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Chapter 7 Continuous Distributions Notes page 137.

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Chapter 7 Chapter 7 Continuous Distributions Notes page 137
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Page 1: Chapter 7 Continuous Distributions Notes page 137.

Chapter 7Chapter 7Continuous

Distributions

Notes page 137

Page 2: Chapter 7 Continuous Distributions Notes page 137.

Continuous random Continuous random variablesvariables

•Are numerical variables whose values fall within a range or interval

•Are measurements•Can be described by density curves

Page 3: Chapter 7 Continuous Distributions Notes page 137.

Density curvesDensity curves• Is always on or aboveon or above the

horizontal axis• Has an area exactly equal to oneequal to one

underneath it• Often describes an overall

distribution• Describe what proportionsproportions of the

observations fall within each range of values

Page 4: Chapter 7 Continuous Distributions Notes page 137.

Unusual density Unusual density curvescurves

•Can be any shape•Are generic continuous distributions

•Probabilities are calculated by finding the finding the area under the curvearea under the curve

Page 5: Chapter 7 Continuous Distributions Notes page 137.

1 2 3 4 5

.5

.25

P(X < 2) =

25.

225.2

How do you find the area of a triangle?

Page 6: Chapter 7 Continuous Distributions Notes page 137.

1 2 3 4 5

.5

.25

P(X = 2) =

0

P(X < 2) =

.25

What is the area of a line

segment?

Page 7: Chapter 7 Continuous Distributions Notes page 137.

In continuous distributions, P(P(XX < 2) & P( < 2) & P(XX << 2)2) are the same answer.

Hmmmm…

Is this different than

discrete distributions?

Page 8: Chapter 7 Continuous Distributions Notes page 137.

1 2 3 4 5

.5

.25

P(X > 3) =

P(1 < X < 3) =

Shape is a trapezoid –

How long are the bases?

2

21 hbbArea

.5(.375+.5)(1)=.4375

.5(.125+.375)(2) =.5

b2 = .375

b1 = .5

h = 1

Page 9: Chapter 7 Continuous Distributions Notes page 137.

Area of Trapezoid

2

21 hbbArea

The bases are always the 2 parallel sides.

Page 10: Chapter 7 Continuous Distributions Notes page 137.

1 2 3 4

0.25

0.50 P(X > 1) =.75

.5(2)(.25) = .25

(2)(.25) = .5

Page 11: Chapter 7 Continuous Distributions Notes page 137.

1 2 3 4

0.25

0.50P(0.5 < X < 1.5) =

.28125

.5(.25+.375)(.5) = .15625

(.5)(.25) = .125

Page 12: Chapter 7 Continuous Distributions Notes page 137.

Homework:

Page 140


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