7. continuous distributions

CONTINOUS DISTRIBUTIONS

NORMAL, EXPONENTIAL AND UNIFORM

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Normal Distribution

• This is the most important continuous distribution.

– Many distributions can be approximated by a normal distribution.

– The normal distribution is the cornerstone distribution of statistical inference.

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• A random variable X with mean and variance is normally distributed if its probability density function is given by

...71828.2eand...14159.3where

xe2

1)x(f

2x

)2/1(

...71828.2eand...14159.3where

xe2

1)x(f

2x

)2/1(

Normal Distribution

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The Shape of the Normal Distribution

The normal distribution is bell shaped, and symmetrical around

Why symmetrical? Let = 100. Suppose x = 110.

2210

)2/1(100110

)2/1(e

21

e2

1)110(f

Now suppose x = 9022

10)2/1(

10090)2/1(

e2

1e

21

)90(f

11090

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The effects of The effects of and and

How does the standard deviation affect the shape of f(x)?

= 2

=3 =4

= 10 = 11 = 12How does the expected value affect the location of f(x)?

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• Two facts help calculate normal probabilities:– The normal distribution is symmetrical.– Any normal distribution can be transformed into

a specific normal distribution called…

“STANDARD NORMAL DISTRIBUTION”

Example:

The amount of time it takes to assemble a computer is normally distributed, with a mean of 50 minutes and a standard deviation of 10 minutes. What is the probability that a computer is assembled in a time between 45 and 60 minutes?

Finding Normal Probabilities

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STANDARD NORMAL DISTRIBUTION

• NORMAL DISTRIBUTION WITH MEAN 0 AND VARIANCE 1.

• IF X~N( , 2), THEN

NOTE: Z IS KNOWN AS Z SCORES.

• “ ~ “ MEANS “DISTRIBUTED AS”

~ (0,1)X

Z N

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• Solution– If X denotes the assembly time of a computer, we

seek the probability P(45<X<60).– This probability can be calculated by creating a new

normal variable the standard normal variable.

x

xXZ

x

xXZ

E(Z) = = 0 V(Z) = 2 = 1

Every normal variablewith some and, canbe transformed into this Z.

Therefore, once probabilities for Zare calculated, probabilities of any normal variable can be found.

Finding Normal Probabilities

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• Example - continued

P(45<X<60) = P( < < )45 X 60- 50 - 5010 10

= P(-0.5 < Z < 1)

To complete the calculation we need to compute the probability under the standard normal distribution.

Finding Normal Probabilities

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z 0 0.01 ……. 0.05 0.060.0 0.5000 0.5040 0.5199 0.52390.1 0.5398 0.5438 0.5596 0.5636

. . . . .

. . . . .1.0 0.8413 0.8438 0.8531 0.8554

. . . . .

. . . . .1.2 0.8849 0.8869 ……. 0.8944 0.8962

. . . . . .

. . . . . .

Standard normal probabilities have been calculated and are provided in a table .

The tabulated probabilities correspondto the area less Z = z0.

Using the Standard Normal TableUsing the Standard Normal Table

Z = 0 Z = z0

P(Z<z0)

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• Example - continued

P(45<X<60) = P( < < )45 X 60- 50 - 5010 10

= P(-.5 < Z < 1)

z0 = 1z0 = -.5

We need to find the shaded area

Finding Normal Probabilities

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P(Z<1)-P(Z<-0.5)

P(45<X<60) = P( < < )45 X 60- 50 - 5010 10

z 0 0.1 ……. 0.05 0.060.0 0.5000 0.5040 0.5199 0.52390.1 0.5398 0.5438 0.5596 0.5636

. . . . .

. . . . .1.0 0.8413 0.8438 0.8531 0.8554

. . . . .

• Example - continued

= P(-.5<Z<1) =

.3413

Finding Normal Probabilities

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• The symmetry of the normal distribution makes it possible to calculate probabilities for negative values of Z using the table as follows:

-z0 +z00

P(-z0<Z<0) = P(0<Z<z0)

Finding Normal Probabilities

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z 0 0.1 ……. 0.05 0.060.0 0.5000 0.5040 0.5199 0.52390.1 0.5398 0.5438 0.5596 0.5636

. . . . .

. . . . .0.5 0.6915 …. …. ….

. . . . .

• Example - continued

Finding Normal Probabilities

.3413

P(Z<-0.5)=1-P(Z>-0.5)=1-0.6915=0.3085By Symmetry

P(Z<0.5)

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z 0 0.1 ……. 0.05 0.060.0 0.5000 0.5040 0.5199 0.52390.1 0.5398 0.5438 0.5596 0.5636

. . . . .

. . . . .0.5 0.6915 …. …. ….

. . . . .

• Example - continued

Finding Normal Probabilities

.3413

P(-.5<Z<1) = P(Z<1)-P(Z<-.5) = .8413-.3085 + = .5328

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AREAS UNDER THE STANDARD NORMAL DENSITY

P(Z<1.24)=.8925

Z0 1

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AREAS UNDER THE STANDARD NORMAL DENSITY

P(1<Z<2)=.9772-.8413=.1359

1 2

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EXAMPLES

• P( Z < 0.94 ) = P( Z < 0.94 )

= 0.8264

0.940

0.8264

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EXAMPLES

• P( Z > 1.76 ) = 1- P( Z < 1.76 )

= 1 – 0.9608 = 0.0392

1.760

0.0392

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EXAMPLES

• P( -1.56 < Z < 2.13 ) =

= P( Z < 2.13 )-P( Z < -1.56 )

= 0.9834 -(1-0.9406) = 0.9240

-1.56 2.13

0.9240

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STANDARDIZATION FORMULA

• If X~N( , 2), then the standardized value Z of any ‘X-score’ associated with calculating probabilities for the X distribution is:

• The standardized value Z of any ‘X-score’ associated with calculating probabilities for the X distribution is:

(Converse Formula)

XZ

.x z

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• Sometimes we need to find the value of Z for a given probability.

• We use the notation zA to express a Z value for which P(Z > zA) = A

Finding Values of Z

zA

A

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EXAMPLES

• Let X be rate of return (X) on a proposed investment. Mean is 0.30 and standard deviation is 0.1.

a) P(X>.55)=?b) P(X<.22)=?c) P(.25<X<.35)=?d) 80th Percentile of X is?e) 30th Percentile of X is?

Standardization formula

Converse Formula

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EXAMPLE

• LET X ~ N( 10, 2=81 ). IF P(X < a) = 0.95, FIND the value of “a”. (95th percentile)

80524964511064519

10

9509

10

6451

..a.a

.aX

ZPaXP

.

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EXAMPLE– The mean time it took all competitors to run

the 100 meters was 12.92 seconds. Assuming a standard deviation of 1.3 seconds and a normal distribution in times, what percentage of the competitors finished the race in under 10.5 seconds?

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EXAMPLE• Assume that the length of time, X between

charges of a pocket calculator is normally distributed with mean 100 hours and a standard deviation of 15 hours. Find the probability that the calculator will last between 80 and 120 hours between charges?

EXAMPLE• A certain brand of flood lamps has a lifetime

that is normally distributed with a mean of 3,750 hours and a standard deviation of 300 hours.

a. What proportion of these lamps will last for more than 4,000 hours?

b. What lifetime should the manufacturer advertise for these lamps in order that only 2% of the lamps will burn out before the advertised lifetime?

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The Normal Approximation to the Binomial Distribution

• The normal distribution provides a close approximation to the Binomial distribution when n (number of trials) is large and p (success probability) is close to 0.5.

• The approximation is used only when

np 5

n(1-p) 5

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The Normal Approximation to the Binomial Distribution

• If the assumptions are satisfied, the Binomial random variable X can be approximated by normal distribution with mean = np and 2 = np(1-p).

• In probability calculations, the continuity correction must be used. For example, if X is Binomial random variable, then

P(X a) approximated by normal X in prob. asP(X<a+0.5) but P(X < a) should be the same.

P(X a) approximated by normal X in prob. asP(X>a-0.5) but P(X > a) should be the same.

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EXAMPLE

• Probability of getting 16 heads in 40 flips of a balanced coin. Find the approximate probability of getting 16 heads.

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EXAMPLE

• Suppose that 10% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped). Consider a random sample of 200 shafts, and let X denote the number among these that are nonconforming and can be reworked. What is the (approximate) probability that X is

a) At most 30?b) Less than 30?c) Between 15 and 25 (inclusive)?

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Exponential Distribution• The exponential distribution can be used to model

– the length of time between telephone calls– the length of time between arrivals at a service station– the lifetime of electronic components.

• When the number of occurrences of an event follows the Poisson distribution, the time between occurrences follows the exponential distribution.

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A random variable is exponentially distributed if its probability density function is given by

f(x) = ex, x>=0. is the distribution parameter (>0).

E(X) = 1/ V(X) = (1/

Exponential Distribution

The cumulative distribution function isF(x) =1ex, x0

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0

0.5

1

1.5

2

2.5f(x) = 2e-2x

f(x) = 1e-1x

f(x) = .5e-.5x

0 1 2 3 4 5

Exponential distribution for = .5, 1, 2

0

0.5

1

1.5

2

2.5

a bP(a<x<b) = e-a eb

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• Finding exponential probabilities is relatively easy:– P(X > a) = e–a.– P(X < a) = 1 – e –a

– P(a< X < b) = e – a) – e – b)

Exponential Distribution

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• Example– The lifetime of an alkaline battery is

exponentially distributed with = .05 per hour.

– What is the mean and standard deviation of the battery’s lifetime?

– Find the following probabilities:• The battery will last between 10 and 15 hours.• The battery will last for more than 20 hours.

Exponential Distribution

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• Solution– The mean = standard deviation =

1/1/.05 = 20 hours.– Let X denote the lifetime.

• P(10<X<15) = e-.05(10) – e-.05(15) = .1341• P(X > 20) = e-.05(20) = .3679

Exponential Distribution

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• Example The service rate at a supermarket checkout

is 6 customers per hour. – If the service time is exponential, find the

following probabilities:• A service is completed in 5 minutes,• A customer leaves the counter more than 10

minutes after arriving• A service is completed between 5 and 8 minutes.

Exponential Distribution

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• Solution– A service rate of 6 per hour =

A service rate of .1 per minute ( = .1/minute).– P(X < 5) = 1-e-x = 1 – e-.1(5) = .3935– P(X >10) = e-x = e-.1(10) = .3679– P(5 < X < 8) = e-.1(5) – e-.1(8) = .1572

Exponential Distribution

EXAMPLE

• A used car salesman in a small town states that, on the average, it takes him 5 days to sell a car. Assume that the probability distribution of the length of time between sales is exponentially distributed.

a.What is the probability that he will have to wait at least 8 days before making another sale?

b.What is the probability that he will have to wait between 6 and 10 days before making another sale?

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Exponential Distribution

The failure time (in years) of an electronic digital display is an exponential random variable with = 1/5.

• Find P(X 4).

• Find P(X>8).

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Exponential Distribution• The key property of the exponential random

variable is that it is memoryless. That is,P(X > s+t | X > t) = P(X > s) for all s and t 0.

Example: Suppose that a number of miles that a car can run before its battery wears out is exponentially distributed with an average value of 10,000 miles. If a person desires to take a 5,000 mile trip, what is the probability that she will be able to complete her trip without having to replace her car battery?

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Exponential Distribution

• If pdf of lifetime of fluorescent lamp is exponential with =10, find the life for 95% reliability?

The reliability function = R(t) = 1-F(t) = e-t

R(t) = 0.95 t = ?

What is the mean time to failure?

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– A random variable X is said to be uniformly distributed if its density function is

– The expected value and the variance are

12)ab(

)X(V2

baE(X)

.bxaab

1)x(f

2

Uniform Distribution

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• Example The daily sale of gasoline is uniformly distributed between 2,000 and 5,000 gallons. Find the probability that sales are:

– Between 2,500 and 3,500 gallons– More than 4,000 gallons– Exactly 2,500 gallons

2000 5000

1/3000

f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000]

x2500 3000

P(2500X3000) = (3000-2500)(1/3000) = .1667

Uniform Distribution

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EXAMPLE

• If the speed of a car is equally likely between 60 and 100 kms,

a) find the average speed of the car.

b) find the probability that the speed of the car is between 60 and 70.

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Other Continuous Distributions

– Student t distribution– Chi-squared distribution– F distribution– Gamma distribution– Beta distribution