Class2 Charts

Quantitative Method for Multi-dimensional Management and Group Decision Making Continuous Random Variables Normal Distribution 2-1

Learning Objectives 1. Define continuous random variable 2. Describe normal random variables 3. Calculate probabilities for continuous random variables

2-2

Data Types Data

Quantitative

Discrete

2-3

Qualitative

Continuous

Continuous Random Variables

2-4

Continuous Random Variables 1. Random variable  

A numerical outcome of an experiment Weight of a student (e.g., 115, 156.8, etc.)

2. Continuous random variable   

Whole or fractional number Obtained by measuring Infinite number of values in interval  Too

2-5

many to list like discrete variable

Continuous Random Variable Examples Experiment

Random Variable

Possible Values

Weigh 100 people

Weight

45.1, 78, ...

Measure part life

Hours

900, 875.9, ...

Ask food spending

Spending

54.12, 42, ...

Measure time between arrivals

Inter-arrival 0, 1.3, 2.78, ... time

2-6

Continuous Probability Density Function 1. Mathematical formula 2. Shows all values, x, & frequencies, f(x) 

3.

f(x) is not probability

z

Properties f ( x )dx = 1 (Area under curve)

f ( x ) ≥ 0, a ≤ x ≤ b 2-7

Frequency (Value, Frequency)

f(x)

a

b Value

x

Continuous Random Variable Probability Probability is area under curve!

z

P (c ≤ x ≤ d) =

d

c

f ( x ) dx

f(x)

c 2-8

© 1984-1994 T/Maker Co.

d

X

Continuous Probability Distribution Models Continuous Probability Distribution

Uniform

2-9

Normal

t-Student

Other

Normal Distribution

2 - 10

Importance of Normal Distribution 1. Describes many random processes or continuous phenomena 2. Can be used to approximate discrete probability distributions 

Example: binomial

3. Basis for classical statistical inference

2 - 11

Normal Distribution 1. ‘Bell-shaped’ & symmetrical

f(X )

2. Mean, median, mode are equal 3. ‘Middle spread’ is 1.33 σ 4. Random variable has infinite range 2 - 12

X Mean Median Mode

Probability Density Function 1 f (x) = e σ 2π f(x) σ π x µ 2 - 13

= = = = =

FI F I HKH K

1 − 2

x−µ 2 σ

Frequency of random variable x Population standard deviation 3.14159; e = 2.71828 Value of random variable (-∞ < x < ∞) Population mean

Effect of Varying Parameters (µ & σ ) f(X) B A

C X

2 - 14

Normal Distribution Probability Probability is area under curve!

P (c ≤ x ≤ d ) =

f(x )

c 2 - 15

d

x

z d

c

f ( x ) dx ?

Normal Distribution Probability Probability is area under curve!

I’ll use tables!

f(x )

c 2 - 16

d

x

Normal Distribution Probability Tables Normal distributions differ by mean & standard deviation.

Each distribution would require its own table.

f(X)

X

That’s an infinite number! 2 - 17

Standardize the Normal Distribution X −µ Z= σ

Normal Distribution

Standardized Normal Distribution

σ

σ =1

µ 2 - 18

X

µ =0 One table!

Z

Standardizing Example

Normal Distribution

X − µ 6.2 − 5 Z= = = .12 σ 10

Standardized Normal Distribution

σ = 10

µ =5 2 - 19

σ =1

6.2

X

µ =0

.12

Z

Obtaining the Probability Standardized Normal Probability Table (Portion) Z

.00

.01

.02

σ =1

0.0 .0000 .0040 .0080

.0478

0.1 .0398 .0438 .0478 0.2 .0793 .0832 .0871

µ =0

0.3 .1179 .1217 .1255

Probabilities 2 - 20

.12

Z

Shaded area exaggerated

Example P(3.8 ≤ X ≤ 5) X − µ 3.8 − 5 Z= = = − .12 σ 10 Normal Distribution

Standardized Normal Distribution

σ = 10

σ =1 .0478

3.8 2 - 21

µ =5

X

-.12

Shaded area exaggerated

µ =0

Z

Example P(2.9 ≤ X ≤ 7.1)

Normal Distribution

X − µ 2.9 − 5 Z= = = − .21 σ 10 X − µ 7.1 − 5 Z= = = .21 σ 10

Standardized Normal Distribution

σ = 10

σ =1 .1664 .0832 .0832

2.9 2 - 22

5 7.1

X

-.21

Shaded area exaggerated

0 .21

Z

Example P(X ≥ 8) X −µ 8−5 Z= = = .30 σ 10 Normal Distribution

Standardized Normal Distribution

σ = 10

σ =1 .5000 .1179

µ =5 2 - 23

8

X

µ =0

Shaded area exaggerated

.30

.3821

Z

Example P(7.1 ≤ X ≤ 8)

Normal Distribution

X − µ 7.1 − 5 Z= = = .21 σ 10 X −µ 8−5 Z= = = .30 σ 10

Standardized Normal Distribution

σ = 10

σ =1 .1179 .0832

µ =5 2 - 24

7.1

8

X

µ =0

Shaded area exaggerated

.21

.30

.0347

Z

Normal Distribution Thinking Challenge You work in Quality Control for GE. Light bulb life has a normal distribution with µ = 2000 hours & σ = 200 hours. What’s the probability that a bulb will last A. between 2000 & 2400 hours? B. less than 1470 hours? 2 - 25

Alone

Group Class

Solution* P(2000 ≤ X ≤ 2400) X − µ 2400 − 2000 Z= = = 2.0 σ 200 Normal Distribution

Standardized Normal Distribution

σ = 200

σ =1

.4772 µ = 2000 2 - 26

2400

X

µ =0

2.0

Z

Solution* P(X ≤ 1470) X − µ 1470 − 2000 Z= = = − 2.65 σ 200 Normal Distribution

Standardized Normal Distribution

σ = 200

σ =1 .5000 .4960

.0040 1470 2 - 27

µ = 2000

X

-2.65

µ =0

Z

Finding Z Values for Known Probabilities Standardized Normal Probability Table (Portion)

What is Z given P(Z) = .1217? .1217

σ =1

Z

.00

.01

0.2

0.0 .0000 .0040 .0080 0.1 .0398 .0438 .0478

µ =0 Shaded area exaggerated 2 - 28

?

Z

0.2 .0793 .0832 .0871

0.3 .1179 .1217 .1255

Finding Z Values for Known Probabilities Standardized Normal Probability Table (Portion)

What is Z given P(Z) = .1217? .1217

σ =1

Z

.00

.01

0.2

0.0 .0000 .0040 .0080 0.1 .0398 .0438 .0478

µ =0 Shaded area exaggerated 2 - 29

.31

Z

0.2 .0793 .0832 .0871

0.3 .1179 .1217 .1255

Finding X Values for Known Probabilities Normal Distribution

Standardized Normal Distribution

σ = 10

σ =1 .1217

µ =5

2 - 30

?

X

.1217

µ =0

Shaded areas exaggerated

.31

Z

Finding X Values for Known Probabilities Normal Distribution

Standardized Normal Distribution

σ = 10

σ =1 .1217

µ =5

?

X

.1217

µ =0

a fa f

.31

X = µ + Z ⋅ σ = 5 + .31 10 = 8.1 2 - 31

Shaded areas exaggerated

Z