continous_probability_distribution

Statistics for Business Analysis

Day 5Session-I

CONTINUOUS PROBABILITY DISTRIBUTIONS

Created by: Prabhat MittalEmail-Id: [email protected]

Learning Objectives

� To learn the concept of continuous random variable

� To compute probabilities from the normal distribution

� To use the normal probability plot to determine whether a set of data is approximately normally distributed

� To compute probabilities from the uniform distribution

� To compute probabilities from the exponential distribution

� To compute probabilities from the normal distribution to approximate probabilities from the binomial distribution


Probability Distributions

ContinuousProbability

Distributions

Binomial

Hypergeometric

Poisson


DiscreteProbability

Distributions

Normal

Uniform

Exponential


Continuous Probability Distributions

� A continuous random variable is a variable that can assume any value on a continuum (can assume an uncountable number of values)

� thickness of an item

� time required to complete a task

� temperature of a solution

� height, in inches

� These can potentially take on any value, depending only on the ability to measure accurately.


The Normal Distribution


Normal

Uniform

Exponential


Distributions


The Normal Distribution

� ‘Bell Shaped’

� Symmetrical

� Mean, Median and Modeare Equal

Location is determined by the mean, µ

Spread is determined by the standard deviation, σ

The random variable has an infinite theoretical range: + ∞∞∞∞ to −−−− ∞∞∞∞

Mean = Median = Mode

X

f(X)

µ

σ


By varying the parameters µ and σ, we obtain different normal distributions

Many Normal Distributions


The Normal Distribution Shape

X

f(X)

µ

σ

Changing µ shifts the distribution left or right.

Changing σ increases or decreases the spread.


The Normal Probability Density Function

� The formula for the normal probability density function with X ~ Normal(µ, σ2) is

Where e = the mathematical constant approximated by 2.71828

π = the mathematical constant approximated by 3.14159

µ = the population mean

σ = the population standard deviation

X = any value of the continuous variable

2µ)/σ](1/2)[(Xe2π

1f(X) −−

=

σ


Empirical Rules

µ ± 1σ encloses about

68% of X’s

f(X)

Xµ µ+1σµ-1σ

What can we say about the distribution of values around the mean? There are some general rules:

σσ

68.26%


The Empirical Rule

� µ ± 2σ covers about 95% of X’s

� µ ± 3σ covers about 99.7% of X’s

xµ

2σ 2σ

xµ

3σ 3σ

95.44% 99.73%

(continued)


The StandardizedNormal Distribution

� Also known as the “Z” distribution

� Mean is 0

� Standard Deviation is 1

Z

f(Z)

0

1

Values above the mean have positive Z-values, values below the mean have negative Z-values


The Standardized Normal

� Any normal distribution (with any mean and standard deviation combination) can be transformed into the standardized normaldistribution (Z)

� Need to transform X units into Z units


Translation to the Standardized Normal Distribution

� Translate from X to the standardized normal (the “Z” distribution) by subtracting the meanof X and dividing by its standard deviation:

σ

µXZ

−=

The Z distribution always has mean = 0 and standard deviation = 1


The Standardized Normal Probability Density Function

� The formula for the standardized normal probability density function is

Where e = the mathematical constant approximated by 2.71828

π = the mathematical constant approximated by 3.14159

Z = any value of the standardized normal distribution

2(1/2)Ze2π

1f(Z) −

=


Example

� If X is distributed normally with mean of 100and standard deviation of 50, the Z value for X = 200 is

� This says that X = 200 is two standard deviations (2 increments of 50 units) above the mean of 100.

2.050

100200

σ

µXZ =

−=

−=


Comparing X and Z units

Z

100

2.00

200 X

Note that the distribution is the same, only the scale has changed. We can express the problem in original units (X) or in standardized units (Z)

(µ = 100, σ = 50)

(µ = 0, σ = 1)


Finding Normal Probabilities

Probability is the area under thecurve!

a b X

f(X) P a X b( )≤

Probability is measured by the area under the curve

≤

P a X b( )<<=

(Note that the probability of any individual value is zero)


f(X)

Xµ

Probability as Area Under the Curve

0.50.5

The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below

1.0)XP( =∞<<−∞

0.5)XP(µ =∞<<0.5µ)XP( =<<−∞


The Standardized Normal Table

� The Cumulative Standardized Normal table in the textbook (Appendix table E.2) gives the probability less than a desired value for Z (i.e., from negative infinity to Z)

Z0 2.00

0.9772Example:

P(Z < 2.00) = 0.9772


The Standardized Normal Table

The value within the table gives the probability from Z = −−−− ∞∞∞∞

up to the desired Z value

.9772

2.0P(Z < 2.00) = 0.9772

The row shows the value of Z to the first decimal point

The column gives the value of Z to the second decimal point

2.0

.

.

.

(continued)

Z 0.00 0.01 0.02 …

0.0

0.1


General Procedure for Finding Probabilities

� Draw the normal curve for the problem in terms of X

� Translate X-values to Z-values

� Use the Standardized Normal Table

To find P(a < X < b) when X is distributed normally:

Suppose X is normal with µ = 8.0 & σ =5.0

� Find P(X < 8.6)

X

8.6

8.0


� Suppose X is normal with mean 8.0 and standard deviation 5.0. Find P(X < 8.6)

Z0.120X8.68

µ = 8

σ = 10

µ = 0

σ = 1

(continued)

Finding Normal Probabilities

0.125.0

8.08.6

σ

µXZ =

−=

−=

P(X < 8.6) P(Z < 0.12)


Z

0.12

Z .00 .01

0.0 .5000 .5040 .5080

.5398 .5438

0.2 .5793 .5832 .5871

0.3 .6179 .6217 .6255

Solution: Finding P(Z < 0.12)

.5478.02

0.1 .5478

Standardized Normal Probability Table (Portion)

0.00

= P(Z < 0.12)

P(X < 8.6)


� Now Find P(X > 8.6)…

(continued)

Z

0.12

0Z

0.12

0.5478

0

1.000 1.0 - 0.5478 = 0.4522

P(X > 8.6) = P(Z > 0.12) = 1.0 - P(Z ≤ 0.12)

= 1.0 - 0.5478 = 0.4522

Upper Tail Probabilities


Probability Between Two Values

� Suppose X is normal with mean 8.0 and standard deviation 5.0. Find P(8 < X < 8.6)

P(8 < X < 8.6)

= P(0 < Z < 0.12)

Z0.120

X8.68

05

88

σ

µXZ =

−=

−=

0.125

88.6

σ

µXZ =

−=

−=

Calculate Z-values:


Z

0.12

Solution: Finding P(0 < Z < 0.12)

0.0478

0.00

= P(0 < Z < 0.12)

P(8 < X < 8.6)

= P(Z < 0.12) – P(Z ≤ 0)

= 0.5478 - .5000 = 0.0478

0.5000

Z .00 .01

0.0 .5000 .5040 .5080

.5398 .5438

0.2 .5793 .5832 .5871

0.3 .6179 .6217 .6255

.02

0.1 .5478



� Suppose X is normal with mean 8.0 and standard deviation 5.0.

� Now Find P(7.4 < X < 8)

X

7.48.0

Probabilities in the Lower Tail


Probabilities in the Lower Tail

Now Find P(7.4 < X < 8)…

X7.4 8.0

P(7.4 < X < 8)

= P(-0.12 < Z < 0)

= P(Z < 0) – P(Z ≤ -0.12)

= 0.5000 - 0.4522 = 0.0478

(continued)

0.0478

0.4522

Z-0.12 0

The Normal distribution is symmetric, so this probability is the same as P(0 < Z < 0.12)


Numerical Problems

Ref. # 5-38 Page No.270: Given that a random variable, X, has a normal distribution with mean 6.4 and standard deviation 2.7, find

a. P(4.0 < X <5.0)

b. P(X > 2.0)

c. P(X < 7.2)

d. P(X < 3.0)

Ans.

a. .1150

b. .9484

c. .6165

d. .8960


Numerical Problems

Ref. # 5-46 Page No.271: Glenn Howell, VP of personnel for the Standard Insurance Company, has developed a new training program, that is entirely self-paced. New employees work various stages at their own pace; completion occurs when the material is learned. Howell’s program has been especially effective in speeding up the raining process, as an employee’s salary during training is only 67% of that earned upon completion of the program. In the last several years, average completion time of the program was 44 days, and the standard deviation was 12 days.

a. What is the probability that an employee will finish the program in 33 to 42 days?

b. What is the probability of finishing the program in fewer than 30 days?

c. Fewer than 25 or more than 60 days?

Ans.

a. .2537

b. .1210

c. .1489


Numerical Problems

Ref. # 5-48 Page No.271: R. V. Poppin, the concession stand manager for the local hockey rink, just had 2 cancellations on his crew. This means that if more than 72000 people come to tonight’s hockey game, the lines for hot dogs will constitute a disgrace to Mr. Poppin and will harm business at future games. Mr. Poppin knows from experience that the number of people who come to the game is normally distributed with mean 67000 and standard deviation 4000 people.

a. What is the probability that there will be more than 72000 people?b. Suppose Mr. Poppin can hire two temporary employees to make sure

business won’t be harmed in the future at an additional cost of $200. if be believes the future harm to business of having more than 72000 fans at the game would be $5000, should he hire the employees? Explain (Assume there will be no harm if 72000 or fewer fans show up, and that the harm due to too many fans doesn’t depend on how many more than 72000 show up?Ans.

a. 0.1056b. Yes, the $200 cost is less than $528 expected loss to future business.

(Since P(x>72000) = 0.1056 than expected loss $5000*.1056 = $528)


Numerical Problems

Ref. # 5-49 Page No.272: Maurine Lewis, an editor for a large publishing company, calculate that it requires 11 months on average to complete the publication process from manuscript to finished book, with a standard deviation of 2.4 months. She believes that the normal distribution well describes the distribution of publication times. Out of 19 books she will handle this year, approximately how many will complete the process in less than a year?

Ans.

• 13 as P (x<12) = .6615


� Steps to find the X value for a known probability:

1. Find the Z value for the known probability

2. Convert to X units using the formula:

Finding the X value for a Known Probability

ZσµX +=


Finding the X value for a Known Probability

Example:

� Suppose X is normal with mean 8.0 and standard deviation 5.0.

� Now find the X value so that only 20% of all values are below this X

X? 8.0

0.2000

Z? 0

(continued)


Find the Z value for 20% in the Lower Tail

� 20% area in the lower tail is consistent with a Z value of -0.84Z .03

-0.9 .1762 .1736

.2033

-0.7 .2327 .2296

.04

-0.8 .2005


.05

.1711

.1977

.2266

…

…

…

…

X? 8.0

0.2000

Z-0.84 0

1. Find the Z value for the known probability


2. Convert to X units using the formula:

Finding the X value

80.3

0.5)84.0(0.8

ZσµX

=

−+=

+=

So 20% of the values from a distribution

with mean 8.0 and standard deviation 5.0 are less than 3.80


Numerical Problems

Ref. # 5-39 Page No.270: In a normal distribution with a standard deviation of 5.0, the probability that an observation selected at random exceeds 21 is 0.14

a. Find the mean of the distribution

b. Find the value below which 4% of the values in the distribution lie.

Ans.

Find the value of Z such that P(X >21) = 0.14

or P(X < 21) = 0.86

The value of Z from the table +1.08 (+ sign because the area is on right of mu)

a. µ = X- Zσ = 21- 1.08*5.0 = 15.6

b. X= 6.85 P (X< 6.85 ) = 4%


Numerical Problems

Ref. # 5-44 Page No.271: Jarrid Medical, Inc., is developing a compact kidney dialysis machine, but its chief engineer, Mike Crowe, is having trouble controlling the variability of the rate at which fluid moves through the device. Medical standards require that the hourly flow be 4 liter, plus or minus 0.1 liter, 80% pf the time. Mr. Crowe, in testing the prototype, has found that 68% of the time, the hourly flow is within 0.08 liter of 4.02 liters. Does the prototype satisfy the medical standards?

Ans.

Given P( 4.02-0.08 <X < 4.02+0.08) = 68% & obtain µ & σ

then test the value of P(3.9 < X < 4.1) = 80%???

a. Since above P is 68% with in 4.02 ± 0.8 implying µ = 4.02, σ = 0.8

b. P(3.9 < X < 4.1) = 0.7745 i.e. 77.45% not 80%


Numerical Problems

Ref. # 5-50 Page No.272: The Quickie Sales Corporation has just been given two conflicting estimates of sales for the upcoming quarter. Estimate I says that sales (in millions of dollars) will be normally distributed with µ = 325 and σ = 60. Estimate II says that sales will be normally distributed with µ = 300 and σ = 50. The board of directors finds that each estimated appears to be equally believable a priori. In order to determine which estimate should be used fro future predictions, the board of directors has decided to meet again at the end of the quarter to use updated sales information to make a statement about the credibility of each estimate.

a. Assuming that estimate I is accurate, what is the probability that Quickie will have quarterly sales in excess of $350 million?

b. Rework part (a) assuming that estimate II is correct.c. At the end of the quarter, the board of directors finds that Quickie Sales Corp.

has had sales in excess of $350 million. Given this updated information, what is the probability that Estimate I was originally the accurate one? (Use BayesTheorem)

d. Rework part © for Estimate II.

Ans. a. 0.3385b. 0.1587c. P(E1:x > 350) = ?


Solution Ctd…

0.6800.07930.1692

0.1692

.5)(0.1587)(0.5)(0.3385)(0

.5)(0.3385)(0

E2)P(E2)|P(DE1)P(E1)|P(D

E1)P(E1)|P(D350) X :D|1P(E

=+

=

+=

+=>

0.320.07930.1692

0.0793

.5)(0.1587)(0.5)(0.3385)(0

.5)(0.1587)(0

E2)P(E2)|P(DE1)P(E1)|P(D

E2)P(E2)|P(D350) X :D|P(E2

=+

=

+=

+=>


Evaluating Normality

� Not all continuous random variables are normally distributed

� It is important to evaluate how well the data set is approximated by a normal distribution


Evaluating Normality

� Construct charts or graphs

� For small- or moderate-sized data sets, do stem-and-leaf display and box-and-whisker plot look symmetric?

� For large data sets, does the histogram or polygon appear bell-shaped?

� Compute descriptive summary measures

� Do the mean, median and mode have similar values?

� Is the interquartile range approximately 1.33 σ?

� Is the range approximately 6 σ?

(continued)


Assessing Normality

� Observe the distribution of the data set

� Do approximately 2/3 of the observations lie within mean 1 standard deviation?

� Do approximately 80% of the observations lie within mean 1.28 standard deviations?

� Do approximately 95% of the observations lie within mean 2 standard deviations?

� Evaluate normal probability plot

� Is the normal probability plot approximately linear with positive slope?

(continued)

±

±

±


The Normal Probability Plot

� Normal probability plot

� Arrange data into ordered array

� Find corresponding standardized normal quantile

values

� Plot the pairs of points with observed data values on

the vertical axis and the standardized normal quantile

values on the horizontal axis

� Evaluate the plot for evidence of linearity


A normal probability plot for data from a normal distribution will be

approximately linear:

30

60

90

-2 -1 0 1 2 Z

X

The Normal Probability Plot(continued)


Normal Probability Plot

Left-Skewed Right-Skewed

Rectangular

30

60

90

-2 -1 0 1 2 Z

X

(continued)

30

60

90

-2 -1 0 1 2 Z

X

30

60

90

-2 -1 0 1 2 Z

X Nonlinear plots indicate a deviation from normality


Numerical Problems

Example: The following ordered array (from left to right) depicts the amount of money (in dollars) withdrawn from a cash machine by 25customers at a local bank:

40, 50, 50, 50, 50, 70, 70, 80, 80, 90,100, 100 90,100, 100 90,100, 100, 110, 110, 120, 120, 130, 140, 140, 150, 160, 160, 200

Decide whether or not the data appear to be approximately normally distributed by

• Evaluating the actual versus theoretical properties.

a. Constructing a normal probability plot.

Ans.

Look at the MS-Excel worksheet


0.2019Kurtosis

0.3849Skewness

160Range

37.38Standard

deviation

50Interquartile range

105midhinge

120Midrange

100mode

100Median

106.8Mean

25N =

VAR001

300

200

100

0

VAR001

200.0175.0150.0125.0100.075.050.0

Histogram

Fre

quency

10

8

6

4

2

0

Std. Dev = 38.16

Mean = 106.8

N = 25.00

200 Largest

135 Quartile 3

100 Median

80 Quartile 1

40 Smallest

Check for Normality


Check for Normality

.38 (App. Equal)Skewness = 0

160 (Not equal)224.31976 times of SD

50 (App. Equal)49.72421.33 times of SD

24 (App. Equal)(32,182)19/20 of observation

in mean± 2S.d.

19 (App. Equal)(59,155)4/5 of observation in

mean± 1.28S.d.

18 (App. Equal)(69,144)2/3 of observation in

mean± S.d.

median =mode ≠ mean

median =mode = mean

ResultsCheck for

normalityNormal P-P Plot of VAR00001

Observed Cum Prob

1.00.75.50.250.00

Exp

ect

ed

Cu

m P

rob

1.00

.75

.50

.25

0.00


Normal Approximation to the Binomial Distribution

� The binomial distribution is a discrete distribution, but the normal is continuous

� The normal Approximation to the Binomial is very convenient because it enables us to solve the problem without extensive tables of the Bin. Distribution.

� To use the normal to approximate the binomial, accuracy is improved if you use a correction for continuity adjustment

� Example:

� X is discrete in a binomial distribution, so P(X = 4) can be approximated with a continuous normal distribution by finding

P(3.5 < X < 4.5)



� The closer p is to 0.5, the better the normal approximation to the binomial

� The larger the sample size n, the better the normal approximation to the binomial

� General rule:� The normal distribution can be used to approximate

the binomial distribution if

np ≥ 5

and

n(1 – p) ≥ 5

(continued)



� The mean and standard deviation of the binomial distribution are

µ = np

� Transform binomial to normal using the formula:

(continued)

p)np(1

npX

σ

µXZ

−

−=

−=

p)np(1σ −=


Using the Normal Approximation to the Binomial Distribution

� If n = 1000 and p = 0.2, what is P(X ≤ 180)?

� Approximate P(X ≤ 180) using a continuity correction adjustment:

P(X ≤ 180.5)

� Transform to standardized normal:

� So P(Z ≤ -1.54) = 0.0618

1.540.2))(1(1000)(0.2

)(1000)(0.2180.5

p)np(1

npXZ −=

−

−=

−

−=

X180.5 200-1.54 0 Z

Statistics for Business Analysis

Day 5Session-II

CONTINUOUS PROBABILITY DISTRIBUTIONS


The Uniform Distribution


Distributions


Normal

Uniform

Exponential


The Uniform Distribution

� The uniform distribution is a

probability distribution that has equal

probabilities for all possible

outcomes of the random variable

� Also called a rectangular distribution


The Continuous Uniform Distribution:

otherwise 0

bXaifab

1≤≤

−

where

f(X) = value of the density function at any X value

a = minimum value of X

b = maximum value of X

The Uniform Distribution(continued)

f(X) =


Properties of the Uniform Distribution

� The mean of a uniform distribution is

� The standard deviation is

2

baµ

+=

12

a)-(bσ

2

=


Uniform Distribution Example

Example: Uniform probability distributionover the range 2 ≤ X ≤ 6:

2 6

0.25

f(X) = = 0.25 for 2 ≤ X ≤ 66 - 21

X

f(X)

42

62

2

baµ =

+=

+=

1547.112

2)-(6

12

a)-(bσ

22

===


Uniform Distribution Example

Example: Using the uniform probability distribution to find P(3 ≤ X ≤ 5):

2 6

0.25

P(3 ≤ X ≤ 5) = (Base)(Height) = (2)(0.25) = 0.5

X

f(X)

(continued)

3 54


The Exponential Distribution


Distributions


Normal

Uniform

Exponential



� Often used to model the length of time between two occurrences of an event (the time between arrivals)

� Examples:

� Time between trucks arriving at an unloading dock

� Time between transactions at an ATM Machine

� Time between phone calls to the main operator



Xλe1X)time P(arrival −−=<

� Defined by a single parameter, its mean λ(lambda)

� The probability that an arrival time is less than some specified time X is

where e = mathematical constant approximated by 2.71828

λ = the population mean number of arrivals per unit

X = any value of the continuous variable where 0 < X < ∞


Exponential Distribution Example

Example: Customers arrive at the service counter at the rate of 15 per hour. What is the probability that the arrival time between consecutive customers is less than three minutes?

� The mean number of arrivals per hour is 15, so λ = 15

� Three minutes is 0.05 hours

� P(arrival time < .05) = 1 – e-λX = 1 – e-(15)(0.05) = 0.5276

� So there is a 52.76% probability that the arrival time between successive customers is less than three minutes

continous_probability_distribution

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