝑝)
Normal Approximation
Geometric Mean
1
𝑛! = 𝑛 × 𝑛 − 1 × 𝑛 − 2 × ⋯ ×
3×2×1
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Geometric Probability
𝑃 𝑋 = 𝑛 = (1 − 𝑝)𝑛−1 𝑝
Binomial Coefficient
𝑛
𝑘
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𝑛
𝑘
(1−𝑝)
𝑝2
Probability that is takes more than n
trials to see the first success
𝑃 𝑥 > 𝑛 = (1 − 𝑝)𝑛
CALCULATOR KEY STROKES
n = number of trials
p = probability of success
r = number of success
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Binomial probability P(X = r) of exactly r
successes in n independent trials, with
probability of success p for a single trial. If r
is omitted, gives a list of all probabilities from
0 to n
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geometpdf(p,n)
binompdf(n,p,r)
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Binomial cumulative probability P(X ≤ r) of r
or fewer successes in n independent trials,
with probability of success p for a single trial.
If r is omitted, gives a list of all cumulative
probabilities from 0 to n
binomcdf(n,p,r)
Geometric probability P(X = n) that the first
success occurs on the nth trial in a series of
independent trials, with probability of success
p for a single trial
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Geometric cumulative probability P(X ≤ n)
that the first success occurs on or before the
nth trial in a series of independent trials, with
probability of success p for a single trial
geometcdf(p,n)
EXAMPLE PROBLEM: BINOMIAL PROBABILITY
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A board game has a spinner on a circle that has five equal sectors, numbered 1, 2, 3, 4,
and 5, respectively. If a player has four spins, find the probability that the player spins an
even number no more than two times on those four spins.
EXAMPLE PROBLEM: BINOMIAL PROBABILITY
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There are 2 even numbers out of the 5 numbers.
No more than = at most.
EXAMPLE PROBLEM: GEOMETRIC PROBABILITY
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In a standard deck of 52 cards, there are 12 face cards. So the probability of drawing a
face card from a full deck is 12/52 = 0.231.
1. If you draw cards with replacement (that is, you replace the card in the deck before drawing
the next card), what is the probability that the first face card you draw is the 10th card?
EXAMPLE PROBLEM: GEOMETRIC PROBABILITY
1. P(X = 10) =(1 − 𝑝)𝑛−1 𝑝= (1 – 0.231)9 (0.231) = 0.022.
On the TI-83/84: geometpdf(0.231,10)= 0.0217)
HELPFUL HINTS
Binomial
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To determine P(X = x)
Use binompdf(n, p, x): where n is the number of
observations, p is the probability of success.
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To determine P(X ≤ x)
Use binomcdf(n, p, x): where n is the number of
observations, p is the probability of success.
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To determine P(X > x)
Use 1-binomcdf(n, p, x): where n is the number of
observations, p is the probability of success.
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To determine P(X < x)
Use binomcdf(n, p, x-1): where n is the number of
observations, p is the probability of success.
Geometric
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The geometric distribution is a special case
of the negative binomial distribution. It deals
with the number of trials required for a single
success. Thus, the geometric distribution is
negative binomial distribution where the
number of successes (r) is equal to 1.