Probability Distribution, Mean, Variance and Standard Deviation
Last Thursday, June 14, 2012, we learned about probability distribution and mean, variance, standard deviation and expectation.
Its easy to construct a probability distribution.
Ex. The proability distribution of tossing 3 coins if X is the no. of tails is:
Outcome, (X) 0 1 2 3
Probability, P(X) 1/8 3/8 3/8 1/8
In finding the mean random variable with a discrete probability distribution, we use our knowledge in constructing a probability distribution.
Using the probability distribution we find the sum of the products of the outcomes and probabilities.
Formula for finding the mean:
µ = Ʃ[X · P(X)]
Using the probability distribution we find the sum of products of the sqaures of the outcomes and probabilities and subtract the square of the mean
Formula for finding the variance:
σ^2 = Ʃ[ (X^2 · P(x)] - µ^2
The Standard Deviation is the square root of variance
Formula for finding the standard deviation
σ = √(σ^2 )
Applications:
The following distribution shows the number enrolled in CPR classes offered by the local fire department. FInd the mean, variance and standard deviation.
Outcome, (X) 12 13 14 15 16
Probability, P(X) 0.15 0.20 0.38 0.18 0.09
µ = (12 · 0.15)+(13 · 0.20)+(14 · 0.38)+(15 · 0.18)+(16 · 0.09)
=13.86
σ^2 = (12)(12) · 0.15 + (13)(13) · 0.20 + (14)(14) · 0.38 + (15)(15) · 0.18
+ (16)(16) · 0.09 - (13.86)(13.86)
= 1.3204
σ = √1.3204
= 1.149
Another concept related to the mean for a probability distribution is the concept of expected value or expectation. Expected value is used in various types of games of chance, in insurance, and in other areas, such as decision theory.
E(X) = Ʃ X · P(X)
Application
A lottery offers one $1000 prize, one $500 prize and five $100 prizes. One thousand tickets are sold at $3 each. Find the expectation if a person buys one ticket.
Gain, (X) $997 $497 $97 -$3
Probability, P(X) 1/1000 1/1000 5/1000 993/1000
E(X) = 997 · 1/1000 + 497 · 1/1000 + 97 · 5/1000 -3 ·
993/1000
= - $1
by: Salva, Julie Clarice
Sampuego, Guendelyn Marie
Its easy to construct a probability distribution.
Ex. The proability distribution of tossing 3 coins if X is the no. of tails is:
Outcome, (X) 0 1 2 3
Probability, P(X) 1/8 3/8 3/8 1/8
In finding the mean random variable with a discrete probability distribution, we use our knowledge in constructing a probability distribution.
Using the probability distribution we find the sum of the products of the outcomes and probabilities.
Formula for finding the mean:
µ = Ʃ[X · P(X)]
Using the probability distribution we find the sum of products of the sqaures of the outcomes and probabilities and subtract the square of the mean
Formula for finding the variance:
σ^2 = Ʃ[ (X^2 · P(x)] - µ^2
The Standard Deviation is the square root of variance
Formula for finding the standard deviation
σ = √(σ^2 )
Applications:
The following distribution shows the number enrolled in CPR classes offered by the local fire department. FInd the mean, variance and standard deviation.
Outcome, (X) 12 13 14 15 16
Probability, P(X) 0.15 0.20 0.38 0.18 0.09
µ = (12 · 0.15)+(13 · 0.20)+(14 · 0.38)+(15 · 0.18)+(16 · 0.09)
=13.86
σ^2 = (12)(12) · 0.15 + (13)(13) · 0.20 + (14)(14) · 0.38 + (15)(15) · 0.18
+ (16)(16) · 0.09 - (13.86)(13.86)
= 1.3204
σ = √1.3204
= 1.149
Another concept related to the mean for a probability distribution is the concept of expected value or expectation. Expected value is used in various types of games of chance, in insurance, and in other areas, such as decision theory.
E(X) = Ʃ X · P(X)
Application
A lottery offers one $1000 prize, one $500 prize and five $100 prizes. One thousand tickets are sold at $3 each. Find the expectation if a person buys one ticket.
Gain, (X) $997 $497 $97 -$3
Probability, P(X) 1/1000 1/1000 5/1000 993/1000
E(X) = 997 · 1/1000 + 497 · 1/1000 + 97 · 5/1000 -3 ·
993/1000
= - $1
by: Salva, Julie Clarice
Sampuego, Guendelyn Marie
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