Your 4 th homework is assigned. It is due on 12 th of Feb, 11:59 pm.

• Your 4 th homework is assigned. It is due on 12th of Feb, 11:59 pm.

a) Adult not working during summer vacation.

b)

•The experiment consist of 10 identical trials. A trial for this experiment is an individual.

•There are only two possible outcomes: work or do not work

•The probability remains same for each individual (trial)

•Individuals are independent

c) 0.35

d) 0.2522

e) 0.2616

By using TI-84:

d) P(X = 3) = = P(X ≤ 3) – P(X ≤ 2)

binomcdf(10,.35,3) - binomcdf(10,.35,2)

e) P(X ≤ 2) binomcdf(10,.35,2)

Thinking Challenge

• The communications monitoring company Postini has reported that 91% of e-mail messages are spam. Suppose your inbox contains 25 messages.

• What are the mean and standard deviation of the number of real messages you should expect to find in your inbox?

• What is the probability that you will find only 1 or 2 real messages?

Thinking Challenge

• What are the mean and standard deviation of the number of real messages you should expect to find in your inbox? (2.25, 1.43)

• What is the probability that you will find only 1 or 2 real messages?

(P(X=1)+P(X=2)= 0.5117)

Content

1. Two Types of Random Variables

2. Probability Distributions for Discrete Random Variables

3. The Binomial Distribution

4. Hypergeometric Distributions

5. Probability Distributions for Continuous Random Variables

6. The Normal Distribution

7. Uniform Distribution

4.4

Hypergeometric Distribution

Characteristics of a Hypergeometric

Random Variable

1. The experiment consists of randomly drawing n elements without replacement from a set of N elements, r of which are S’s (for success) and (N – r) of which are F’s (for failure).

2. The hypergeometric random variable x is the number of S’s in the draw of n elements.

Hypergeometric Probability Distribution Function

where . . .

[x = Maximum [0, n – (N – r), …, Minimum (r, n)]

p x

r

x

N r

n x

N

n

µ nr

N 2

r N r n N n N 2 N 1

Hypergeometric Probability Distribution Function

N = Total number of elements

r = Number of S’s in the N elements

n = Number of elements drawn

x = Number of S’s drawn in the n elements

a. Hypergeometric

b. Binomial

Thinking Challenge

A carton of 12 eggs has 4 rotten eggs and 8 good eggs. Three eggs are chosen at random from the carton to make a three-egg omelet.

Let X = the number of rotten eggs chosen. What is the probability that the sample will consist of one rotten egg and two good eggs, that is, what is P(X = 1)?

(a) 81/220 (b) 192/220 (c) 112/220 (d) 56/220

Thinking Challenge

• 5 cards are picked from a deck of 52 cards without replacement. What is the probability that 2 of the selected cards will be Ace?

a) b) c) d)

• • • •

4.5

Probability Distributions for Continuous Random Variables

Continuous Probability Density Function

The graphical form of the probability distribution for a continuous random variable x is a smooth curve that might appear as below:


This curve, a function of x, is denoted by the symbol f(x) and is variously called a probability density function (pdf), a frequency function, or a probability distribution.•The areas under a probability distribution correspond to probabilities for x. The area A beneath the curve between two points a and b is the probability that x assumes a value between a and b.


P(a<x<b)=P(a≤x≤b) sinceP(x=a)=P(x=b)=0

4.6

The Normal Distribution

Importance of Normal Distribution

1. Describes many random processes or continuous phenomena

2. Can be used to approximate discrete probability distributions

• Example: binomial

3. Basis for classical statistical inference

Normal Distribution

1. ‘Bell-shaped’ & symmetrical

2. Mean, median, mode are equal •x

•f•(•x•)

Mean Median Mode

Probability Density Function

whereµ = Mean of the normal random variable x = Standard deviationπ = 3.1415 . . .e = 2.71828 . . . P(x < a) is obtained from a table of normal

probabilities

f (x) 1

2e

1

2

x

2

Effect of Varying Parameters ( & )

Normal Distribution Probability

c dx

f(x)

•Probability is area under curve!

P(c x d) f (x)c

d

dx?

Standard Normal Distribution •The standard normal distribution is a normal distribution with µ = 0 and = 1. A random variable with a standard normal distribution, denoted by the symbol z, is called a standard normal random variable.

The Standard Normal Table:P(0 < z < 1.96)

z = 0

= 1

1.96

Z .04 .05

1.8 .4671 .4678 .4686

.4738 .4744

2.0 .4793 .4798 .4803

2.1 .4838 .4842 .4846

.06

1.9 .4750

Standard Normal Probability Table (Portion)

Probabilities

.4750

The Standard Normal Table:P(–1.26 z 1.26)

z = 0

= 1

–1.26

Standard Normal Distribution

.3962

1.26

.3962 P(–1.26 ≤ z ≤ 1.26)

= .3962 + .3962

= .7924

The Standard Normal Table:P(z > 1.26)

z


1.26

P(z > 1.26)

= .5000 – .3962

= .1038.3962

.500

= 1

= 0

.500

The Standard Normal Table:P(–2.78 z –2.00)

–2.78 z–2.00

.4973

.4772


P(–2.78 ≤ z ≤ –2.00)

= .4973 – .4772

= .0201

= 1

= 0

The Standard Normal Table:P(z > –2.13)

z–2.13


P(z > –2.13)

= .4834 + .5000

= .9834

.5000.4834

= 1

= 0

•The rest of the quizzes will be held in the lecture on Wednesdays.

•The updated dates for quizzes are as below.

•I will take attendance and will give extra credit to the students who have attended at least 8 of the lectures till the end of the semester.

Non-standard Normal Distribution

x

f(x)

Normal distributions differ by mean & standard deviation.

Each distribution would require its own table.

That’s an infinite number of tables!

Property of Normal Distribution

•If x is a normal random variable with mean μ and standard deviation , then the random variable z, defined by the formula

has a standard normal distribution. The value z describes the number of standard deviations between x and µ.

z x µ

Standardize theNormal Distribution

Normal Distribution

x

One table!

= 1

z


z x

= 0

Finding a Probability Corresponding to a Normal

Random Variable1. Sketch normal distribution, indicate mean, and shade

the area corresponding to the probability you want.

2. Convert the boundaries of the shaded area from x values to standard normal random variable z

z x µ

Show the z values under corresponding x values.

3. Use Table in Appendix D to find the areas corresponding to the z values. Use symmetry when necessary.

• Of course, the TI does it all: normalcdf(a,b,μ,σ) returns the probability P(a < x < b) with x distributed N(μ, σ).

Non-standard Normal μ = 5, σ = 10: P(5 < x < 6.2)

z

= 1

.12


.0478

Normal Distribution

x = 5

= 10

6.2 = 0

z

x

6.2 5

10.12

Non-standard Normal μ = 5, σ = 10: P(3.8 x 5)

z = 0-.12

Standardized Normal Distribution

Normal Distribution

x = 53.8

.0478

z

x

3.8 5

10 .12

= 10 = 1

Non-standard Normal μ = 5, σ = 10: P(2.9 x 7.1)

0-.21 z.21


52.9 7.1 x

Normal Distribution

.1664

.0832.0832

z

x

2.9 5

10 .21

z

x

7.1 5

10.21

= 10 = 1

Non-standard Normal μ = 5, σ = 10: P(x 8)

x = 5

= 10

8

Normal Distribution

= 1

z = 0 .30


.3821.5000

.1179

z

x

8 5

10.30

Non-standard Normal μ = 5, σ = 10: P(7.1 X 8)

= 1

.30 z.21


= 10

87.1 x

Normal Distribution

.1179 .0347.0832

z

x

7.1 5

10.21

z

x

8 5

10.30

= 5 = 0

Normal Distribution Thinking Challenge

You work in Quality Control for GE. Light bulb life has a normal distribution with = 2000 hours and = 200 hours. What’s the probability that a bulb will last

A. between 2000 and 2400 hours?

B. less than 1470 hours?

Solution* P(2000 x 2400)


z = 0 2.0

Normal Distribution

x= 2000

= 200

2400

= 1

.4772

z

x

2400 2000

2002.0

Solution* P(x 1470)

z= 0

= 1

–2.65


x = 2000

= 200

1470

Normal Distribution

.0040 .4960

.5000

z

x

1470 2000

200 2.65

Finding z-Values for Known Probabilities

What is z, given P(z) = .1217?

z= 0

= 1

•?

.1217

Standard Normal Probability Table (Portion)

Z .00 0.2

0.0 •.0000 .0040 .0080

0.1 .0398 .0438 .0478

0.2 .0793 .0832 .0871

.1179 .1255

.01

0.3 .1217

.31

Finding x Values for Known Probabilities

Normal Distribution

•x= 5

= 10

•?

.1217


z= 0

= 1

.31

.1217

8.1

x z 5 .31 10

Using a TI: invNorm(p, μ, σ) returns the 100pth percentile of N(μ, σ).

Normal Distribution Thinking Challenge

At one university, the students are given z-scores at the end of each semester rather than the traditional GPAs. The mean and the standard deviation of all students’ cumulative GPAs, on which the z-scores are based, are 2.7 and .5, respectively.

a) Translate each of the following z-scores to the corresponding GPA: z=2.0, z=-1, z=0.5, z=-2.5.

b) Students with z-scores below -1.6 are put on probation. What is the corresponding probationary GPA?

c) The president of the university wishes to graduate the top 20% of the students with cum laude honors and the top 2.5% with summa cum laude honors. Under the assumption that the distribution is exactly normal, by using the Table Z in the appendix, determine the limits be set in terms of original GPAs.

•

Solution

a) 3.7, 2.2, 2.95, 1.45

b) 1.9

c)P(z > 0.84)=0.20 P(z > 1.96)=0.025

So,

for cum laude: 0.84=(x-2.7)/0.5x= 3.12;

for summa cum laude: 1.96=(x-2.7)/0.5

x= 3.68

4.7

Descriptive Methods forAssessing Normality

Determining Whether the Data Are from an Approximately

Normal Distribution1. Construct either a histogram or stem-and-leaf

display for the data and note the shape of the graph. If the data are approximately normal, the shape of the histogram or stem-and-leaf display will be similar to the normal curve.


Normal Distribution

2. Compute the intervals and determine the percentage of measurements falling in each. If the data are approximately normal, the percentages will be approximately equal to 68%, 95%, and 100%, respectively; from the Empirical Rule (68%, 95%, 99.7%).

x s, x 2s, and x 3s,


Normal Distribution

3. Find the interquartile range, IQR, and standard deviation, s, for the sample, then calculate the ratio IQR/s. If the data are approximately normal, then IQR/s ≈ 1.3.

3 1Q QIQR

s s


Normal Distribution

4. Examine a normal probability plot for the data. If the data are approximately normal, the points will fall (approximately) on a straight line.

Observed value

Exp

ecte

d z

–sco

re

Normal Probability Plot

A normal probability plot for a data set is a scatterplot with the ranked data values on one axis and their corresponding expected z-scores from a standard normal distribution on the other axis. [Note: Computation of the expected standard normal z-scores are beyond the scope of this lecture. Therefore, we will rely on available statistical software packages to generate a normal probability plot.]

Using Stat Cranch for Normal probability plot

4.8

Other Continuous Distributions:

Uniform

Uniform Distribution•Continuous random variables that appear to have equally likely outcomes over their range of possible values possess a uniform probability distribution.•Suppose the random variable x can assume values only in an interval c ≤ x ≤ d. Then the uniform frequency function has a rectangular shape.

Probability Distribution for a Uniform Random Variable x

Mean:

f (x)

1

d cc x dProbability density function:

d c

12

c d

2Standard Deviation:

P a x b b a d c , c a b d

Uniform Distribution Example

You’re production manager of a soft drink bottling company. You believe that when a machine is set to dispense 12 oz., it really dispenses between 11.5 and 12.5 oz. inclusive. Suppose the amount dispensed has a uniform distribution. What is the probability that less than 11.8 oz. is dispensed?

SODA

Uniform Distribution Solution

P(11.5 x 11.8) = (Base)/(Height)

= (11.8 – 11.5)/(1) = .30

11.5 12.5

f(x)

•x11.8

1 1

12.5 11.51

1.01

d c

1.0

• Consider a powder specimen that has exactly 10 anthrax spores. Suppose that the number of anthrax spores in the sample detected by the new method follows an approximate uniform distribution between 0 and 10.

a. Find the probability that 8 or fewer anthrax spores are detected in the powder specimen.

b. Find the probability that between 2 and 5 anthrax spores are detected in the powder specimen.

c. Find the expected number of anthrax spores.

d. Find the standard deviation of anthrax spores.

Review Q1

Review Q2

• The output from a statistical computer program indicates that the mean and standard deviation of a data set consisting of 200 measurements are $1500 and $300, respectively. Suppose the frequency distribution of the data set has normal distribution. What of percentage of data will be between $750 and $1200?

Review Q3• Almost all companies utilize some type of year-end performance review

for their employees. Human Resources(HR) at the university of Texas health Science Center provides guidelines for supervisors rating their subordinates. For example raters are advised to examine their ratings for a tendency to be too lenient or too harsh. According to HR, “if you have this tendency, consider using a normal distribution-10% of employees (rated) exemplary, 20% distinguished, 40% competent, 20% marginal, and 10% unacceptable”.

• Suppose you are rating an employee’s performance on a scale of 1(lowest) to 100(highest). Also assume the ratings follow normal distribution with a mean of 50 and a standard deviation of 15.

a) What is the lowest rating you should give to an “exemplary” employee if you follow the Univ. of Texas HR guidelines?

b) What is the lowest rating you should give to an “competent” employee if you follow the Univ. of Texas HR guidelines?

Review Q4

• Suppose that 5 out of 13 liver transplants done at a hospital will fail within a year. Consider a random sample of 4 of these 13 patients. What is the probability that 1 of these patients will result in failed transplants?

Key IdeasProperties of Probability Distributions

Discrete Distributions

1. p(x) ≥ 0

2.

Continuous Distributions

1. P(x = a) = 0

2. P(a < x < b) = area under curve between a and b

p x 1all x

Key Ideas

Methods for Assessing Normality

1. Histogram

Key Ideas


2. Stem-and-leaf display

1 7

2 3389

3 245677

4 19

5 2

Key Ideas


3. (IQR)/S ≈ 1.3

4. Normal probability plot

Your 4 th homework is assigned. It is due on 12 th of Feb, 11:59 pm.

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