continuous distributions chapter 6 msis 111 prof. nick dedeke

ContinuousDistributions

Chapter 6MSIS 111 Prof. Nick Dedeke

Learning Objectives

Appreciate the importance of the normal distribution.Recognize normal distribution problems, and know how to solve them.Decide when to use the normal distribution to approximate binomial distribution problems, and know how to work them.Decide when to use the exponential distribution to solve problems in business, and know how to work them.

Probabilities

If a man had six children and he wanted to go a Red Sox game with one of them, what is the likelihood that he will choose Buba, his first son? A. 1 B. 6 C. 1/6 D. I have no idea!

Probability Example 1If a man had six children and he wanted to go a Red Sox game with one of them, what is the likelihood that he will choose Buba?Probability of event A = Number of ways event A could occur Number of events in the sample space

Events that could occur (EVENTS SAMPLE SPACE) He takes Buba He takes Maria He takes Bakuba He takes Melinda He takes Susan He takes Elisha

Probability (A= He takes Buba) = 1/6

Probability Example 2If a man had ten bills in his pocket. Two $20 bills, three $100 bills and five $ 50 bills. He wanted to buy a baseball for $20. What is the probability that if he drew a bill out of his pocket it would be a $50 bill?Probability of event A = Number of ways event A could occur Number of events in the sample space

Events that could occur (EVENTS SAMPLE SPACE) He draws out first $20 bill He draws out second $20 bill He draws out a $50 bill (five possibilities) He draws out a $100 bill (three possibilities)

Probability (A= It is a $50 ) = 5/(5+3+2) = 0.5

Probability Example 3If a man had ten bills in his pocket. Two $20 bills, three $100 bills and five $ 50 bills. He wanted to buy a baseball for $20. What is the probability that if he drew a bill out of his pocket it would be a $20 bill?Probability of event A = Number of ways event A could occur Number of events in the sample space

Events that could occur (EVENTS SAMPLE SPACE) He draws out a $20 bill (two possibilities) He draws out a $50 bill (five possibilities) He draws out a $100 bill (three possibilities)

Probability (A= It is a $20 ) = 2/(5+3+2) = 0.2

Probability (Freq. Table) Example 3

If a man had ten bills in his pocket. Two $20 bills, three $100 bills and five $ 50 bills. He wanted to buy a baseball for $20. What is the probability that if he drew a bill out of his pocket it would be a $50 bill?Probability of event A = Number of ways event A could occur

Number of events in the sample space Xi Fi Rel. Freq or P(x)$20 2 2/10 = 0.2$50 5 5/10 = 0.5$100 3 3/10 = 0.3N 10 1

Probability (x = Bill is a $20 ) = 0.2 Probability (x = Bill is $50 or less) = 0.2 + 0.5 = 0.7 Probability (x = Bill is > $50) = 1 - Probability (x = Bill is $50 or less) = 1 – 0.7 = 0.3

Probability Distributionand Areas & Curve Example 5

Xi Fi Rel. Freq or P(x)

$20 2 2/10 = 0.2

$50 5 5/10 = 0.5

$100 3 3/10 = 0.3

10 1.0

Probability (B= Bill is $50 or less) = 0.2 + 0.5 = 0.7P(x)

1.0

$100 $50 $20

0.5 Probability distribution of an event X is the listing of all possible events and their probabilities (in table of graphical form). If we put the results in form ofan equation, we will have a probability density function.

Continuous Variable Problem

Xi (Prices) Fi

$20 57

$25 Not measured

$35 Not measured

$40 234

$60 123

Relative frequency approach not advisable

Normal Distribution

Probably the most widely known and used of all distributions is the normal distribution.It fits many human characteristics, such as height, weight, length, speed, IQ scores, scholastic achievements, and years of life expectancy, among others.Many things in nature such as trees, animals, insects, and others have many characteristics that are normally distributed.It is used for measured data not counted ones

Normal Distribution

Many variables in business and industry are also normally distributed. For example variables such as the annual cost of household insurance, the cost per square foot of renting warehouse space, and managers’ satisfaction with support from ownership on a five-point scale, amount of fill in soda cans, etc.Because of the many applications, the normal distribution is an extremely important distribution.

Normal Distribution

Discovery of the normal curve of errors is generally credited to mathematician and astronomer Karl Gauss (1777 – 1855), who recognized that the errors of repeated measurement of objects are often normally distributed.Thus the normal distribution is sometimes referred to as the Gaussian distribution or the normal curve of errors.

Properties of the Normal Distribution

The normal distribution exhibits the The normal distribution exhibits the following characteristics:following characteristics:It is a continuous distribution.It is symmetric about the mean.It is asymptotic to the horizontal axis.It is unimodal.It is a family of curves.Area under the curve is 1. It is bell-shaped.

Graphic Representation of the Normal Distribution

Probability Density of the Normal Distribution

Note:The probabilitydistribution changes ifeither mean or SD changes.

Family of Normal Curves

Standardized Normal Distribution

Since there is an infinite number of combinations for and , then we can generate an infinite family of curves.Because of this, it would be impractical to deal with all of these normal distributions.Fortunately, a mechanism was developed by which all normal distributions can be converted into a single distribution called the z z distributiondistribution.This process yields the standardized standardized normal distributionnormal distribution (or curve).


The conversion formula for any x value of a given normal distribution is given below. It is called the z-score.

A z-score gives the number of standard deviations that a value x, is above or below the mean.

xz xz


If x is normally distributed with a mean of and a standard deviation of , then the z-score will also be normally distributed with a mean of 0 and a standard deviation of 1.Since we can covert to this standard normal distribution, tables have been generated for this standard normal distribution which will enable us to determine probabilities for normal variables. The tables in the text are set up to give the probabilities between z = 0 and some other z value, z0 say, which is depicted on the next slide.

Z TableSecond Decimal Place in Z

Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.00 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.03590.10 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.07530.20 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.11410.30 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517

0.90 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.33891.00 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.36211.10 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.38301.20 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015

2.00 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817

3.00 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.49903.40 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.49983.50 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998

Applying the Z Formula

X is normally distributed with = 485, and = 105 P X P Z( ) ( . ) .485 600 0 1 10 3643

For X = 485,

Z =X -

485 485

1050

For X = 600,

Z =X -

600 485

1051 10.

Z 0.00 0.01 0.02

0.00 0.0000 0.0040 0.00800.10 0.0398 0.0438 0.0478

1.00 0.3413 0.3438 0.3461

1.10 0.3643 0.3665 0.3686

1.20 0.3849 0.3869 0.3888

What is the probability that a measured value would be less than or equal to 600 but greater than or equal to 485?


7123.)56.0()550(

100= and 494,= with ddistributenormally is X

ZPXP

56.0100

494550-X=Z

550 = XFor

0.5 + 0.2123 = 0.71230.5 + 0.2123 = 0.7123

What is the probability that a measured value would be less than or equal to 550?

Exercise: Applying the Z Formula

_____)()700(


ZPXP

Z

700 = XFor

Use chart on page 788 to find ProbabilityUse chart on page 788 to find Probability


94.1100

494300-X=Z

300 = XFor

8292.)06.194.1()600300(


ZPXP

0.4738+ 0.3554 = 0.82920.4738+ 0.3554 = 0.8292

06.1100

494600-X=Z

600 = XFor



-X

=Z

300 = XFor

___)(___)600300(


ZPXP

Use chart on page 788 to find Use chart on page 788 to find probabilityprobability

=Z

600 = XFor


0197.)06.2()700(


ZPXP

06.2100

494700-X=Z

700 = XFor

0.5 – 0.4803 = 0.01970.5 – 0.4803 = 0.0197


)600(


XP

-X=Z

600 = XFor



94.1100

494300-X=Z

300 = XFor

8292.)06.194.1()600300(


ZPXP

0.4738+ 0.3554 = 0.82920.4738+ 0.3554 = 0.8292

06.1100

494600-X=Z

600 = XFor


For X = 350

X-Z= ??

X is normally distributed with =494, and =100

(350 450) (?? ??) ??P X P Z

For X = 650

X-Z= ??

Exercise 1The weekly demand for bicycles has a mean of 5,000 and a standard deviation of 600. Assume normal distribution. What is the probability that the firm will

sell between 3,000 and 6,000 (>= 3,000 and <=6,000) in a week?

Exercise 2The distribution of customer spending has a mean of $25 and standard deviation of $8.

What is the probability that a randomly selected customer spends less than $35 at the store?

What is the probability that a randomly selected customer spends between $15 and $35?

What is the probability that a randomly selected customer spends more than $10 at the store?

What is the $ amount such that 75% of all customers spend no more than this amount?

What is the $ amount such that 80% of all customers spend at least this amount?

Exercise 2 ResponseThe distribution of customer spending has a mean of $25

and standard deviation of $8. What is the probability that a randomly selected customer spends

less than $35 at the store? What is the probability that a randomly selected customer spends

between $15 and $35? What is the probability that a randomly selected customer spends

more than $10 at the store?


( 35) ( 1.25) 0.3944 0.5 0.8944

(15 35) ( 1.25 1.25) 0.3944 0.3944 0.7888

(10 ) ( 1.875 ) 0.5 0.4693 0.0307

P X P Z

P X P Z

P X P Z

Exercise 2 ResponseWhat is the $ amount such that 75% of all customers spend

no more than this amount? What is the $ amount such that 80% of all customers spend

at least this amount?


( ??) (?? ) 0.5 0.25

_ _ _ 0.25 _ _ _?? 0.67

250.67

88*(0.67) 25 $30.36

P X P Z

from the table corresponds to

X X

X

0.67 is positive because its above mean

Exercise 2 ResponseWhat is the $ amount such that 80% of all customers spend

at least this amount?

2) X is normally distributed with =25, and =8

( ??) ( ??) 0.5 0.3

_ _ _ (0 ??) 0.3_ _ _?? 0.84

_ _?? 0.84 _( _ _ )

250.84

825 8*(0.67) $18.28

P X P Z

from the table P Z corresponds to

from graph Below the mean

X X

X

0.84 is negative because its below mean

Exercise 3A furniture store has to decide if the current range of prices of its goods is appropriate for its strategy. The mean amount that customers spend in the store is $ 3,500/sale. The standard deviation is $ 550. Assuming that the distribution is normal, the firm wants its prices to cover persons about 45% of the customers around the mean. The current price range of goods is $4200 and $ 2500. Does the price range match the expectation?

Exercise 3 ResponseA furniture store has to decide if the current range of prices of its goods is appropriate for its strategy. The mean amount that customers spend in the store is $ 3,500. The standard deviation is $ 550. Assuming that the distribution is normal, the firm wants its prices to cover persons about 45% of the customers around the mean. The current price range of goods is $4200 and $ 2500.

Does the price range match the expectation? X is normally distributed with =3500, and =550

(2500 4200) ( 1.818 1.2727)

2500 35001.818

5504200 3500

1.2727550

P X P Z

Z

Z

= 0.4656 + 0.3980 = 0.8636 (86.36%). Scope too wide!

Exercise 4A furniture store decided to change its price range of its goods. The mean amount that customers spend in the store is $ 3,500. The standard deviation is $ 550. Assuming that the distribution is normal, the firm wants its prices to cover persons about 45% of the customers around the mean. The new price range of goods is $3800

and $ 2800. Does the price range match the expectation?

Exercise 4 ResponseA furniture store decided to change its price range of its goods. The mean amount that customers spend in the store is $ 3,500. The standard deviation is $ 550. Assuming that the distribution is normal, the firm wants its prices to cover persons about 45% of the customers around the mean. The current price range of goods is $3800 and $ 2800. Does the price range match the

expectation? X is normally distributed with =3500, and =550

(3200 3800) ( 0.5454 1.2727)

3200 35000.5454

5503800 3500

0.5454550

P X P Z

Z

Z

= 0.2054 + 0.2054 = 0.4108 (41.08%). Scope is fine!

Exercises

continuous distributions chapter 6 msis 111 prof. nick dedeke

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