President UniversityErwin SitompulPBST 8/1 Lecture 8 Probability and Statistics Dr.-Ing. Erwin Sitompul President University

President University Erwin Sitompul PBST 8/1

Lecture 8

Probability and Statistics

Dr.-Ing. Erwin SitompulPresident University

http://zitompul.wordpress.com

2 0 1 3


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


Homework 7AProbability 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)

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President UniversityErwin SitompulPBST 8/1 Lecture 8 Probability and Statistics Dr.-Ing. Erwin Sitompul President University