Download Discrete Probability Distributions (Chapter 5)

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Discrete Probability Distributions (Chapter 5)
Discrete Probability Distributions
Requirements for a Discrete Probability Distribution
1. Each individual probability is between 0 and 1 inclusive. That is, 0 ≤ P ( x) ≤ 1
2. The sum of all the probabilities is 1. That is,
∑ P( x) = 1 .
How to Find the Mean (Expected Value) and standard deviation of a Discrete Probability:
µ=
Σ [ x ⋅ P( x)]
σ = Σ( x − µ ) 2 ⋅ P( x)
Example: Find the mean and standard deviation for the following probability distribution.
X P(X)
0
0.3
1
0.2
3
0.5
X P(X)
X·P(X)
0
0.3
0
1
0.2
0.2
3
0.5
1.5
X P(X)
0
0.3
1
0.2
3
0.5
𝝁 = 𝜮[𝑿 ∙ 𝑷(𝑿)] = 𝟏. 𝟕
(𝑿 − 𝝁)𝟐
(𝟎 − 𝟏. 𝟕)𝟐 = 𝟐. 𝟖𝟗
(𝑿 − 𝝁)𝟐 ∙ 𝑷(𝑿)
2.89∙ 𝟎. 𝟑 = 0.867
(𝟑 − 𝟏. 𝟕)𝟐 = 𝟏. 𝟔𝟗
1.69∙ 𝟎. 𝟓 = 0.845
(𝟏 − 𝟏. 𝟕)𝟐 = 𝟎. 𝟒𝟗
0.49∙ 𝟎. 𝟐 = 0.098
𝜮�(𝑿 − 𝝁)𝟐 ∙ 𝑷(𝑿)� = 𝟏. 𝟖𝟏
µ=
Σ [ x ⋅ P( x)] =
1.7 and σ = Σ( x − µ ) 2 ⋅ P( x) = 1.81 =1.35
USING TI-84
STATS  Edit…  Enter X values in L1 and P(X) in L2  2ND + QUIT  STAT  CALC  1-VAR Stats  List: L1, FreqList: L2
USING TI-83
STATS  Edit…  Enter X values in L1 and P(X) in L2  2ND + QUIT  STAT  CALC  1-VAR Stats L1,L2
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Binomial Probability Distribution
A probability experiment that satisfies the following four requirements is said to be a binomial experiment.
1. The experiment is repeated for a fixed number of trials, where each trial is independent of other trials.
2. There are only two possible outcomes of interest for each trial. The outcomes can be classified as a success or as a failure.
3. The probability of a success is the same for each trial.
4. The random variable X counts the number of successful trials.
Binomial Probability Formula
The Mean (Expected Value ) & Standard Deviation for Binomial Probability
The probability of exactly x successes in n trials is
P( x)
=
Cx p x q n− x
n=
n!
p x q n− x
(n − x)! x !
µ= n ⋅ p
σ=
n⋅ p⋅q
n = Number of Trials
p = Probability of success
q = 1 − p (Probability of failure)
x = 0, 1, 2, 3,…,n
Example:
Sixty percent of American adults access the Internet wirelessly, according to a 2010 report by the Pew Research Center. Suppose we take a random sample
of three Americans.
A) Find the mean or expected number of Americans adults who access the internet wirelessly.
B) Calculate the standard deviation of the number of American adults who access the internet wirelessly.
C) Find the probability that exactly 2 two Americans access the internet wirelessly.
Solution:
n=3
p = 0.60
q =1 − p =0.40
x = 0, 1, 2, 3,…,n
A)Mean : µ = n ⋅ p = (3)(0.60) =1.8
B)Standard Deviation : σ =
C)P (=
x 2)
=
2
3
C2 (0.6) (0.4)
n⋅ p⋅q =
3− 2
(3)(0.60)(0.40) ≈ 0.8485
≈ 0.432
USING TI-84
2DN + DISTR  binompdf(  Enter the following- trials: n, P: , x Value:
USING TI-83
2DN+ DISTR  binompdf( n, p, x)