Chap 06(1)

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Business Statistics I

Chapter 6

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Calendar

10/18 Chapter 6

10/23 Finish Chapter 6

Homework Chapter 5 - DELAYED

10/25 Homework Chapter 6

Review Chapter 4-6

10/30 Mid-Term 2 on Chapters 4-6

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Continuous Variables

Chapter 5 introduced Discrete variables:a finite number of values or an infinite

sequence of values

Chapter 6 covers Continuous variables:

any number in an interval or collection

of intervals

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Probability Distributions

Discrete Distributions provide theprobability for any particular value

Probability Density Functions, also

denoted by f(x), do not directlyprovide the probability at that point,

but the area under its curve

provides the probability

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Uniform Probability Distribution

Similar to the discrete uniform probabilitydistribution function:

=

= 0

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Area as a Measure of Probability

Ex: Uniform flight times from 120 to140 minutes - Fig 6.1 (p 235)

For the probability that the flight is

between 120 and 130 minutes,

thats half of the rectangle:

HxW = (1/20) x (130-120) = 0.50

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Area as a Measure of Probability

Ex: Same flight time distribution

Q: What is the probability that theflight takes 128 to 136 minutes?

A: HxW = (1/20) x (136-128) = 0.40

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Area as a Measure of Probability

Just like the discrete probabilitydistributions, two rules:

f(x) 0

Total area under the graph = 1.0

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Normal Probability Distribution

The normal curve, or bell-shaped curve,describes many naturally occurring

events

Notice that its symmetric

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Normal Probability Distribution

Formula is on p 239

Two parameters describe a normal

distribution: = mean

(see bottom p 239)

= standard deviation

(see p 240)

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Normal Probability Distribution

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Example 1

P 242: What is the probability that z 1.00 ?

See the problem as a picture

See the solution from the table

See Table 1, pp 978-979

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Example 2

P 243: What is the probability that-0.50 z 1.25 ?

See the problem as a picture

Find the solution from the tables

f(-0.50) = 0.3085 f(1.25) = 0.8944

Answer: 0.8944 0.3085 = 0.5859

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Transformation of Data into a

Standard Normal Curve

z =

= +

Then use Table 1, pp 978-979

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Example 3

P 246: What is the probability that tires willlast 40,000 miles?

P(x 40,000) = ?

Process: See the problem as a picture

Transform to a standard normal dist

Find the solution from the tables

Solution: =

=

4, ,

, = 0.70Table p 979: P(0.70) = 0.7580

P(x 40,000) = 1 0.7580 = 0.2420

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Example 4

P 247: How many miles represent 10% or less?P(x) 10% ?

Process: See the problem as a picture

Find the solution from the tablesTransform to a standard normal dist

Solution: Table p 978: P(x) 0.1000; x = -1.28

= + = 5,000 1.28 + 36,500

= 30,100

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Chapter 6 Exclusions

We do not cover Sections 6.3 or 6.4

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Questions?