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7

Chapter

Continuous Distributions

Describing a Continuous Distribution Uniform Continuous Distribution Normal Distribution Standard Normal Distribution Normal Approximation to the Binomial Normal Approximation to the Poisson Exponential Distribution Triangular Distribution McGraw-Hill/Irwin

Copyright Š 2009 by The McGraw-Hill Companies, Inc. All rights reserved.


Continuous Variables Events as Intervals • •

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Discrete Variable – each value of X has its own probability P(X). Continuous Variable – events are intervals and probabilities are areas underneath smooth curves. A single point has no probability.


Describing a Continuous Distribution PDFs and CDFs •

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Probability Density Function (PDF) – For a continuous random variable, the PDF is an equation that shows the height of the curve f(x) at each possible value of X over the range of X.


Describing a Continuous Distribution PDFs and CDFs Continuous PDF’s: • Denoted f(x) • Must be nonnegative • Total area under curve = 1 • Mean, variance and shape depend on the PDF parameters • Reveals the shape of the distribution 7-4

Normal PDF


Describing a Continuous Distribution PDFs and CDFs Continuous CDF’s: • Denoted F(x) • Shows P(X < x), the cumulative proportion of scores • Useful for finding probabilities

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Describing a Continuous Distribution Probabilities as Areas Continuous probability functions are smooth curves. • Unlike discrete distributions, the area at any single point = 0. • The entire area under any PDF must be 1. • Mean is the balance point of the distribution. 7-6


Describing a Continuous Distribution Expected Value and Variance

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Uniform Continuous Distribution Characteristics of the Uniform Distribution •

If X is a random variable that is uniformly distributed between a and b, its PDF has constant height. • Denoted U(a,b) • Area = base x height = (b-a) x 1/(b-a) = 1

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Uniform Continuous Distribution Characteristics of the Uniform Distribution

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Uniform Continuous Distribution Characteristics of the Uniform Distribution â&#x20AC;˘ â&#x20AC;˘

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The CDF increases linearly to 1. CDF formula is (x-a)/(b-a)


Uniform Continuous Distribution Example: Anesthesia Effectiveness â&#x20AC;˘

â&#x20AC;˘ â&#x20AC;˘

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An oral surgeon injects a painkiller prior to extracting a tooth. Given the varying characteristics of patients, the dentist views the time for anesthesia effectiveness as a uniform random variable that takes between 15 minutes and 30 minutes. X is U(15, 30) a = 15, b = 30, find the mean and standard deviation.


Uniform Continuous Distribution Example: Anesthesia Effectiveness a + b 15 + 30 m= = = 22.5 minutes 2 2 (b – a)2 = (30 – 15)2 = 4.33 minutes s= 12 12 Find the probability that the anesthetic takes between 20 and 25 minutes. P(c < X < d) = (d – c)/(b – a) P(20 < X < 25) = (25 – 20)/(30 – 15) = 5/15 = 0.3333 or 33.33% 7-12


Uniform Continuous Distribution Example: Anesthesia Effectiveness P(20 < X < 25)

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Normal Distribution Characteristics of the Normal Distribution •

• • • •

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Normal or Gaussian distribution was named for German mathematician Karl Gauss (1777 – 1855). Defined by two parameters, m and s Denoted N(m, s) Domain is – ∞< X < + ∞ Almost all area under the normal curve is included in the range m – 3s < X < m + 3s


Normal Distribution Characteristics of the Normal Distribution

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Normal Distribution Characteristics of the Normal Distribution â&#x20AC;˘

Normal PDF f(x) reaches a maximum at m and has points of inflection at m + s

Bell-shaped curve

Figure 7.9 7-16


Normal Distribution Characteristics of the Normal Distribution â&#x20AC;˘

Normal CDF

Figure 7.9 7-17


Normal Distribution Characteristics of the Normal Distribution â&#x20AC;˘

All normal distributions have the same shape but differ in the axis scales. m = 42.70mm s = 0.01mm

Diameters of golf balls

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m = 70 s = 10

CPA Exam Scores


Normal Distribution What is Normal? A normal random variable should: • Be measured on a continuous scale. • Possess clear central tendency. • Have only one peak (unimodal). • Exhibit tapering tails. • Be symmetric about the mean (equal tails).

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Standard Normal Distribution Characteristics of the Standard Normal â&#x20AC;˘

Since for every value of m and s, there is a different normal distribution, we transform a normal random variable to a standard normal distribution with m = 0 and s = 1 using the formula:

x â&#x20AC;&#x201C; m z= s â&#x20AC;˘

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Denoted N(0,1)


Standard Normal Distribution Characteristics of the Standard Normal

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Standard Normal Distribution Characteristics of the Standard Normal â&#x20AC;˘

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Standard normal PDF f(x) reaches a maximum at 0 and has points of inflection at +1. Shape is unaffected by the transformation. It is still a bellshaped curve. Figure 7.11

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Standard Normal Distribution Characteristics of the Standard Normal â&#x20AC;˘

Standard normal CDF

Figure 7.11 7-23


Standard Normal Distribution Characteristics of the Standard Normal • • • •

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A common scale from -3 to +3 is used. Entire area under the curve is unity. The probability of an event P(z1 < Z < z2) is a definite integral of f(z). However, standard normal tables or Excel functions can be used to find the desired probabilities.


Standard Normal Distribution Normal Areas from Appendix C-1

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Appendix C-1 allows you to find the area under the curve from 0 to z.

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For example, find P(0 < Z < 1.96):

Figure 7.12


Standard Normal Distribution

Table 7.4 7-26


Standard Normal Distribution Normal Areas from Appendix C-1 â&#x20AC;˘

Now find P(Z < 1.96): .5000

.5000 - .4750 = .0250

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Standard Normal Distribution Normal Areas from Appendix C-1 • •

Now find P(-1.96 < Z < 1.96). Due to symmetry, P(-1.96 < Z) is the same as P(Z < 1.96). .9500

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So, P(-1.96 < Z < 1.96) = .4750 + .4750 = .9500 or 95% of the area under the curve.


Standard Normal Distribution Basis for the Empirical Rule â&#x20AC;˘ â&#x20AC;˘

Approximately 95% of the area under the curve is between + 2s Approximately 99.7% of the area under the curve is between + 3s

68.26%

Figure 7.15 7-29


Standard Normal Distribution Normal Areas from Appendix C-2 â&#x20AC;˘

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Appendix C-2 allows you to find the area under the curve from the left of z (similar to Excel). For example,

.9500

P(Z < 1.96)

P(Z < -1.96) Figure 7.15

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P(-1.96 < Z < 1.96)


Standard Normal Distribution

Table 7.5

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Standard Normal Distribution Normal Areas from Appendices C-1 or C-2 • •

Appendices C-1 and C-2 yield identical results. Use whichever table is easiest.

Finding z for a Given Area • • •

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Appendices C-1 and C-2 be used to find the z-value corresponding to a given probability. For example, what z-value defines the top 1% of a normal distribution? This implies that 49% of the area lies between 0 and z.


Standard Normal Distribution Finding z for a Given Area â&#x20AC;˘

Look for an area of .4900 in Appendix C-1:

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Without interpolation, the closest we can get is z = 2.33

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Standard Normal Distribution Finding z for a Given Area â&#x20AC;˘

Some important Normal areas:

Table 7.7

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Standard Normal Distribution Finding Normal Areas with Excel

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Standard Normal Distribution Finding Normal Areas with Excel

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Standard Normal Distribution Finding Normal Areas with Excel

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Standard Normal Distribution Finding Normal Areas with Excel

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Standard Normal Distribution Finding Areas by using Standardized Variables • Suppose John took an economics exam and scored 86 points. The class mean was 75 with a standard deviation of 7. What percentile is John in (i.e., find P(X < 86)? 86 – 75 x – m = 11/7 = 1.57 zJohn = = 7 s •

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So John’s score is 1.57 standard deviations about the mean.


Standard Normal Distribution Finding Areas by using Standardized Variables â&#x20AC;˘ P(X < 86) = P(Z < 1.57) = .9418 (from Appendix C-2) â&#x20AC;˘ So, John is approximately in the 94th percentile.

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Figure 7.18


Standard Normal Distribution Inverse Normal

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â&#x20AC;˘

You can manipulate the transformation formula to find the normal percentile values (e.g., 5th, 10th, 25th, etc.): x = m + zs

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Here are some common percentiles


Standard Normal Distribution Using Excel Without Standardizing • •

• •

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Excel’s NORMDIST and NORMINV function allow you to evaluate areas without standardizing. For example, let m = 2.040 cm and s = .001 cm, what is the probability that a given steel bearing will have a diameter between 2.039 and 2.042cm? In other words, P(2.039 < X < 2.042) Excel only gives left tail areas, so break the formula into two, find P(X < 2.039) and P(X < 2.042), then subtract them to find the desired probability:


Standard Normal Distribution Using Excel Without Standardizing P(X < 2.042) = .9773

P(X < 2.039) = .1587

P(2.039 < X < 2.042) = .9773 - .1587 = .8186 or 81.9% 7-43


Normal Approximation to the Binomial When is Approximation Needed? • • •

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Binomial probabilities are difficult to calculate when n is large. Use a normal approximation to the binomial. As n becomes large, the binomial bars become more continuous and smooth.


Normal Approximation to the Binomial When is Approximation Needed? â&#x20AC;˘

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Rule of thumb: when np > 10 and n(1-p) > 10, then it is appropriate to use the normal approximation to the binomial. In this case, the binomial mean and standard deviation will be equal to the normal m and s, respectively. m = np s = np(1-p)

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Normal Approximation to the Binomial Example Coin Flips •

• • • •

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If we were to flip a coin n = 32 times and p = .50, are the requirements for a normal approximation to the binomial met? Are np > 5 and n(1-p) > 10? np = 32 x .50 = 16 n(1-p) = 32 x (1 - .50) = 16 So, a normal approximation can be used. When translating a discrete scale into a continuous scale, care must be taken about individual points.


Normal Approximation to the Binomial Example Coin Flips • • •

For example, find the probability of more than 17 heads in 32 flips of a fair coin. This can be written as P(X > 18). However, “more than 17” actually falls between 17 and 18 on a discrete scale. Figure 7.26

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Normal Approximation to the Binomial Example Coin Flips •

• •

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Since the cutoff point for “more than 17” is halfway between 17 and 18, we add 0.5 to the lower limit and find P(X > 17.5). This addition to X is called the Continuity Correction. At this point, the problem can be completed as any normal distribution problem.


Normal Approximation to the Binomial Continuity Correction â&#x20AC;˘

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The table below shows some events and their cutoff point for the normal approximation.


Normal Approximation to the Poisson When is Approximation Needed? â&#x20AC;˘

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The normal approximation to the Poisson works best when l is large (e.g., when l exceeds the values in Appendix B). Set the normal m and s equal to the Poisson mean and standard deviation.

m=l s= l

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Normal Approximation to the Poisson Example Utility Bills •

• •

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On Wednesday between 10A.M. and noon customer billing inquiries arrive at a mean rate of 42 inquiries per hour at Consumers Energy. What is the probability of receiving more than 50 calls? l = 42 which is too big to use the Poisson table. Use the normal approximation with m = l = 42 s = l = 42 = 6.48074


Normal Approximation to the Poisson Example Utility Bills â&#x20AC;˘

â&#x20AC;˘

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To find P(X > 50) calls, use the continuitycorrected cutoff point halfway between 50 and 51 (i.e., X = 50.5). At this point, the problem can be completed as any normal distribution problem.


Exponential Distribution Characteristics of the Exponential Distribution â&#x20AC;˘

â&#x20AC;˘

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If events per unit of time follow a Poisson distribution, the waiting time until the next event follows the Exponential distribution. Waiting time until the next event is a continuous variable.


Exponential Distribution Characteristics of the Exponential Distribution

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Exponential Distribution Characteristics of the Exponential Distribution

Probability of waiting more than x

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Probability of waiting less than x


Exponential Distribution Example Customer Waiting Time •

• • •

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Between 2P.M. and 4P.M. on Wednesday, patient insurance inquiries arrive at Blue Choice insurance at a mean rate of 2.2 calls per minute. What is the probability of waiting more than 30 seconds (i.e., 0.50 minutes) for the next call? Set l = 2.2 events/min and x = 0.50 min P(X > 0.50) = e–lx = e–(2.2)(0.5) = .3329 or 33.29% chance of waiting more than 30 seconds for the next call.


Exponential Distribution Example Customer Waiting Time

P(X > 0.50)

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P(X < 0.50)


Exponential Distribution Inverse Exponential â&#x20AC;˘

â&#x20AC;˘

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If the mean arrival rate is 2.2 calls per minute, we want the 90th percentile for waiting time (the top 10% of waiting time). Find the x-value that defines the upper 10%.


Exponential Distribution Inverse Exponential • • • •

• 7-59

P(X < x) = .90 or P(X > x) = .10 So, e–lx = .10 -lx = ln(.10) = -2.302585 x = 2.302585/l = 2.302585/2.2 = 1.0466 min. 90% of the calls will arrive within 1.0466 minutes (62.8 seconds).


Exponential Distribution Inverse Exponential â&#x20AC;˘

Quartiles for Exponential with l = 2.2

Table 7.11

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Exponential Distribution Mean Time Between Events • • •

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Exponential waiting times are described as Mean time between events (MTBE) = 1/l 1/MTBE = l = mean events per unit of time In a hospital, if an event is patient arrivals in an ER, and the MTBE is 20 minutes, then l = 1/20 = 0.05 arrivals per minute (or 3/hour).


Exponential Distribution Using Excel â&#x20AC;˘

In Excel, use =EXPONDIST(x,l,1) to return the left-tail area P(X < x).

Relation Between Exponential and Poisson

Table 7.12

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Triangular Distribution Characteristics of the Triangular Distribution • A simple distribution that can be symmetric or skewed. • Ranges from a to b and has a mode or “peak” at c • Denoted T(a,b,c)

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Triangular Distribution Characteristics of the Triangular Distribution

Table 7.13

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Triangular Distribution Special Cases: Symmetric Triangular • • • •

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A symmetric triangular distribution is centered at 0 Lower limit is identical to the upper limit except for the sign, with mode 0. Mean m = 0, standard deviation s = b/ 6 =2.45 This distribution closely resembles a standard normal distribution N(0,1) Generate random triangular data in Excel by summing RAND()+RAND()


Applied Statistics in Business & Economics

End of Chapter 7

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Chap7-Applied Stats in Bus & Eco - Doane/Seward-2E