President UniversityErwin SitompulPBST 7/1 Dr.-Ing. Erwin Sitompul President University Lecture 7...

President University Erwin Sitompul PBST 7/1

Dr.-Ing. Erwin SitompulPresident University

Lecture 7

Probability and Statistics

http://zitompul.wordpress.com


Chapter 6

Some Continuous Probability Distributions

Chapter 6 Some Continuous Probability Distributions


Continuous Uniform Distribution |Uniform Distribution| The density function of the continuous

uniform random variable X on the interval [A, B] is

Chapter 6.1 Continuous Uniform Distribution

1,

( ; , )

0, elsewhere

A x BB Af x A B

The uniform density function for a random variable on the interval [1, 3]

The mean and variance of the uniform distribution are2

2 ( )

2 12

A B B A and


Continuous Uniform DistributionChapter 6.1 Continuous Uniform Distribution

Suppose that a large conference room for a certain company can be reserved for no more than 4 hours. However, the use of the conference room is such that both long and short conference occur quite often. In fact, it can be assumed that length X of a conference has a uniform distribution on the interval [0, 4].(a) What is the probability density function?(b) What is the probability that any given conference lasts at least 3

hours?

(a)

(b)

1, 0 4

4( )

0, elsewhere

xf x

4

3

13

4P X dx

1

4


Normal Distribution Normal distribution is the most important continuous probability

distribution in the entire field of statistics. Its graph, called the normal curve, is the bell-shaped curve which

describes approximately many phenomena that occur in nature, industry, and research.

The normal distribution is often referred to as the Gaussian distribution, in honor of Karl Friedrich Gauss, who also derived its equation from a study of errors in repeated measurements of the same quantity.

Chapter 6.2 Normal Distribution

The normal curve


Normal Distribution A continuous random variable X having the bell-shaped distribution

as shown on the figure is called a normal random variable.

21

21( ; , ) ,

2

x

n x e x

where π = 3.14159... and e = 2.71828...


The density function of the normal random variable X, with mean μ and variance σ2, is


Normal Curve

μ1 < μ2, σ1 = σ2 μ1 = μ2, σ1 < σ2

μ1 < μ2, σ1 < σ2



Normal CurveChapter 6.2 Normal Distribution

σ σ

x

f(x)

μ

The mode, the point where the curve is at maximum

Symmetry about a vertical axis through the mean μ

Point of inflection

Concave downward

Concave upward

Approaches zero asymptotically

Total area under the curve and above the horizontal axis is equal to 1


Area Under the Normal Curve The area under the curve bounded by two ordinates x = x1 and

x = x2 equals the probability that the random variable X assumes a value between x = x1 and x = x2.

Chapter 6.3 Areas Under the Normal Curve

22 2

1 1

1

21 2

1( ) ( ; , )

2

xx x

x x

P x X x n x dx e dx


Area Under the Normal Curve As seen previously, the normal curve is dependent on the mean μ

and the standard deviation σ of the distribution under investigation.

The same interval of a random variable can deliver different probability if μ or σ are different.


Same interval, but different probabilities for two different normal curves


Area Under the Normal Curve The difficulty encountered in solving integrals of normal density

functions necessitates the tabulation of normal curve area for quick reference.

Fortunately, we are able to transform all the observations of any normal random variable X to a new set of observation of a normal random variable Z with mean 0 and variance 1.


XZ

2

2

1

1

21 2

1( )

2

xx

x

P x X x e dx

22

1

21

2

z z

z

e dz

2

1

( ;0,1)z

z

n z dz 1 2( )P z Z z


Area Under the Normal Curve The distribution of a normal random variable with mean 0 and

variance 1 is called a standard normal distribution.



Table A.3 Normal Probability TableChapter 6.3 Areas Under the Normal Curve


InterpolationChapter 6.3 Areas Under the Normal Curve

a

( )f a

b

( )f b

Interpolation is a method of constructing new data points within the range of a discrete set of known data points.

Examine the following graph. Two data points are known, which are (a,f(a)) and (b,f(b)).

If a value of c is given, with a < c < b, then the value of f(c) can be estimated.

If a value of f(c) is given, with f(a) < f(c) < f(b), then the value of c can be estimated.

?c

( )?f c

( ) ( ) ( ) ( )c a

f c f a f b f ab a

( ) ( )

( ) ( )

f c f ac a b a

f b f a


InterpolationChapter 6.3 Areas Under the Normal Curve

P(Z < 1.172)?

Answer: 0.8794 P(Z < z) = 0.8700, z = ?

1.126


Area Under the Normal CurveChapter 6.3 Areas Under the Normal Curve

Given a standard normal distribution, find the area under the curve that lies (a) to the right of z = 1.84 and (b) between z = –1.97 and z = 0.86.

(a)

(b)

( 1.84) 1 ( 1.84)P Z P Z

1 0.9671

0.0329

( 1.94 0.86) ( 0.86) ( 1.94)P Z P Z P Z

0.8051 0.0244

0.7807



Given a standard normal distribution, find the value of k such that (a) P ( Z > k ) = 0.3015, and (b) P ( k < Z < –0.18 ) = 0.4197.

(a)

(b)

( ) 1 ( )P Z k P Z k

( ) 1 ( )P Z k P Z k

0.52k

( 0.18) ( 0.18) ( )P k Z P Z P Z k

1 0.3015 0.6985

( ) ( 0.18) ( 0.18)P Z k P Z P k Z

0.4286 0.4197 0.0089

2.37k



Given a random variable X having a normal distribution with μ = 50 and σ = 10, find the probability that X assumes a value between 45 and 62.

11

xz

(45 62) ( 0.5 1.2)P X P Z

22

xz

( 1.2) ( 0.5)P Z P Z

0.8849 0.3085

0.5764

45 500.5

10

62 501.2

10



Given that X has a normal distribution with μ = 300 and σ = 50, find the probability that X assumes a value greater than 362.

xz

( 362) ( 1.24)P X P Z

1 0.8925

1 ( 1.24)P Z

0.1075

362 3001.24

50



Given a normal distribution with μ = 40 and σ = 6, find the value of x that has (a) 45% of the area to the left, and (b) 14% of the area to the right.

(a) ( ) 0.45P Z z 0.45 0.44830.13 0.12 ( 0.13)

0.4522 0.4483z

0.1256

x z

0.13

0.4483

0.12

0.4522

?

0.45

40 ( 0.1256)(6) 39.2464



Given a normal distribution with μ = 40 and σ = 6, find the value of x that has (a) 45% of the area to the left, and (b) 14% of the area to the right.

(b) ( ) 0.14 1 ( )P z Z P Z z

( ) 1 0.14 0.86P Z z

1.08z

x z 40 (1.08)(6) 46.48


Applications of the Normal DistributionChapter 6.4 Applications of the Normal Distribution

A certain type of storage battery lasts, on average, 3.0 years, with a standard deviation of 0.5 year. Assuming that the battery lives are normally distributed, find the probability that a given battery will last less than 2.3 years.

( 1.4) 0.0808P Z

xz

2.3 3.0

1.40.5

8.08%



In an industrial process the diameter of a ball bearing is an important component part. The buyer sets specifications on the diameter to be 3.0 ± 0.01 cm. All parts falling outside these specifications will be rejected. It is known that in the process the diameter of a ball bearing has a normal distribution with mean 3.0 and standard deviation 0.005. On the average, how many manufactured ball bearings will be scrapped?

11

xz

22

xz

(2.99 3.01) ( 2 2)P X P Z

( 2) ( 2)P Z P Z 0.9772 0.0228

0.9544

4.56% rejected

2.99 3.02

0.005

3.01 3.02

0.005

95.44% accepted



A certain machine makes electrical resistors having a mean resistance of 40 Ω and a standard deviation of 2 Ω. It is assumed that the resistance follows a normal distribution.What percentage of resistors will have a resistance exceeding 43 Ω if:(a) the resistance can be measured to any degree of accuracy.(b) the resistance can be measured to the nearest ohm only.

43 401.5

2z

( 43) ( 1.5)P X P Z 1 ( 1.5)P Z 1 0.9332 0.0668 6.68%(a)

(b)

( 43.5) ( 1.75)P X P Z 1 ( 1.75)P Z 1 0.9599 0.0401 4.01%

43.5 401.75

2z

As many as 6.68%–4.01% = 2.67% of the resistors will be accepted although the value is greater than 43 Ω due to measurement limitation



The average grade for an exam is 74, and the standard deviation is 7. If 12% of the class are given A’s, and the grade are curved to follow a normal distribution, what is the lowest possible A and the highest possible B?

( ) 0.12P Z z

( ) 1 ( )P Z z P Z z 1 0.12 0.88

1.175z

x z Lowest possible A is 83 Highest possible B is 82

74 (1.175)(7) 82.225


Normal Approximation to the BinomialChapter 6.5 Normal Approximation to the Binomial

The probabilities associated with binomial experiments are readily obtainable from the formula b(x;n, p) of the binomial distribution or from the table when n is small.

For large n, making the distribution table is not practical anymore. Nevertheless, the binomial distribution can be nicely approximated

by the normal distribution under certain circumstances.



If X is a binomial random variable with mean μ = np and variance σ2 = npq, then the limiting form of the distribution of

X npZ

npq

as n ∞, is the standard normal distribution n(z;0, 1).

Normal approximation of b(x; 15, 0.4) Each value of b(x; 15, 0.4) is

approximated by P(x–0.5 < X < x+0.5)



Normal approximation of

(4;15,0.4)b9

7

( ;15,0.4)x

b x

( 4) (4;15,0.4)P X b

( 4) (3.5 4.5)P X P X

np

npq

( 1.32 0.79)P Z 0.1214

4 1115 4 (0.4) (0.6)C

0.1268

(15)(0.4)(0.6) 1.897

(15)(0.4) 6

9

7

(7 9) ( ;15,0.4)x

P X b x

0.3564

(7 9) (6.5 9.5)P X P X

(0.26 1.85)P Z 0.3652

and



The degree of accuracy, that is how well the normal curve fits the binomial histogram, will increase as n increases.

If the value of n is small and p is not very close to 1/2, normal curve will not fit the histogram well, as shown below.

( ;6,0.2)b x ( ;15,0.2)b x

The approximation using normal curve will be excellent when n is large or n is small with p reasonably close to 1/2.

As rule of thumb, if both np and nq are greater than or equal to 5, the approximation will be good.



Let X be a binomial random variable with parameters n and p. For large n, X has approximately a normal distribution with μ = np and σ2 = npq = np(1–p) and

0

( ) ( ; , )x

k

P X x b k n p

0.5area under normal curve to the left of x

( 0.5)P X x

( 0.5)xP Z

and the approximation will be good if np and nq = n(1–p) are greater than or equal to 5.



The probability that a patient recovers from a rare blood disease is 0.4. If 100 people are known to have contracted this disease, what is the probability that less than 30 survive?

np

npq (100)(0.4)(0.6) 4.899

(100)(0.4) 40 100, 0.4n p

29

0

( 30) ( ;100,0.4)x

P X b x

( 30) ( 29.5)P X P X

( 2.143)P Z

29.5 402.143

4.899z

0.01608 After interpolation

1.608% Can you calculate the

exact solution?



A multiple-choice quiz has 200 questions each with 4 possible answers of which only 1 is the correct answer. What is the probability that sheer guess-work yields from 25 to 30 correct answers for 80 of the 200 problems about which the student has no knowledge?

np

npq 314 4(80)( )( ) 3.873

14(80)( ) 20 1

80,4

n p

3014

25

(25 30) ( ;80, )x

P X b x

(24.5 30.5)P X

(1.162 2.711)P Z

( 2.711) ( 1.162)P Z P Z

0.9966 0.8774 0.1192

1

24.5 201.162,

3.873z

2

30.5 202.711

3.873z



PU Physics entrance exam consists of 30 multiple-choice questions each with 4 possible answers of which only 1 is the correct answer. What is the probability that a prospective students will obtain scholarship by correctly answering at least 80% of the questions just by guessing?

np

npq 314 4(30)( )( ) 2.372

14(30)( ) 7.5 1

30,4

n p

3014

24

( 24) ( ;30, )x

P X b x

1 ( 23.5)P X

1 ( 6.745)P Z

0

23.5 7.56.745

2.372z

It is practically impossible to get scholarship just by pure luck in the entrance exam


Gamma and Exponential DistributionsChapter 6.6 Gamma and Exponential Distributions

There are still numerous situations that the normal distribution cannot cover. For such situations, different types of density functions are required.

Two such density functions are the gamma and exponential distributions.

Both distributions find applications in queuing theory and reliability problems.

1

0

( ) xx e dx

The gamma function is defined by

for α > 0.



|Gamma Distribution| The continuous random variable X has a gamma distribution, with parameters α and β, if its density function is given by

11, 0

( )( )

0, elsewhere

xx e xf x

where α > 0 and β > 0.

|Exponential Distribution| The continuous random variable X has an exponential distribution, with parameter β, if its density function is given by

where β > 0.

1, 0

( )

0, elsewhere

xe xf x



Gamma distributions for certain values of the parameters α and β

The gamma distribution with α = 1 is called the exponential distribution



The mean and variance of the gamma distribution are2 2

The mean and variance of the exponential distribution are2 2 and

and


ApplicationsChapter 6.7 Applications of the Gamma and Exponential Distributions

Suppose that a system contains a certain type of component whose time in years to failure is given by T. The random variable T is modeled nicely by the exponential distribution with mean time to failure β = 5. If 5 of these components are installed in different systems, what is the probability that at least 2 are still functioning at the end of 8 years?

5

8

1( 8)

5tP T e dt

5

2

( 2) ( ;5,0.2)x

P X b x

8 5 0.2e 1

0

1 ( ;5,0.2)x

b x

1 0.7373 The probability whether

the component is still functioning at the end of 8 years

The probability whether at least 2 out of 5 such component are still functioning at the end of 8 years

0.2627



Suppose that telephone calls arriving at a particular switchboard follow a Poisson process with an average of 5 calls coming per minute. What is the probability that up to a minute will elapse until 2 calls have come in to the switchboard?

20

1( )

xxP X x xe dx

15

0

( 1) 25 xP X xe dx

1 5, 2

5(1)1 (1 5) 0.96e

β is the mean time of the event of calling

α is the quantity of the event of calling



Based on extensive testing, it is determined that the average of time Y before a washing machine requires a major repair is 4 years. This time is known to be able to be modeled nicely using exponential function. The machine is considered a bargain if it is unlikely to require a major repair before the sixth year.(a) Determine the probability that it can survive without major repair

until more than 6 years.(b) What is the probability that a major repair occurs in the first

year?

4

6

1( 6)

4tP Y e dt

(a)

(b)

Only 22.3% survives until more than 6 years without major reparation

4

1

1( 1) 1

4tP Y e dt

22.1% will need major

reparation after used for 1 year

6 4 0.223e

1 41 0.221e

14

0

1

4te dt


Chi-Squared DistributionChapter 6.8 Chi-Squared Distribution

Another very important special case of the gamma distribution is obtained by letting α = v/2 and β = 2, where v is a positive integer.

The result is called the chi-squared distribution, with a single parameter v called the degrees of freedom.

The chi-squared distribution plays a vital role in statistical inference. It has considerable application in both methodology and theory.

Many chapters ahead of us will contain important applications of this distribution.


Chi-Squared DistributionChapter 6.8 Chi-Squared Distribution

|Chi-Squared Distribution| The continuous random variable X has a chi-squared distribution, with v degrees of freedom, if its density function is given by

2 12

1, 0

2 ( 2)( )

0, elsewhere

v xv

x e xvf x

where v is a positive integer.

The mean and variance of the chi-squared distribution are2 2v v and


Lognormal DistributionChapter 6.9 Lognormal Distribution

The lognormal distribution is used for a wide variety of applications.

The distribution applies in cases where a natural log transformation results in a normal distribution.



|Lognormal Distribution| The continuous random variable X has a lognormal distribution if the random variable Y = ln(X) has a normal distribution with mean μ and standard deviation σ. The resulting density function of X is

2 2ln( ) (2 )1, 0

2( )

0, 0

xe xxf x

x

The mean and variance of the chi-squared distribution are2 2 22 2( ) Var( ) ( 1)E X e X e e and



Concentration of pollutants produced by chemical plants historically are known to exhibit behavior that resembles a log normal distribution. This is important when one considers issues regarding compliance to government regulations. Suppose it is assumed that the concentration of a certain pollutant, in parts per million, has a lognormal distribution with parameters μ = 3.2 and σ = 1. What is the probability that the concentration exceeds 8 parts per million?

( 8) 1 ( 8)P X P X

ln(8) 3.2( 8)

1P X F

F denotes the cumulative distribution function of the standard normal distribution

a. k. a. the area under the normal curve

( 1.12) 0.1314F


Homework 7Probability and Statistics

1. Suppose the current measurements in a strip of wire are assumed to follow a normal distribution with a mean of 10 milliamperes and a variance of 4 milliamperes2. (a) What is the probability that a measurement will exceed 13 milliamperes? (b) Determine the value for which the probability that a current measurement is below this value is 98%. (Mo.E4.13-14 p.113)

2. A lawyer commutes daily from his suburban home to midtown office. The average time for a one-way trip is 24 minutes, with a standard deviation of 3.8 minutes. Assume the distribution of trip times to be normally distributed. (a) If the office opens at 9:00 A.M. and the lawyer leaves his house at 8:45 A.M. daily, what percentage of the time is he late for work? (b) Find the probability that 2 of the next 3 trips will take at least 1/2 hour.

(Wa.6.15 s.186)

3. (a) Suppose that a sample of 1600 tires of the same type are obtained at random from an ongoing production process in which 8% of all such tires produced are defective. What is the probability that in such sample 150 or fewer tires will be defective?

(Sou18. CD6-13)

(b) If 10% of men are bald, what is the probability that more than 100 in a random sample of 818 men are bald?

President UniversityErwin SitompulPBST 7/1 Dr.-Ing. Erwin Sitompul President University Lecture 7...

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