statistics- random variables working with uncertain numbers

8/3/2019 Statistics- Random Variables Working With Uncertain Numbers

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Chapter 7

Random Variables: Working with

Uncertain Numbers


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Random Variable

A specification or description of a numerical result

from a random experiment

The number is the observation of the random variable

The random variable is the meaningof the number

The observed value is 17 for the random variable

Last weeks warranty returns (number of customers)

Examples

Todays stock market close

The number of defective parts produced today Next quarters sales

Summaries: Q = mean (expected value)

W = standard deviation


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Examples ofRandom Variables

Random Standard

Variable Mean Deviation

X= $1.40 $1.40 $0

Y= $1.50 $0.50

Z= $1.90 $13.30

$1 prob 0.5

$2 prob 0.5{$0 prob 0.98$95 prob 0.02{


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Discrete Random Variable

Can list all possible outcomes

Probability Distribution

The list of values and probabilities. Use it to compute!

Mean (expected value) of random variableXQ =Sum ofValue v Prob ofValue =E(X) =XP(X)

Gives a typical or central value of the random variable

Standard deviation of random variableX

W = Sum of (ValueQ)2 v Probability ofValue

=

Tells about how far from expected this random variable will be

Q )(P)( 2 XX


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Example: Investment Payoffs

Probability distribution of investment payoffs

Q =0 v 0.98 + 95 v 0.02 = $1.90

The expected payoff is $1.90

A compromise between $0 (most of the time) and $95 (rarely)

W = (01.90)2

v 0.98 + (951.90)2

v 0.02 = $13.30 Actual payoffs approximately $13.30 above or below expected

A compromise between being $1.90below average (most of

the time) and $93.10 above average (rarely)

Payoff (Value) Probability$0 0.98

$95 0.02


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Example: Profit Scenarios

0.0

0.1

0.2

0.3

0.4

0.5

-5 0 5 10

Probability

Profit ($millions)

Lousy

OK

Good

Great

Expected profit: $3.65 millionStandard deviation:

$4.40 million

Fig 7.1.1

Scenario Profit ($millions) Probability

0.20

0.40

0.25

0.15

$10

5

1

4

Great

Good

OK

Lousy


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Binomial Distribution

A special type of discrete random variable e.g., Interview 50 random customers

How many like the new product? 0, 1, 2, , 49, or50

e.g., What percent of stocks went up yesterday?

Xis binomial if it is the number of occurrences ofsome event, out ofn trials, provided that

The probability T is the same for all trials, and The trials are independent of one another

so that each trial brings new, independent information

Binomial proportion or percent: p =X/n This is the relative frequency of the event

e.g., ifX=35 ofn =50people like product, p = 35/50 = 0.70


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Binomial Probabilities

A discrete distribution

Skewed ifT is close to 0

Symmetric ifT is 0.5 (or close)

Approximately normal

Skewed ifT is close to 1

0

0.1

0.2

0.3

0 5 10 15 20 25

0

0.1

0.2

0.3

0 5 10 15 20 25

T=0.05, n=25

T=0.5, n=25

T=0.9, n=25

0

0.2

0.4

0 5 10 15 20 25


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Binomial Mean and Std. Deviation

Shortcut to find mean and standard deviation

quickly for binomial random variables (Xorp)

No need to compute the probability distribution

work directly from n and T

Mean

Std. Dev.

Proportion or

Percent, p=X/n

Qp

= T

Wp= T(1T)/n

Number of

Occurrences,X

QX= nT

WX= nT(1T)


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Example: Sampling Defective Parts

Draw a random sample ofn=200 from the days production

Suppose that T =7% of this production is defective

Number of

Defects,X

QX= nT= 200v0.07 = 14

Expect to see 14

defects, on average

WX

= nT(1T)

= 200v0.07v0.93

= 3.61

Typically expect approx.

3.61 more or less than 14

Mean

Std. Dev.

Proportion orPercent

Defective, p=X/n

Qp = T = 0.07

Expect to see 7%

defective, on average

Wp

= T(1T)/n

= 0.07v0.93/200

= 0.018

Typically expect approx.

1.8% more or less than 7%


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Computing Binomial Probabilities

Binomial probability thatXequals a

Example: probability that exactly a= 2 of yourn= 6

major customers will call tomorrow (assuming that

T=0.25 is the probability that each one will call)

ana

ana

ana

ana

nana

n

a

naX

TTvvvvvvvv

vvvv!

TT

!

TT

!!

)1()](...321][...321[

...321

)1(

)!(!

!

)1()(P

297.0316406.00625.015

)25.01(25.02

6)2(P 262

!vv!

!!

X


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Normal Distribution

A special continuous distribution (notdiscrete)

For every mean Q and (positive) standard

deviation W there is a normal distribution The mean Q moves the curve left and right

The standard deviation W widens and narrows the curve

Q

WW


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Probability: Area Under the Curve

Probability of observing a value between a and b

is area under the curve

Note: total area = 1Probability = 0.50

Probability = 0.95

(two std. devs.)

Probability = 0.68

(one std. dev.)

a b a b

More likely

Less likely

WW

Q

WW

Q


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Standard Normal Distribution

Normal with mean Q = 0 and std. deviation W = 1

Standard normal probability table

Gives probability that a standard normal is less than a

given valueExamples

-3 -2 -1 0 1 2 3

Value =0.5

Probability

=0.3085

-3 -2 -1 0 1 2 3

Value1

0

1

2

Probability0.1587

0.5

0.8413

0.9772


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Finding Normal Probabilities

Example: Sales are forecast as $80 million (mean) with a standarddeviation of$10 million. Find the probability that sales will exceed

$86 million, assuming a normal distribution

Figure out the question

Find Prob(X>

86) where Q=

80 and W=

10 Standardize (Subtract Q, divide by W to get std. normal)

Prob

= Prob

Draw Picture

Use tables, find answer: 10.7257 = 0.274

-3 -2 -1 0 1 2 3

0.60

XQ

W

8680

10>( )

Standardnormal >0.60( )


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Normal Approx. to the Binomial

Idea: to make it easierto compute binomialprobabilities

If

n is large, and

T is not too close to 0 or1,

Then

Binomial probabilities forXare close to normal

probabilities with Q =Q

X= nT W =W

X= nT(1T)

Similarly forp=X/n with

Q =Qp= T W =W

p= T(1T)/n


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Example: an Opinion Poll

Suppose T= 55% approve presidents performance We interview n=400 chosen at random

What are the chances that fewer than 200 will say they

approve? That is, find Prob (X< 200)

When you round any number less than 199.5 to the nearest

whole number, the result will be fewer than 200

The mean is QX= nT =400v0.55 =220

The standard deviation is WX= nT(1T) =9.9499

We need to compute the probability that a normalrandom

variable with this mean and this standard deviation is less than

199.5

Prob =Prob = 0.020

XQX

W

199.5220

9.9499


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Example: Poll (continued)

Still assuming 55% approve and we interview 400at random

What are the chances that more than 58% will say they

approve?

Convert from percentage p to number of peopleX

Note that 58% of400people is 0.58v400 = 232people

Numbers that round to more than 232 are 232.5 and above

We need to compute the probability that a normalrandom

variable with mean QX=220 and standard deviation W

X=

9.9499 is more than 232.5

Prob =Prob = 0.10

About 10% of the time we will find more than 58%

XQX

W

232.5220

9.9499>( ) Standard

normal>1.26( )

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