continuous distributions chapter 6 msis 111 prof. nick dedeke
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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).
Standardized Normal Distribution
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
Standardized Normal Distribution
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.
Standardized Normal Distribution
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?
Applying the Z Formula
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(
100= and 494,= with ddistributenormally is X
ZPXP
Z
700 = XFor
Use chart on page 788 to find ProbabilityUse chart on page 788 to find Probability
Applying the Z Formula
94.1100
494300-X=Z
300 = XFor
8292.)06.194.1()600300(
100= and 494,= with ddistributenormally is X
ZPXP
0.4738+ 0.3554 = 0.82920.4738+ 0.3554 = 0.8292
06.1100
494600-X=Z
600 = XFor
Use chart on page 788 to find ProbabilityUse chart on page 788 to find Probability
Exercise: Applying the Z Formula
-X
=Z
300 = XFor
___)(___)600300(
100= and 500,= with ddistributenormally is X
ZPXP
Use chart on page 788 to find Use chart on page 788 to find probabilityprobability
=Z
600 = XFor
Applying the Z Formula
0197.)06.2()700(
100= and 494,= with ddistributenormally is X
ZPXP
06.2100
494700-X=Z
700 = XFor
0.5 – 0.4803 = 0.01970.5 – 0.4803 = 0.0197
Exercise: Applying the Z Formula
)600(
100= and 500,= with ddistributenormally is X
XP
-X=Z
600 = XFor
Use chart on page 788 to find ProbabilityUse chart on page 788 to find Probability
Applying the Z Formula
94.1100
494300-X=Z
300 = XFor
8292.)06.194.1()600300(
100= and 494,= with ddistributenormally is X
ZPXP
0.4738+ 0.3554 = 0.82920.4738+ 0.3554 = 0.8292
06.1100
494600-X=Z
600 = XFor
Exercise: Applying the Z Formula
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?
X is normally distributed with =25, and =8
( 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?
X is normally distributed with =25, and =8
( ??) (?? ) 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
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