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Probability Distributions

• Random Variable• A numerical outcome of a random experiment• Can be discrete or continuous• Generically, x

• Probability Distribution• The pattern of probabilities associated with all of the random

variables for a specific experiment• Can be a table, formula, or graph• Generically, f(x)• Examples

• Binomial (but won’t cover here) • Uniform• Normal or bell-shaped distribution

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Birth of a Distribution

4.00-14.9915.00-25.00

Class Width = 10Cyberland Wages

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Birth of a Distribution

Class Width = 5

4

Birth of a Distribution

Class Width = 2

5

Birth of a Distribution

Class Width = 1

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Birth of a Distribution

Class Width = Very Small

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Uniform Distribution

x

f(x)

1 / (

b-a)

a b

Area = 1

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Normal Distribution

Bell-shaped, symmetrical distribution

f(x)

x

2

2

2

)x(

e2

1)x(f

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Normal Distributions

= 5

=2

=3

=1

-2 12

10

Normal Distributions

Same ,Different

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Normal Distributions

68.26%

+-

12

Normal Distributions

95.44%

+2-2

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Normal Distributions

99.72%

+3-3

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Standard Normal Distribution

z

x

xxz

0

z = 1

z = 0

If x has a normal distribution…

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t Distribution

3.50

but has thicker tails

-3.5

Specific thickness depends on degrees

of freedom

Looks like a normal distribution,

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t Distribution

3.50-3.5

Specific thickness depends on degrees of freedom

5 d.f.10 d.f.30 d.f.

100 d.f. d.f (normal)

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Find the Probabilities

1. P(z > 2.36)

2. P(t > -1.02) with 5 degrees of freedom

3. P(-0.95 < z < 1.93)

4. P(-0.95 < t < -0.07) with 100 degrees of freedom

5. Find z* such that P(z < z*) = 0.719

6. Find z0.025 such that P(z > z0.025) = 0.025

7. Find t0.025 such that P(t > t0.025) = 0.025 with 5 degrees of freedom

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z/2

0-z/2

Standard Normal Distribution (z)

z/2

P(z < -z/2)) = /2

P(z > z/2) = /2

P(-z/2 < z < z/2) = 1 - /2

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z/2 for =0.05

0-z0.025

Standard Normal Distribution (z)

z0.025

P(z < ) = 0.025

P(z > ) = 0.025

P( < z < ) = 0.95

? ?

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t/2 for =0.05, df=5

0-t0.025

t distribution with 5 degrees of freedom

t0.025

P(t < ) = 0.025

P(t > ) = 0.025

P( < t < ) = 0.95

? ?

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2 Distribution

0

Specific skewness depends on degrees

of freedom

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2 Distribution

0

Specific skewness depends on degrees

of freedom

5 d.f 10 d.f 15 d.f

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2 Distribution

018.307

P(2 > 18.307) = 0.05

P(2 < 18.307) = 0.95

10 d.f

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F Distribution

0

Specific skewness depends on a pair of degrees of freedom

(df1, df2)

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F Distribution

03.02

P(F > 3.02) = 0.05

P(F < 3.02) = 0.95

9 and 10 d.f

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Probability Distributions

Different shapes and df’s, but

SAME LOGIC !

Normal & t 2

F

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In Excel

• To find probability above a value x• =1-NORMSDIST(x)• =TDIST(x,df,1) [1=1-tail]• =CHIDIST(x,df) • =FDIST(x,df1,df2)

• To find value with p% above (e.g., 0.05)• =NORMSINV(p)• =TINV(p,df) • =CHIINV(p,df) • =FINV(p,df1,df2)

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Word Problem

• From past experience, the management of a well-known fast food restaurant estimates that the number of weekly customers at a particular location is normally distributed, with a mean of 5000 and a standard deviation of 800 customers.• What is the probability that on a given week the

number of customers will be between 4760 and 5800?

• What is the probability of a week with more than 6500 customers?

• For 90% of the weeks, the number of customers should exceed what amount?