STA 218: Statistics for Management and Economicsmath.unm.edu/~alvaro/sta218-p1.pdf · Picturing distributions with graphs. Describing distributions with numbers. Normal distributions

Picturing distributions with graphs.Describing distributions with numbers.

Normal distributions.Introduction to Probability

Continuous probabiilty distributions.Sampling Distributions

STA 218: Statistics for Management andEconomics

Al Nosedal.University of Toronto

Fall 2014

Al Nosedal. University of Toronto STA 218: Statistics for Management and Economics




1 Picturing distributions with graphs.

2 Describing distributions with numbers.

3 Normal distributions.

4 Introduction to Probability

5 Continuous probabiilty distributions.

6 Sampling Distributions





My momma always said: ”Life was like a box of chocolates. Younever know what you’re gonna get.”

Forrest Gump.





Definitions

Statistics is the science of data.

Individuals are the objects described by a set of data.Individuals may be people, but they may also be animals orthings.

A variable is any characteristic of an individual. A variable cantake different values for different individuals.





Definitions








Definitions








Descriptive Statistics

Most of the statistical information in newspapers, magazines,company reports, and other publications consists of data that aresummarized and presented in a form that is easy for the reader tounderstand. Such summaries of data, which may be tabular,graphical, or numerical, are referred to as descriptive statistics.





Statistical Inference

Many situations require information about a large group ofelements. But, because of time, cost, and other considerations,data can be collected from only a small portion of the group. Thelarger group of elements in a particular study is called thepopulation, and the smaller group is called the sample. As one ofits major contributions, Statistics uses data from a sample to makeestimates and test hypotheses about the characteristics of apopulation through a process referred to as statistical inference.





Categorical and Quantitative Variables

A categorical variable places an individual into one of severalgroups or categories.

A quantitative variable takes numerical values for whicharithmetic operations such as adding and averaging makessense. The values of a quantitative variable are usuallyrecorded in a unit of measurement such as seconds orkilograms.





Categorical and Quantitative Variables

A categorical variable places an individual into one of severalgroups or categories.

A quantitative variable takes numerical values for whicharithmetic operations such as adding and averaging makessense. The values of a quantitative variable are usuallyrecorded in a unit of measurement such as seconds orkilograms.





Example. Fuel economy

Here is a small part of a data set that describes the fuel economy(in miles per gallon) of model year 2010 motor vehicles:

Make and Model Type Transmission Cylinders

Aston Martin Vantage Two-seater Manual 8Honda Civic Subcompact Automatic 4Toyota Prius Midsize Automatic 4

Chevrolet Impala Large Automatic 6





Example. Fuel economy (cont.)

Here is a small part of a data set that describes the fuel economy(in miles per gallon) of model year 2010 motor vehicles:

Make and Model City mpg Highway mpg Carbon footprint

Aston Martin Vantage 12 19 13.1Honda Civic 25 36 6.3Toyota Prius 51 48 3.7

Chevrolet Impala 18 29 8.3

The carbon footprint measures a vehicle’s impact on climatechange in tons of carbon dioxide emitted annually.a) What are the individuals in this data set?b) For each individual, what variables are given? Which of thesevariables are categorical and which are quantitative?





Fuel economy (solution)

a) The individuals are the car makes and models.b) For each individual, the variables recorded are Vehicle Type(categorical), Transmission Type (categorical), Number of cylinders(quantitative), City mpg (quantitative), Highway mpg(quantitative), and Carbon footprint (tons, quantitative).





Distribution of a variable

The distribution of a variable tells us what values it takes and howoften it takes these values.The values of a categorical variable are labels for the categories.The distribution of a categorical variable lists the categories andgives either the count or the percent of individuals that fall in eachcategory.





Example. Never on Sunday?

Births are not, as you might think, evenly distributed across thedays of the week. Here are the average numbers of babies born oneach day of the week in 2008:

Day Births

Sunday 7,534Monday 12,371Tuesday 13,415

Wednesday 13,171Thursday 13,147

Friday 12,919Saturday 8,617





Example. Never on Sunday? (cont.)

Present these data in a well-labeled bar graph. Would it also becorrect to make a pie chart? Suggest some possible reasons whythere are fewer births on weekends.Solution.It would be correct to make a pie chart but a pie chart would makeit more difficult to distinguish between the weekend days and theweekdays. Some births are scheduled (e.g., induced labor), andprobably most are scheduled for weekdays.





Example. Never on Sunday? Bar chart.

Sun Mon Tue Wed Thu Fri Sat

Births

02000

4000

6000

8000

10000

12000

14000





Example. Never on Sunday? Bar chart.

R code to make a bar chart:

# Step 1. Entering Data;

births=c(7534,12371,13415,13171,13147,12919,8617);

names=c("Sun","Mon","Tue","Wed","Thu","Fri","Sat");

# Step 2. Making bar chart;

barplot(births,names.arg=names,ylab="Births");





Example. Never on Sunday? Pie chart.

Sun

MonTue

Wed

ThuFri

Sat





Example. Never on Sunday? Pie chart.

R code to make a pie chart:


births=c(7534,12371,13415,13171,13147,12919,8617);

names=c("Sun","Mon","Tue","Wed","Thu","Fri","Sat");

# Step 2. Making pie chart;

pie(births,names,col=c(1:7));





Example. What color is your car?

The most popular colors for cars and light trucks vary by regionand over time. In North America white remains the top colorchoice, with black the top choice in Europe and silver the topchoice in South America. Here is the distribution of the top colorsfor vehicles sold globally in 2010.

Color Popularity (%)

Silver 26Black 24White 16Gray 16Red 6Blue 5

Beige, brown 3Other colors





What color is your car? (cont.)

a) Fill in the percent of vehicles that are in other colors.b) Make a graph to display the distribution of color popularity.





Solution

a) Other = 100− (26 + 24 + 16 + 16 + 6 + 5 + 3) = 4.





Graph

silver black white gray red blue brown other

Popularity

05

1015

2025





Example. What color is your car? Bar chart.

R code to make a bar chart:


popularity=c(26,24,16,16,6,5,3,4);

color=c("silver","black","white","gray",

"red","blue","brown","other");

# Step 2. Making bar chart;

barplot(popularity,names.arg=color,ylab="Popularity");





Summarizing Quantitative Data

A common graphical representation of quantitative data is ahistogram. This graphical summary can be prepared for datapreviously summarized in either a frequency, relative frequency, orpercent frequency distribution. A histogram is constructed byplacing the variables of interest on the horizontal axis and thefrequency, relative frequency, or percent frequency on the verticalaxis.





Example

Consider the following data14 21 23 21 16 19 22 25 16 1624 24 25 19 16 19 18 19 21 1216 17 18 23 25 20 23 16 20 1924 26 15 22 24 20 22 24 22 20.a. Develop a frequency distribution using classes of 12-14, 15-17,18-20, 21-23, and 24-26.b. Develop a relative frequency distribution and a percentfrequency distribution using the classes in part (a).c. Make a histogram.





Example (solution)

Class Frequency Relative Freq. Percent Freq.

12 -14 2 2/40 0.0515 - 17 8 8/40 0.2018 - 20 11 11/40 0.27521 - 23 10 10/40 0.2524 - 26 9 9/40 0.225





Modified classes (solution)

Class Frequency Relative Freq. Percent Freq.

12 ≤ x < 15 2 2/40 0.0515 ≤ x < 18 8 8/40 0.2018 ≤ x < 21 11 11/40 0.27521 ≤ x < 24 10 10/40 0.2524 ≤ x < 27 9 9/40 0.225





Histogram of Frequencies

Histogram of data

data

Frequency

15 20 25

02

46

810

12

2

8

11

10

9





Histogram of Frequencies (R code)


data=c(14,21,23,21,16,19,22,25,16,16,

24,24,25,19,16,19,18,19,21,12,

16,17,18,23,25,20,23,16,20,19,

24,26,15,22,24,20,22,24,22,20);

# Step 2. Making histogram;

classes=c(12,15,18,21,24,27);

hist(data,breaks=classes,col="gray",

right=FALSE,labels=TRUE);





Symmetric and Skewed Distributions

A distribution is symmetric if the right and left sides of thehistogram are approximately mirror images of each other. Adistribution is skewed to the right if the right side of the histogram(containing the half of the observations with larger values) extendsmuch farther out than the left side. It is skewed to the left if theleft side of the histogram extends much farther out than the rightside.





Symmetric Distribution

Symmetric

0 1 2 3 4 5 6

050

100

150

200





Distribution Skewed to the Right

Skewed to the right

0 2 4 6 8 10 12 14

0100

200

300





Distribution Skewed to the Left

Skewed to the left

0.4 0.5 0.6 0.7 0.8 0.9 1.0

0100

200

300





Examining a histogram

In any graph of data, look for the overall pattern and for strikingdeviations from that pattern.You can describe the overall pattern of a histogram by its shape,center, and spread.An important kind of deviation is an outlier, and individual valuethat falls outside the overall pattern.





Quantitative Variables: Stemplots

To make a stemplot:1. Separate each observation into a stem, consisting of all but thefinal (rightmost) digit, and a leaf, the final digit. Stems may haveas many digits as needed, but each leaf contains only a single digit.2. Write the stems in a vertical column with the smallest at thetop, and draw a vertical line at the right of this column.3. Write each leaf in the row to the right of its stem, in increasingorder out from the stem.





Example: Making a stemplot

Construct stem-and-leaf display (stemplot) for the following data:70 72 75 64 58 83 80 82 76 75 68 65 57 78 85 72.





Solution

5 7 86 4 5 87 0 2 2 5 5 6 88 0 2 3 5





Making a stemplot (R code)


data=c(70,72,75,64,58,83,80,82,76,75,68,65,57,78,85,72);

# Step 2. Making stem plot;

stem(data);





Health care spending.

The table below shows the 2009 health care expenditure per capitain 35 countries with the highest gross domestic product in 2009.Health expenditure per capita is the sum of public and privatehealth expenditure (in international dollars, based onpurchasing-power parity, or PPP) divided by population. Healthexpenditures include the provision of health services, for health butexclude the provision of water and sanitation. Make a stemplot ofthe data after rounding to the nearest $100 (so that stems arethousands of dollars, and leaves are hundreds of dollars). Split thestems, placing leaves 0 to 4 on the first stem and leaves 5 to 9 onthe second stem of the same value.Describe the shape, center, and spread of the distribution. Whichcountry is the high outlier?





Table

Country Dollars Country Dollars Country Dollars

Argentina 1387 India 132 Saudi Arabia 1150Australia 3382 Indonesia 99 South Africa 862Austria 4243 Iran 685 Spain 3152Belgium 4237 Italy 3027 Sweden 3690

Brazil 943 Japan 2713 Switzerland 5072Canada 4196 Korea, South 1829 Thailand 345China 308 Mexico 862 Turkey 965

Denmark 4118 Netherlands 4389 U. A. E. 1756Finland 3357 Norway 5395 U. K. 3399France 3934 Poland 1359 U. S. A. 7410

Germany 4129 Portugal 2703 Venezuela 737Greece 3085 Russia 1038





Table, after rounding to the nearest $ 100










Table, rounded to units of hundreds










Stemplot

0 1 1 3 30 7 7 9 9 91 0 0 2 4 41 8 822 7 73 0 1 2 4 4 43 7 94 1 1 2 2 2 445 1 45667 4





Shape, Center and Spread

This distribution is somewhat right-skewed, with a single highoutlier (U.S.A.). There are two clusters of countries. The center ofthis distribution is around 27 ($2700 spent per capita), ignoringthe outlier. The distribution’s spread is from 1 ($100 spent percapita) to 74 ($7400 spent per capita).





Time Plots

A time plot of a variable plots each observation against the time atwhich it was measured. Always put time on the horizontal scale ofyour plot and the variable you are measuring on the vertical scale.Connecting the data points by lines helps emphasize any changeover time.When you examine a time plot, look once again for an overallpattern and for strong deviations from the pattern. A commonoverall pattern in a time plot is a trend, a long-term upward ordownward movement over time. Some time plots show cycles,regular up-and-down movements over time.





Example. The cost of college

Below you will find data on the average tuition and fees charged toin-state students by public four-year colleges and universities forthe 1980 to 2010 academic years. Because almost any variablemeasured in dollars increases over time due to inflation (the fallingbuying power of a dollar), the values are given in ”constant dollars”adjusted to have the same buying power that a dollar had in 2010.a) Make a time plot of average tuition and fees.b) What overall pattern does your plot show?c) Some possible deviations from the overall pattern are outliers,periods when changes went down (in 2010 dollars), and periods ofparticularly rapid increase. Which are present in your plot, andduring which years?





Table

Year Tuition ($) Year Tuition ($) Year Tuition ($)

1980 2119 1991 3373 2002 49611981 2163 1992 3622 2003 55071982 2305 1993 3827 2004 59001983 2505 1994 3974 2005 61281984 2572 1995 4019 2006 62181985 2665 1996 4131 2007 64801986 2815 1997 4226 2008 65321987 2845 1998 4338 2009 71371988 2903 1999 4397 2010 76051989 2972 2000 44261990 3190 2001 4626





Time plot

1980 1985 1990 1995 2000 2005 2010

2000

3000

4000

5000

6000

7000

Time Plot of Average Tuition and Fees (1980-2010)

Year

Avera

ge Tu

ition (

$)





Time plot (R code)


data<-read.table(file="collegecost.txt",header=TRUE);

names(data);

# Step 2. Making time plot;

plot(data$year,data$tuition,type="l",col="red",

xlab="Year",ylab="Average Tuition ($)");

title("Time Plot of Average Tuition and Fees

(1980-2010)");

points(data$year,data$tuition,col="blue");





Answers to b) and c)

b) Tuition has steadily climbed during the 30-year period, withsharpest absolute increases in the last 10 years.c) There is a sharp increase from 2000 to 2010.





Problem

How much do people with a bachelor’s degree (but no higherdegree) earn? Here are the incomes of 15 such people, chosen atrandom by the Census Bureau in March 2002 and asked how muchthey earned in 2001. Most people reported their incomes to thenearest thousand dollars, so we have rounded their responses tothousands of dollars. 110 25 50 50 55 30 35 30 4 32 50 30 32 7460.How could we find the ”typical” income for people with abachelor’s degree (but no higher degree)?





Measuring center: the mean

The most common measure of center is the ordinary arithmeticaverage, or mean. To find the mean of a set of observations, addtheir values and divide by the number of observations. If the nobservations are x1, x2, ..., xn, their mean is

x̄ =x1 + x2 + ...+ xn

n

or in more compact notation,

x̄ =n∑

i=1

xi





Income Problem

x̄ = 110+25+50+50+55+30+...+32+74+6015 = 44.466

Do you think that this number represents the ”typical” income forpeople with a bachelor?s degree (but no higher degree)?





Measuring center: the median

The median M is the midpoint of a distribution, the number suchthat half the observations are smaller and the other half are larger.To find the median of the distribution:Arrange all observations in order of size, from smallest to largest.If the number of observations n is odd, the median M is the centerobservation in the ordered list. Find the location of the median bycounting n+1

2 observations up from the bottom of the list.If the number of observations n is even, the median M is the meanof the two center observations in the ordered list. Find the locationof the median by counting n+1

2 observations up from the bottom ofthe list.





Income Problem (Median)

We know that if we want to find the median, M, we have to orderour observations from smallest to largest: 4 25 30 30 30 32 32 3550 50 50 55 60 74 110. Let?s find the location of Mlocation of M = n+1

2 = 15+12 = 8

Therefore, M = x8 = 35 (x8= 8th observation on our ordered list).





Measuring center: Mode

Another measure of location is the mode. The mode is defined asfollows. The mode is the value that occurs with greatest frequency.Note: situations can arise for which the greatest frequency occursat two or more different values. In these instances more than onemode exists.





Income Problem (Mode)

Using the definition of mode, we have that:mode1 = 30andmode2 = 50Note that both of them have the greatest frequency, 3.





Mean, Median, and Mode (R code)


income<-c(110,25,50,50,55,30,35,30,4,32,50,30,32,74,60);

# Step 2. Finding Mean;

mean(income);

# Step 3. Finding Median;

median(income);

# Step 3. Finding Mode;

Table<-table(income);

sort(Table);





Example: New York travel times.

Here are the travel times in minutes of 20 randomly chosen NewYork workers:10 30 5 25 40 20 10 15 30 20 15 20 85 15 65 15 60 60 40 45.Compare the mean and median for these data. What general factdoes your comparison illustrate?





Solution

Mean:x̄ = 10+30+5+...+60+60+40+45

20 = 31.25Median:First, we order our data from smallest to largest5 10 10 15 15 15 15 20 20 20 25 30 30 40 40 45 60 60 65 85 .location of M = n+1

2 = 20+12 = 10.5

Which means that we have to find the mean of x10 and x11.M = x10+x11

2 = 20+252 = 22.5





Comparing the mean and the median

The mean and median of a symmetric distribution are closetogether. In a skewed distribution, the mean is farther out in thelong tail than is the median. Because the mean cannot resist theinfluence of extreme observations, we say that it is not a resistantmeasure of center.





The quartiles Q1 and Q3

To calculate the quartiles:Arrange the observations in increasing order and locate the medianM in the ordered list of observations.The first quartile Q1 is the median of the observations whoseposition in the ordered list is to the left of the location of theoverall median.The third quartile Q3 is the median of the observations whoseposition in the ordered list is to the right of the location of theoverall median.





Income Problem (Q1)

Data:4 25 30 30 30 32 32 35 50 50 50 55 60 74 110.From previous work, we know that M = x8 = 35.This implies that the first half of our data has n1 = 7 observations.Let us find the location of Q1:location of Q1 = n1+1

2 = 7+12 = 4.

This means that Q1 = x4 = 30.





Income Problem (Q3)

Data:4 25 30 30 30 32 32 35 50 50 50 55 60 74 110.From previous work, we know that M = x8 = 35.This implies that the first half of our data has n2 = 7 observations.Let us find the location of Q3:location of Q3 = n2+1

2 = 7+12 = 4.

This means that Q3 = 55.





Five-number summary

The five-number summary of a distribution consists of the smallestobservation, the first quartile, the median, the third quartile, andthe largest observation, written in order from smallest to largest.In symbols, the five-number summary is

min Q1 M Q3 MAX .





Income Problem (five-number summary)

Data: 4 25 30 30 30 32 32 35 50 50 50 55 60 74 110. Thefive-number summary for our income problem is given by:

4 30 35 55 110.





Income Problem (R code)


income<-c(110,25,50,50,55,30,35,30,4,32,50,30,32,74,60);

# Step 2. Finding five-number summary;

fivenum(income);





Why R gives a different five-number summary?

location of Q1 = floor((N + 3)/2)/2

location of Q3 = (N + 1)− location of Q1

where N = number of observations in our data set.





Income Problem (again...)

In this case, N = 15

location of Q1 = floor(((N + 3)/2)/2) = 4.5

location of Q3 = (N + 1)− location of Q1 = 11.5

Q1 =30 + 30

2= 30

Q3 =50 + 55

2= 52.5





Boxplot

A boxplot is a graph of the five-number summary.A central box spans the quartiles Q1 and Q3.A line in the box marks the median M.Lines extended from the box out to the smallest and largestobservations.





Boxplot for income data0

2040

6080

100

Incom

e (tho

usan

ds of

dolla

rs)





Boxplot for income data (R code)


income<-c(110,25,50,50,55,30,35,30,4,32,50,30,32,74,60);

# Step 2. Making boxplot;

boxplot(income,col="gray",range=0,ylab="Income

(thousands of dollars)");





Pittsburgh Steelers

The 2010 roster of the Pittsburgh Steelers professional footballteam included 7 defensive linemen and 9 offensive linemen. Theweights in pounds of the defensive linemen were305 325 305 300 285 280 298and the weights of the offensive linemen were338 324 325 304 344 315 304 319 318a) Make a stemplot of the weights of the defensive linemen andfind the five-number summary.b) Make a stemplot of the weights of the offensive linemen andfind the five-number summary.c) Does either group contain one or more clear outliers? Whichgroup of players tends to be heavier?





Solution

a) Defensive linemen.

28 0 529 830 0 5 53132 5





Solution

b) Offensive linemen.

30 4 431 5 8 932 4 533 834 4





Five-number summary

Minimum Q1 Median Q3 MaximumOffensive line (lbs) 304 309.5 319 331.5 344Defensive line (lbs) 280 285 300 305 325

c) Apparently, neither of these two groups contain outliers. Itseems that the offensive line players are heavier.





Measures of Variability: Range

The simplest measure of variability is the range.Range= Largest value - smallest valueRange= MAX - min





Measures of Variability: IQR

A measure of variability that overcomes the dependency onextreme values is the interquartile range (IQR).IQR = third quartile - first quartileIQR = Q3 − Q1





1.5 IQR Rule

Identifying suspected outliers. Whether an observation is an outlieris a matter of judgement: does it appear to clearly stand apartfrom the rest of the distribution? When large volumes of data arescanned automatically, however, we need a rule to pick outsuspected outliers. The most common rule is the 1.5 IQR rule. Apoint is a suspected outlier if it lies more than 1.5 IQR below thefirst quartile Q1 or above the third quartile Q3.





A high income.

In our income problem, we noted the influence of one high incomeof $110,000 among the incomes of a sample of 15 collegegraduates. Does the 1.5 IQR rule identify this income as asuspected outlier?





Solution

Data: 4 25 30 30 30 32 32 35 50 50 50 55 60 74 110.Q1 and Q3 are given by:Q1 =30 and Q3 =55Q3 + 1.5 IQR = 55 + 1.5(25) = 92.5Since 110 > 92.5 we conclude that 110 is an outlier.





Problem

Ebby Halliday Realtors provide advertisements for distinctiveproperties and estates located throughout the United States. Theprices listed for 22 distinctive properties and estates are shownhere. Prices are in thousands.1500 895 719 619 625 4450 22001280 700 619 725 739 799 24951395 2995 880 3100 1699 1120 1250 912.a)Provide a five-number summary.b)The highest priced property, $ 4,450,000, is listed as an estateoverlooking White Rock Lake in Dallas, Texas. Should thisproperty be considered an outlier?





Solution

a) min = 619Q1 = 725M = 1016Q3 = 1699MAX = 4450b) IQR = 1699 - 725 = 974Q3+1.5 IQR = 1699 + 1.5 (974) = 1699 + 1461 = 3160.Since 4450 > 3160, we conclude that 4450 is an outlier.





Measures of Variability: Variance

The variance s2 of a set of observations is an average of thesquares of the deviations of the observations from their mean. Insymbols, the variance of n observations x1, x2, ..., xn is

s2 = (x1−x̄)2+(x2−x̄)2+...+(xn−x̄)2

n−1or, more compactly,s2 = 1

n−1

∑ni=1(xi − x̄)2





Measures of Variability: Standard Deviation

The standard deviation s is the square root of the variance s2:

s =√∑n

i=1(xi−x̄)2

n−1





Example

Consider a sample with data values of 10, 20, 12, 17, and 16.Compute the variance and standard deviation.





Solution

First, we have to calculate the mean, x̄ :x̄ = 10+20+12+17+16

5 = 15.Now, let’s find the variance s2:s2 = (10−15)2+(20−15)2+(12−15)2+(17−15)2+(16−15)2

5−1 .

s2 = 644 = 16.

Finally, let’s find the standard deviation s:s =√

16 = 4.





Solution (R code)


data<-c(10, 20, 12, 17, 16);

# Step 2. Finding variance;

var(data);

# Step 2. Finding standard deviation;

sd(data);





x̄ and s

Radon is a naturally occurring gas and is the second leading causeof lung cancer in the United States. It comes from the naturalbreakdown of uranium in the soil and enters buildings throughcracks and other holes in the foundations. Found throughout theUnited States, levels vary considerably from state to state. Thereare several methods to reduce the levels of radon in your home, andthe Environmental Protection Agency recommends using one ofthese if the measured level in your home is above 4 picocuries perliter. Four readings from Franklin County, Ohio, where the countyaverage is 9.32 picocuries per liter, were 5.2, 13.8, 8.6 and 16.8.a) Find the mean step-by-step.b)Find the standard deviation step-by-step.c)Now enter the data into your calculator and use the mean andstandard deviation buttons to obtain x̄ and s. Do the results agreewith your hand calculations?Al Nosedal. University of Toronto STA 218: Statistics for Management and Economics




Solution

First, we have to calculate the mean, x̄ :x̄ = 5.2+13.8+8.6+16.8

4 = 11.1.Now, let’s find the variance s2:s2 = (5.2−11.1)2+(13.8−11.1)2+(8.6−11.1)2+(16.8−11.1)2

4−1 .

s2 = 80.843 = 26.9466

Finally, let’s find the standard deviation s:s =√

26.9466 = 5.1910.





Choosing a summary

The five-number summary is usually better than the mean andstandard deviation for describing a skewed distribution or adistribution with strong outliers. Use x̄ and s only for reasonablysymmetric distributions that are free of outliers.





Simple Example

Random Experiment: Rolling a fair die 300 times.

Class Expected Frequency Expected Relative Freq

1 ≤ x < 2 50 1/62 ≤ x < 3 50 1/63 ≤ x < 4 50 1/64 ≤ x < 5 50 1/65 ≤ x < 6 50 1/66 ≤ x < 7 50 1/6





Histogram of Expected Frequencies

1 2 3 4 5 6 7

010

2030

4050

Histogram of expected frequencies

frequency





Histogram of Expected Relative Frequencies

1 2 3 4 5 6 7

0.00

0.05

0.10

0.15

Histogram of expected relative frequencies

frequency





Density Curve

A density curve is a curve that is always on or above the horizontalaxis, and has area exactly 1 underneath it.A density curve describes the overall pattern of a distribution. Thearea under the curve and above any range of values is theproportion of all observations that fall in that range.Note. No set of real data is exactly described by a density curve.The curve is an idealized description that is easy to use andaccurate enough for practical use.





Accidents on a bike path

Examining the location of accidents on a level, 5-mile bike pathshows that they occur uniformly along the length of the path. Thefigure below displays the density curve that describes thedistribution of accidents.a) Explain why this curve satisfies the two requirements for adensity curve.b) The proportion of accidents that occur in the first mile of thepath is the area under the density curve between 0 miles and 1mile. What is this area?c) There is a stream alongside the bike path between the 0.8-milemark and the 1.3-mile mark. What proportion of accidents happenon the bike path alongside the stream?d) The bike path is a paved path through the woods, and there isa road at each end. What proportion of accidents happen morethan 1 mile from either road?Al Nosedal. University of Toronto STA 218: Statistics for Management and Economics




Density Curve

0 1 2 3 4 5

0.00

0.05

0.10

0.15

0.20

Density Curve

Distance along bike path (miles)





Solution

a) It is on or above the horizontal axis everywhere, and because itforms a 1/5 × 5 rectangle, the area beneath the curve is 1.





Solution b)

0 1 2 3 4 5

0.00

0.05

0.10

0.15

0.20

Density Curve


proportion = 1 x 0.20 = 0.20





Solution c)

0 1 2 3 4 5

0.00

0.05

0.10

0.15

0.20

Density Curve


proportion = (1.3-0.8) x 0.20 = 0.10





Solution d)

0 1 2 3 4 5

0.00

0.05

0.10

0.15

0.20

Density Curve


proportion = (4-1) x 0.20 = 0.60





Normal Distributions

A Normal Distribution is described by a Normal density curve. Anyparticular Normal distribution is completely specified by twonumbers, its mean µ and standard deviation σ.The mean of a Normal distribution is at the center of thesymmetric Normal curve. The standard deviation is the distancefrom the center to the change-of-curvature points on either side.





Standard Normal Distribution

-3 -2 -1 0 1 2 3

0.00.1

0.20.3

0.4

Normal Distributionmean=0 and standard deviation=1





Two Different Standard Deviations

-15 -10 -5 0 5 10 15

0.00

0.05

0.10

0.15

0.20

std. dev.= 2std. dev.= 5





Two Different Means

-15 -10 -5 0 5 10 15

0.00

0.05

0.10

0.15

0.20

mean = -5mean = 5





The 68-95-99.7 rule

In the Normal distribution with mean µ and standard deviation σ:Approximately 68% of the observations fall within σ of the mean µ.Approximately 95% of the observations fall within 2σ of µ.Approximately 99.7% of the observations fall within 3σ of µ.





Problem

The national average for the verbal portion of the College Board’sScholastic Aptitude Test (SAT) is 507. The College Boardperiodically rescales the test scores such that the standarddeviation is approximately 100. Answer the following questionsusing a bell-shaped distribution and the empirical rule for theverbal test scores.a. What percentage of students have an SAT verbal score greaterthan 607?b. What percentage of students have an SAT verbal score greaterthan 707?c. What percentage of students have an SAT verbal score between407 and 507?d. What percentage of students have an SAT verbal score between307 and 707?





Solution a)

200 400 600 800

0.000

0.001

0.002

0.003

0.004

SAT score

16 %68 %16 %





Solution b)

200 400 600 800

0.000

0.001

0.002

0.003

0.004

SAT score

2.5 %95 %2.5 %





Solution c)

200 400 600 800

0.000

0.001

0.002

0.003

0.004

SAT score

34 % 34 %16 % 16 %





Solution d)

200 400 600 800

0.000

0.001

0.002

0.003

0.004

SAT score

95 %2.5 % 2.5 %





Fruit flies

The common fruit fly Drosophila melanogaster is the most studiedorganism in genetic research because it is small, easy to grow, andreproduces rapidly. The length of the thorax (where the wings andlegs attach) in a population of male fruit flies is approximatelyNormal with mean 0.800 millimeters (mm) and standard deviation0.078 mm. Draw a Normal curve on which this mean and standarddeviation are correctly located.





Solution

0.5 0.6 0.7 0.8 0.9 1.0 1.1

01

23

45

Thorax length

0.800-0.078 0.800+0.078





Fruit flies

The lenght of the thorax in a population of male fruit flies isapproximately Normal with mean 0.800 mm and standarddeviation 0.078 mm. Use the 68-95-99.7 rule to answer thefollowing questions.a) What range of lengths covers almost all (99.7%) of thisdistribution?b) What percent of male fruit flies have a thorax length exceeding0.878 mm?





Solution a) Between 0.566 mm and 1.034 mm

0.5 0.6 0.7 0.8 0.9 1.0 1.1

01

23

45

Thorax length

99.7 %0.800-3(0.078)=0.566 0.800+3(0.078)=1.034





Solution b) 16% of thorax lenghts exceed 0.878 mm

0.5 0.6 0.7 0.8 0.9 1.0 1.1

01

23

45

Thorax length

84 % 16 %





Monsoon rains

The summer monsoon brings 80% of India’s rainfall and isessential for the country’s agriculture. Records going back morethan a century show that the amount of monsoon rainfall variesfrom the year according to a distribution that is approximatelyNormal with mean 582 mm and standard deviation 82 mm. Usethe 68-95-99.7 rule to answer the following questions.a) Between what values do the monsoon rains fall in 95% of allyears?b) How small are the monsoon rains in the dryest 2.5% of all years?





Solution

a) In 95% of all years, monsoon rain levels are between582 - 2(82) and 582 + 2(82) i.e. 688 mm and 1016 mm.b) The driest 2.5% of monsoon rainfalls are less than 688 mm; thisis more than two standard deviations below the mean.





Standard Normal Distribution

The standard Normal distribution is the Normal distribution N(0,1)with mean 0 and standard deviation 1.If a variable x has any Normal distribution N(µ,σ) with mean µand standard deviation σ, then the standardized variable

z =x − µσ

has the standard Normal distribution.





SAT vs ACT

In 2010, when she was a high school senior, Alysha scored 670 onthe Mathematics part of the SAT. The distribution of SAT Mathscores in 2010 was Normal with mean 516 and standard deviation116. John took the ACT and scored 26 on the Mathematicsportion. ACT Math scores for 2010 were Normally distributed withmean 21.0 and standard deviation 5.3. Find the standardizedscores for both students. Assuming that both tests measure thesame kind of ability, who had the higher score?





Solution

Alysha’s standardized score is

zA =670− 516

116= 1.33.

John’s standardized score is

zJ =26− 21

5.3= 0.94.

Alysha’s score is relatively higher than John’s.





Men’s and women’s heights

The heights of women aged 20 to 29 are approximately Normalwith mean 64.3 inches and standard deviation 2.7 inches. Men thesame age have mean height 69.9 inches with standard deviation3.1 inches. What are the z-scores for a woman 6 feet tall and aman 6 feet tall? Say in simple language what information thez-scores give that the original nonstandardized heights do not.





Solution

We need to use the same scale, so recall that 6 feet = 72 inches.A woman 6 feet tall has standardized score

zW =72− 64.3

2.7= 2.85

(quite tall, relatively).A man 6 feet tall has standardized score

zM =72− 69.9

3.1= 0.68.

Hence, a woman 6 feet tall is 2.85 standard deviations taller thanaverage for women. A man 6 feet tall is only 0.68 standarddeviations above average for men.





Using the Normal table

Use table A to find the proportion of observations from a standardNormal distribution that satisfies each of the following statements.In each case, sketch a standard Normal curve and shade the areaunder the curve that is the answer to the question.a) z < −1.42b) z > −1.42c) z < 2.35d) −1.42 < z < 2.35





Solution a) 0.0778

-3 -2 -1 0 1 2 3

0.00.1

0.20.3

0.4

0.0778

z* = -1.42





Solution b) 0.9222

-3 -2 -1 0 1 2 3

0.00.1

0.20.3

0.4

0.92220.0778

z* = -1.42





Solution c) 0.9906

-3 -2 -1 0 1 2 3

0.00.1

0.20.3

0.4

0.9906

z* = 2.35





Solution d) 0.9966 - 0.0778 = 0.9128

-3 -2 -1 0 1 2 3

0.00.1

0.20.3

0.4

0.9906-0.0778 = 0.9128

z* = 2.35z* = -1.42





Monsoon rains

The summer monsoon rains in India follow approximately a Normaldistribution with mean 852 mm of rainfall and standard deviation82 mm.a) In the drought year 1987, 697 mm of rain fell. In what percentof all years will India have 697 mm or less of monsoon rain?b) ”Normal rainfall” means within 20% of the long-term average,or between 683 and 1022 mm. In what percent of all years is therainfall normal?





Solution a)

1. State the problem. Let x be the monsoon rainfall in a givenyear. The variable x has the N(852, 82) distribution. We want theproportion of years with x ≤ 697.2. Standardize. Subtract the mean, then divide by the standarddeviation, to turn x into a standard Normal z .Hence x ≤ 697 corresponds to z ≤ 697−852

82 = −1.89.3. Use the table. From Table A, we see that the proportion ofobservations less than −1.89 is 0.0294. Thus, the answer is 2.94%.





Solution b)

1. State the problem. Let x be the monsoon rainfall in a givenyear. The variable x has the N(852, 82) distribution. We want theproportion of years with 683 < x < 1022.2. Standardize. Subtract the mean, then divide by the standarddeviation, to turn x into a standard Normal z .683 < x < 1022 corresponds to 683−852

82 < z < 1022−85282 , or

−2.06 < z < 2.07.3. Use the table. Hence, using Table A, the area is0.9808− 0.0197 = 96.11%.





The Medical College Admission Test

Almost all medical schools in the United States require students totake the Medical College Admission Test (MCAT). The exam iscomposed of three multiple-choice sections (Physical Sciences,Verbal Reasoning, and Biological Sciences). The score on eachsection is converted to a 15-point scale so that the total score hasa maximum value of 45. The total scores follow a Normaldistribution, and in 2010 the mean was 25.0 with a standarddeviation of 6.4. There is little change in the distribution of scoresfrom year to year.a) What proportion of students taking the MCAT had a score over30?b) What proportion had scores between 20 and 25?





Solution a)

1. State the problem. Let x be the MCAT score of a randomlyselected student. The variable x has the N(25, 6.4) distribution.We want the proportion of students with x > 30.2. Standardize. Subtract the mean, then divide by the standarddeviation, to turn x into a standard Normal z .Hence x > 30 corresponds to z > 30−25

6.4 = 0.78.3. Use the table. From Table A, we see that the proportion ofobservations less than 0.78 is 0.7823. Hence, the answer is1− 0.7823 = 0.2177, or 21.77%.





Solution b)

1. State the problem. Let x be the MCAT score of a randomlyselected student. The variable x has the N(25, 6.4) distribution.We want the proportion of students with 20 ≤ x ≤ 25.2. Standardize. Subtract the mean, then divide by the standarddeviation, to turn x into a standard Normal z .20 ≤ x ≤ 25 corresponds to 20−25

6.4 ≤ z ≤ 25−256.4 , or −0.78 ≤ z ≤ 0.

3. Use the table. Using Table A, the area is0.5− 0.2177 = 0.2833, or 28.33%.





Problem

For borrowers with good credit scores, the mean debt for revolvingand installment accounts is $ 15,015. Assume the standarddeviation is $ 3540 and that debt amounts are normally distributed.a. What is the probability that the debt for a randomly selectedborrower with good credit is more than $ 18,000?b. What is the probability that the debt for a randomly selectedborrower with good credit is less than $ 10,000?c. What is the probability that the debt for a randomly selectedborrower with good credit is between $ 12,000 and $ 18,000?d. What is the probability that the debt for a randomly selectedborrower with good credit is no more than $ 14,000?





Solution

Let X = debt amount.a. P(X > 18000) = P(X−µσ > 18000−15015

3540 ) = P(Z > 0.8432) =1− P(Z < 0.8432) = 1− 0.7995 = 0.2005.b. P(X < 10000) = P(X−µσ < 10000−15015

3540 ) = P(Z < −1.4166) =0.0773.c. P(12000 < X < 18000)= P( 12000−15015

3540 < X−µσ < 18000−15015

3540 )= P(−0.8516 < Z < 0.8432)= P(Z < 0.8432)−P(Z < −0.8516) = 0.7995− 0.1977 = 0.6018.d. P(X ≤ 14000) = P(X−µσ ≤ 14000−15015

3540 ) = P(Z ≤ −0.2867) =0.3897





Using a table to find Normal proportions

Step 1. State the problem in terms of the observed variable x .Draw a picture that shows the proportion you want in terms ofcumulative proportions.Step 2. Standardize x to restate the problem in terms of astandard Normal variable z .Step 3. Use Table A and the fact that the total are under the curveis 1 to find the required area under the standard Normal curve.





Table A

Use Table A to find the value z∗ of a standard Normal variablethat satisfies each of the following conditions. (Use the value of z∗from Table A that comes closest to satisfying the condition.) Ineach case, sketch a standard Normal curve with your value of z∗marked on the axis.a) The point z∗ with 15% of the observations falling below it.b) The point z∗ with with 70% of the observations falling above it.





Solution a) z* = -1.04

-3 -2 -1 0 1 2 3

0.00.1

0.20.3

0.4

0.1492

z* = -1.04





Solution b) z* = -0.52

-3 -2 -1 0 1 2 3

0.00.1

0.20.3

0.4

0.3015 1-0.3015 = 0.6985

z* = -0.52





The Medical College Admission Test

The total scores on the Medical College Admission Test (MCAT)follow a Normal distribution with mean 25.0 and standarddeviation 6.4. What are the median and the first and thirdquartiles of the MCAT scores?





Solution: Finding the median

Because the Normal distribution is symmetric, its median andmean are the same. Hence, the median MCAT score is 25.





Solution: Finding Q1

1. State the problem. We want to find the MCAT score x witharea 0.25 to its left under the Normal curve with mean µ = 25 andstandard deviation σ = 6.4.2. Use the table. Look in the body of Table A for the entry closestto 0.25. It is 0.2514. This is the entry corresponding toz∗ = −0.67. So z∗ = −0.67 is the standardized value with area0.25 to its left.3. Unstandardize to transform the solution from the z∗ back tothe original x scale. We know that the standardized value of theunknown x is z∗ = −0.67.So x itself satisfies

x − 25

6.4= −0.67

Solving this equation for x gives x = 25 + (−0.67)(6.4) = 20.71Al Nosedal. University of Toronto STA 218: Statistics for Management and Economics




Solution: Finding Q3

1. State the problem. We want to find the MCAT score x witharea 0.75 to its left under the Normal curve with mean µ = 25 andstandard deviation σ = 6.4.2. Use the table. Look in the body of Table A for the entry closestto 0.75. It is 0.7486. This is the entry corresponding to z∗ = 0.67.So z∗ = 0.67 is the standardized value with area 0.75 to its left.3. Unstandardize to transform the solution from the z∗ back tothe original x scale. We know that the standardized value of theunknown x is z∗ = 0.67.So x itself satisfies

x − 25

6.4= 0.67

Solving this equation for x gives x = 25 + (0.67)(6.4) = 29.29





Finding a value when given a proportion

1. State the problem.2. Use the table.3. Unstandardize to transform the solution from the z∗ back tothe original x scale.





Problem

The personnel department of ZTel, a large communicationscompany, is reconsidering its hiring policy. Each applicant for a jobmust take a standard exam, and the hire or no-hire decisiondepends at least in part on the result of the exam. The scores ofall applicants have been examined closely. They are approximatelynormally distributed with mean 525 and standard deviation 55.





Problem (cont.)

The current hiring policy occurs in two phases. The first phaseseparates all applicants into three categories: automatic accepts,automatic rejects, and ”maybes”. The automatic accepts are thosewhose test scores are 600 or above. The automatic rejects arethose whose test scores are 425 or below. All other applicants (the”maybes”) are passed on to a second phase where their previousjob experience, special talents, and other factors are used as hiringcriteria.





Problem (cont.)

The personnel manager at ZTel wants to calculate the percentageof applicants who are automatic accepts or rejects, given thecurrent standards. She also wants to know how to change thestandards in order to automatically reject 10% of all applicants andautomatically accept 15% of all applicants.





Solution

Let X be the test score of a typical applicant. Then the distributionof X is N(525, 55). If we find a probability such as P(X ≤ 425),we can interpret this as the probability that a typical applicant isan automatic reject, or we can interpret it as the percentage of allapplicants who are automatic rejects. The probability that atypical applicant is automatically accepted is 0.0863. Similarly, theprobability that a typical applicant is automatically rejected is0.0345. Therefore, ZTel automatically accepts about 8.6% andrejects about 3.5% of all applicants under the current policy.





Solution

To find the new cutoff values that reject 10% and accept 15% ofthe applicants, we need the 10th and 85th percentiles of theN(525, 55) distribution. These are 455 and 582 (rounded to thenearest integer). To accomplish its objective, ZTel needs to raisethe automatic rejection point from 425 to 455 and lower theautomatic acceptance point from 600 to 582.





Solution (R code)

# 1. Automatic rejects;

pnorm(425,mean=525,sd=55);

# 2. Automatic accepts;

1-pnorm(600,mean=525,sd=55);

# 3. Bottom 10% (Automatic rejection point);

qnorm(0.10,mean=525,sd=55);

# 4. Top 15% (Automatic acceptance point);

qnorm(0.85,mean=525,sd=55);





Toy Example

Random experiment: Rolling a fair die.List of all possible outcomes = {1, 2, 3, 4, 5, 6}.Probability of rolling a 6 = 1

6 .





Definitions

The sample space S of a random phenomenon is the set of allpossible outcomes.An event is an outcome or a set of outcomes of a randomphenomenon. That is, an event is a subset of the sample space.Probability is a numerical measure of the likelihood that an eventwill occur.





Another example

In each of the following situations, describe a sample space S forthe random phenomenon.a. A basketball player shoots three free throws. You record thesequence of hits and misses.b. A basketball player shoots three free throws. You record thenumber of baskets she makes.





Solution

H=hit and M=miss.a. S={(H,H,H),(H,H,M),(H,M,H),(H,M,M),(M,H,H),(M,H,M),(M,M,H),(M,M,M)}.

b. S={0,1,2,3}.





Problem

How many teams consisting of two individuals can be selectedfrom a group of five individuals?





Solution

{A,B}, {A,C}, {A,D}, {A,E}, {B,C}, {B,D}, {B,E}, {C,D},{C,E}, {D,E}.

10 teams.





Combinations

The number of combinations of N objects taken n at a time is

CNn =

N!

n!(N − n)!

whereN! = N(N − 1)(N − 2)...(2)(1)n! = n(n − 1)(n − 2)...(2)(1)and by definition, 0! = 1.





Problem

How many different committees consisting of a president and asecretary can be selected from a group of five individuals?





Solution

(A,B),(A,C),(A,D),(A,E) (B,A),(B,C),(B,D),(B,E)(C,A),(C,B),(C,D),(C,E) (D,A),(D,B),(D,C),(D,E)(E,A),(E,B),(E,C),(E,D).

20 committees.





Permutations

The number of permutations on N objects taken n at a time isgiven by

PNn =

N!

(N − n)!





Problem

How many ways can three items be selected from a group of sixitems?





Problem

How many permutations of three items can be selected from agroup of six?





Problem

The Powerball lottery is played twice each week in 28 states, theVirgin Islands, and the District of Columbia. To play Powerball aparticipant must purchase a ticket and then select five numbersfrom the digits 1 through 55 and a Powerball number from thedigits 1 through 42. To determine the winning numbers for eachgame, lottery officials draw five white balls out of a drum with 55white balls, and one red ball out of a drum with 42 red balls. Towin the jackpot, a participant?s numbers must match the numberson the five white balls in any order and the number on the redPowerball.





Problem (cont.)

Eight coworkers at the ConAgra Foods plant in Lincoln, Nebraska,claimed the record $365 million jackpot on February 18, 2006, bymatching the numbers 15-17-43-44-49 and the Powerball number29. A variety of other cash prizes are awarded each time the gameis played. For instance, a prize of $ 200,000 is paid if theparticipant?s five numbers match the numbers on the five whiteballs.a. Compute the number of ways the first five numbers can beselected.b. What is the probability of winning a prize of $200,000 bymatching the numbers on the five white balls? c. What is theprobability of winning the Powerball jackpot?





Solution

a. 55!5!(55?5)! = 3, 478, 761.

b. 13,478,761

c. Number of choices = (3, 478, 761)(42) = 146, 107, 962.Probability of winning = 1

146,107,962 .





Problem

Consider the experiment of rolling a fair pair of dice. Suppose thatwe are interested in the sum of the face values showing on the dice.a. How many sample points are possible?b. List the sample points.c. What is the probability of obtaining a value of 7?d. What is the probability of obtaining a value of 9 or greater?e. Because each roll has six possible even values {2,4,6,8,10 and12} and only five possible odd values {3,5,7,9, and 11}, the diceshould show even values more often than odd values. Do you agreewith this statement? Explain.





Solution

a. 36.b. (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)(2,1),(2,2),(2,3),(2,4),(2,5),(2,6) (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)(4,1),(4,2),(4,3),(4,4),(4,5),(4,6) (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)(6,1),(6,2),(6,3),(6,4),(6,5),(6,6).c. P(7) = P{(6, 1)or(5, 2)or(4, 3)or(3, 4)or(2, 5)or(1, 6)} = 6

36 .d. P(9 or greater) = 10

36 .e. P(even) = P(2) + P(4) + P(6) + P(8) + P(10) + P(12) = 18

36 .P(odd) = P(1) + P(3) + P(5) + P(7) + P(9) + P(11) = 18

36 .





Colors of M & M’s

If you draw an M & M candy at random from a bag of the candies,the candy you draw will have one of the seven colors. Theprobability of drawing each color depends on the proportion ofeach color among all candies made. Here is the distribution formilk chocolate M & M’s:

Color Purple Yellow Red Orange Brown Green Blue

Probability 0.2 0.2 0.2 0.1 0.1 0.1 ?





Colors of M & M’s (cont.)

a) What must be the probability of drawing a blue candy?b) What is the probability that you do not draw a brown candy?c) What is the probability that the candy you draw is either yellow,orange, or red?





Solution

a. Probability of Blue = P(Blue) = 1 - 0.9 = 0.1b. P(Not Brown) = 1 - P(Brown) = 1-0.1 = 0.9c. P(Yellow or Orange or Red) = 0.2 + 0.1 + 0.2 = 0.5





Probability Rules

1. The probability P(A) of any event A satisfies 0 ≤ P(A) ≤ 1.2. If S is the sample space in a probability model, then P(S) = 1.3. For any event A, P(Adoes not occur) = 1?P(A).4. Two events A and B are disjoint if they have no outcomes incommon and so can never occur simultaneously. If A and B aredisjoint, P(A or B) = P(A) + P(B). This is the addition rule fordisjoint events.





Problem

A survey of magazine subscribers showed that 45.8% rented a carduring the past 12 months for business reasons, 54% rented a carduring the past 12 months for personal reasons, and 30% rented acar during the past 12 months for both business and personalreasons.a. What is the probability that a subscriber rented a car during thepast 12 months for business or personal reasons?b. What is the probability that a subscriber did not rent a carduring the past 12 months for either business or personal reasons?





Solution

B = rented a car for business reasons.0 = rented a car for personal reasons.a. Probability of B or O =P(B ∪ O) = 0.458 + 0.54− 0.30 = 0.698.b. P(Neither) = P((B ∪ O)c) = 1− P(B ∪ O) = 1− 0.698.





Addition Law

P(A ∪ B) = P(A) + P(B)− P(A ∩ B)





Conditional probability

Steve and Al are avid tennis players and they enjoy playingmatches against each other. They do, however, have one differenceof opinion on the court. Al likes to have a nice long warm-upsession at the start where they hit the ball back and forth and backand forth. Josh?s ideal warm-up is to bend at the waist to tie hissneakers and to adjust his shorts. Al thinks that when they rushthrough the warm-up, he doesn?t play as well.





Conditional probability (cont.)

The following table shows the outcomes of their last 20 matches,along with the type of warm-up before they started keeping score.Does the type of warm-up have an influence on the outcome of amatch?

Warm-up time Al wins Steve wins Total

Less than 10 min. 4 9 1310 min. or more 5 2 7

Total 9 11 20





Solution

Event A = Al wins the match.Event B = the warm-up time is less than 10 minutes.Event Bc = the warm-up time is 10 minutes or more.P(A given B) = 4

13 ≈ 0.31P(A given Bc) = 5

7 ≈ 0.71It seems that warm-up time has an effect on the outcome of amatch.





Another solution

P(A ∩ B) = 420 P(B) = 13

20 .

P(A∩B)P(B) = 4/20

13/20 = 413

P(A ∩ Bc) = 520 P(Bc) = 7

20 .

P(A∩Bc )P(B) = 5/20

7/20 = 517





Conditional Probability

P(A|B) =P(A ∩ B)

P(B).

P(B|A) =P(A ∩ B)

P(A).





Independent Events

Two events A and B are independent if

P(A|B) = P(A)

or

P(B|A) = P(B).

Otherwise, the events are dependent.





Multiplication Law

P(A ∩ B) = P(B)P(A|B)

P(B ∩ A) = P(A)P(B|A)

Multiplication Law for Independent Events

P(A ∩ B) = P(A)P(B)





Example

Suppose that we have two events, A and B, with P(A) = 0.50,P(B) = 0.60, and P(A ∩ B) = 0.40.a. Find P(A|B).b. Find P(B|A).c. Are A and B independent? Why or why not?





Solution

a. P(A|B) = P(A∩B)P(B) = 0.40

0.60 ≈ 0.66

b. P(B|A) = P(B∩A)P(A) = 0.40

0.50 ≈ 0.80

c. No, P(A ∩ B) = 0.4 6= P(A)P(B) = (0.5)(0.6) = 0.3





Problem

A Morgan Stanley Consumer Research Survey sampled men andwomen and asked whether they preferred to drink plain bottledwater or a sports drink such as Gatorade or Propel Fitness water.Suppose 200 men and 200 women participated in the study, and280 reported they preferred plain bottled water. Of the grouppreferring a sports drink, 80 were men and 40 were women. LetM=the event the consumer is a man W=the event the consumer isa woman B= the event the consumer preferred plain bottled waterS= the event the consumer preferred sports drink.





Problem (cont.)

a. What is the probability a person in the study preferred plainbottled water?b. What is the probability a person in the study preferred a sportsdrink?c. What are the conditional probabilities P(M|S) and P(W |S)?d. What are the joint probabilities P(M ∩ S) and P(W ∩ S)?e. Given a consumer is a man, what is the probability he will prefera sports drink?f. Given a consumer is a woman, what is the probability she willprefer a sports drink?g. Is preference for a sports drink independent of whether theconsumer is a man or a woman?Explain using probability information.





Solution

M W Total

B 120 160 280S 80 40 120

Total 200 200 400





Solution

a. P(B) = 280400 = 0.70

b. P(S) = 120400 = 0.30

c. P(M|S) = P(M∩S)P(S) = 0.2

0.3 = 0.66

d. P(M ∩ S) = 0.2 P(W ∩ S) = 0.1

e. P(S |M) = P(S∩M)P(M) = 0.2

0.5 = 0.40

f. P(S |W ) = P(S∩W )P(W ) = 0.1

0.5 = 0.20

g. P(S |M) 6= P(S).





Problem

A local bank reviewed its credit card policy with the intention ofrecalling some of its credit cards. In the past approximately 5% ofcardholders defaulted, leaving the bank unable to collect theoutstanding balance. Hence, management established a priorprobability of 0.05 that any particular cardholder will default. Thebank also found that the probability of missing a monthly paymentis 0.20% for customers who do not default. Of course, theprobability of missing a monthly payment for those who default is1.





Problem (cont.)

a. Given that a customer missed one or more monthly payments,compute the posterior probability that the customer will default.b. The bank would like to recall its card if the probability that acustomer will default is greater than 0.20. Should the bank recallits card if the customer misses a monthly payment? Why or whynot?





Solution

D = Default, Dc = customer doesn’t default, M = missedpayment.a. P(D|M) = P(D∩M)

P(M) = P(D∩M)P(D∩M)+P(Dc∩M) .

P(D) = 0.05 P(Dc) = 0.95 P(M|Dc) = 0.20 P(M|D) = 1

P(D|M) = (0.05)(1)(0.05)(1)+(0.95)(0.20) = 0.2083

b. Yes, the bank should recall its card if the customer misses amonthly payment.





Example

Suppose that 80 percent of used car buyers are good credit risks.Suppose, further, that the probability is 0.7 that an individual whois a good credit risk has a credit card, but that this probability isonly 0.4 for a bad credit risk. Calculate the probabilitya. a randomly selected car buyer has a credit card.b. a randomly selected car buyer who has a credit card is a goodrisk.c. a randomly selected car buyer who does not have a credit cardis a good risk.





Solution

a. A = selecting a good credit risk.B = selecting an individual with a credit card.P(B) = P(A)P(B|A) + P(Ac)P(B|Ac)P(B) = (0.8)(0.7) + (0.2)(0.4) = 0.64

b. P(A|B) = P(A∩B)P(B) = P(A)P(B|A)

P(B) = (0.8)(0.7)0.64 = 7

8

c. P(A|Bc) = P(A∩Bc )P(Bc ) = P(A)P(Bc |A)

P(Bc ) = (0.8)(0.3)0.36 = 2

3





Bayes’ Theorem

P(A1|B) =P(A1)P(B|A1)

P(A1)P(B|A1) + P(A2)P(B|A2)

P(A2|B) =P(A2)P(B|A2)

P(A1)P(B|A1) + P(A2)P(B|A2)

where A1 ∪ A2 = S and A1 ∩ A2 = ∅.





Uniform Probability Distribution

Uniform probability density function.

f (x) =

{1

b−a a ≤ x ≤ b

0 elsewhere.





Expected Value and Variance

E (X ) =a + b

2

Var(X ) =(b − a)2

12





Problem

The random variable X is known to be uniformly distributedbetween 10 and 20.a. Show the graph of the probability density function.b. Compute P(X < 15).c. Compute P(12 ≤ X ≤ 18).d. Compute E (X ).e. Compute Var(X ).





Solution

b. P(X < 15) = (0.10)(5) = 0.50.c. P(12 ≤ X ≤ 18) = (0.10)(6) = 0.60.d. E (X ) = 10+20

2 = 15.

e. V (X ) = (20−10)2

12 = 8.33.





Example

Delta Airlines quotes a flight time of 24 hours, 5 minutes for itsflights from Cincinnati to Tampa. Suppose we believe that actualflight times are uniformly distributed between 2 hours and 2 hours,20 minutes.a. Show the graph of the probability density function for flighttime.b. What is the probability that the flight will be no more than 5minutes late?c. What is the probability that the flight will be more than 10minutes late?d. What is the expected flight time?





Solution

b. P(X ≤ 130) = 1020 = 0.5

c. P(X > 135) = 520 = 0.25

d. E (X ) = 120+1402 = 130 minutes.





Standard Normal Probability Distribution

f (z) =1√2π

exp−z2

2

Note. Don’t worry about this formula, we are NOT going to use itto do calculations.





Normal Probability Density Function

f (x) =1√

2πσ2e−

(x−µ)2

2σ2

where µ = mean, σ = standard deviation, e = 2.71828, andπ = 3.14159.Note. Don’t worry about this formula, we are NOT going to use itto do calculations.





Example

Given that Z is a standard normal random variable, compute thefollowing probabilities.a. P(Z ≤ −1).b. P(Z ≥ −1).c. P(Z ≥ −1.5).d. P(Z ≥ −2.5).e. P(−3 < Z ≤ 0).





Solution

a. P(Z ≤ −1) = 0.1587.b. P(Z ≥ −1) = 1− P(Z < −1) = 1− 0.1587 = 0.8413.c. P(Z ≥ −1.5) = 1− P(Z < −1.5) = 1− 0.0668 = 0.9332.d. P(Z ≥ −2.5) = 1− P(Z < −2.5) = 1− 0.0062 = 0.9938.e. P(−3 < Z ≤ 0) = P(Z ≤ 0)− P(Z ≤ −3) = 0.5− 0.0013 =0.4987.





Example

Given that Z is a standard normal random variable, compute thefollowing probabilities.a. P(0 ≤ Z ≤ 0.83).b. P(−1.57 ≤ Z ≤ 0).c. P(Z ≥ 0.44).d. P(Z ≥ −0.23).e. P(Z ≤ 1.2).f. P(Z ≤ −0.71).





Solution

a. P(0 ≤ Z ≤ 0.83) = 0.2967.b. P(−1.57 ≤ Z ≤ 0) = 0.4418.c. P(Z ≥ 0.44) = 0.3300.d. P(Z ≥ −0.23) = 0.5910.e. P(Z ≤ 1.2) = 0.8849.f. P(Z ≤ −0.71) = 0.2389.





Example

Given that Z is a standard normal random variable, find z∗ foreach situation.a. The area to the left of z∗ is 0.9750.b. The area between 0 and z∗ is 0.4750.c. The area to the left of z∗ is 0.7291.d. The area to the right of z∗ is 0.1314.e. The area to the left of z∗ is 0.6700.f. The area to the right of z∗ is 0.3300.





Solution

a. z∗ = 1.96b. z∗ = 1.96c. z∗ = 0.61d. z∗ = 1.12e. z∗ = 0.44f. z∗ = 0.44





Example

Given that Z is a standard normal random variable, find z∗ foreach situation.a. The area to the left of z∗ is 0.2119.b. The area between −z∗ and z∗ is 0.9030.c. The area between −z∗ and z∗ is 0.2052.d. The area to the left of z∗ is 0.9948.e. The area to the right of z∗ is 0.6915.





Solution

a. z∗ = -0.80.b. z∗ = 1.66.c z∗ = 0.26.d. z∗ = 2.56.e. z∗ = -0.50.





Problem

For borrowers with good credit scores, the mean debt for revolvingand installment accounts is $ 15,015. Assume the standarddeviation is $ 3540 and that debt amounts are normally distributed.a. What is the probability that the debt for a randomly selectedborrower with good credit is more than $ 18,000?b. What is the probability that the debt for a randomly selectedborrower with good credit is less than $ 10,000?c. What is the probability that the debt for a randomly selectedborrower with good credit is between $ 12,000 and $ 18,000?d. What is the probability that the debt for a randomly selectedborrower with good credit is no more than $ 14,000?





Solution

Let X = debt amount.a. P(X > 18000) = P

(X−µσ > 18000−15015

3540

)= P(Z > 0.8432)

= 1− P(Z ≤ 0.8432) = 1− 0.7995 = 0.2005.

b. P(X < 10000) = P(X−µσ < 10000−15015

3540

)= P(Z < −1.4166)

= 0.0773.





Solution

c.P(12000 < X < 18000) = P

(12000−15015

3540 < X−µσ < 18000−15015

3540

)= P(−0.8516 < Z < 0.8432) = P(Z ≤ 0.8432)−P(Z ≤ −0.8516)= 0.7995− 0.1977 = 0.6018.

d. P(X ≤ 14000) = P(X−µσ < 14000−15015

3540

)= P(Z < −0.2867)

= 0.3897.





Exponential Probability Distribution

Exponential probability density function

f (x) =1

µe−

xµ

for x ≥ 0 and µ > 0, where µ = expected value or mean.





Exponential Probability Distribution (cont.)

Cumulative Probability.

P(X ≤ k) = 1− e−kµ

where k is a number greater than zero.





Example

Consider the following exponential probability density function

f (x) =1

8e−

x8 .

a. Find P(X ≤ 6).b. Find P(X ≤ 4).c. Find P(X ≥ 6).d. Find P(4 ≤ X ≤ 6).





Solution

a. P(X ≤ 6) = 1− e−6/8 = 0.5276.b. P(X ≤ 4)1− e−4/8 = 0.3934.c. P(X ≥ 6) = 1− P(X < 6) = 1− 0.5276 = 0.4724.d. P(4 ≤ X ≤ 6) = P(X ≤ 6)− P(X ≤ 4) = 0.5276− 0.3934 =0.1342.





Problem

The time required to pass through security screening at the airportcan be annoying to travelers. The mean wait time during peakperiods at Cincinnati/Northern Kentucky International Airport is12.1 minutes. Assume the time to pass through security screeningfollows an exponential distribution. a. What is the probability itwill take less than 10 minutes to pass through security screeningduring a peak period?b. What is the probability it will take more than 20 minutes topass through security screening during a peak period?





Problem (cont.)

c. What is the probability it will take between 10 and 20 minutesto pass through security screening during a peak period?d. It is 8:00 a.m. (a peak period) and you just entered the securityline. To catch your plane you must be at the gate within 30minutes. If it takes 12 minutes from the time you clear securityuntil you reach your gate, what is the probability you will miss yourflight?





Solution

Let T be the time required to pass through security screening. Thas an exponential distribution with µ = 12.1.a. P(T < 10) = 1− e−10/12.1 = 0.5623.b. P(T > 20) = 1− P(T ≤ 20) = 1− (1− e−20/12.1) =1− 0.8085 = 0.1915.c. P(10 < T < 20) = P(T < 20)− P(T < 10) =0.8085− 0.5623 = 0.2462.d. P(missing your flight) = P(T > 18) = 1− P(T ≤ 18) =1− (1− e−18/12.1) = 1− 0.7740 = 0.226.





Sampling Distribution

The sampling distribution of a statistic is the distribution ofvalues taken by the statistic in all possible samples of the same sizefrom the same population.





Toy Problem

We have a population with a total of six individuals: A, B, C,D, E and F.

All of them voted for one of two candidates: Bert or Ernie.

A and B voted for Bert and the remaining four people votedfor Ernie.

Proportion of voters who support Bert is p = 26 = 33.33%.

This is an example of a population parameter.





Toy Problem

We are going to estimate the population proportion of peoplewho voted for Bert, p, using information coming from an exitpoll of size two.

Ultimate goal is seeing if we could use this procedure topredict the outcome of this election.





List of all possible samples

{A,B} {B,C} {C,E}{A,C} {B,D} {C,F}{A,D} {B,E} {D,E}{A,E} {B,F} {D,F}{A,F} {C,D} {E,F}





Sample proportion

The proportion of people who voted for Bert in each of thepossible random samples of size two is an example of a statistic.In this case, it is a sample proportion because it is the proportionof Bert’s supporters within a sample; we use the symbol p̂ (read”p-hat”) to distinguish this sample proportion from the populationproportion, p.





List of possible estimates

p̂1={A,B} = {1,1}=100% p̂9 ={B,F} = {1,0}=50%p̂2={A,C} = {1,0}=50% p̂10 ={C,D} = {0,0}=0%p̂3={A,D} = {1,0}=50% p̂11 ={C,E}= {0,0}=0%p̂4 ={A,E}= {1,0}=50% p̂12={C,F} {0,0}=0%p̂5={A,F} = {1,0}=50% p̂13 ={D,E}{0,0}=0%p̂6={B,C} = {1,0}=50% p̂14={D,F}{0,0}=0%p̂7={B,D} = {1,0}=50% p̂15={E,F} {0,0}=0%p̂8 ={B,E} = {1,0}=50%

mean of sample proportions = 0.3333 = 33.33%.standard deviation of sample proportions = 0.3333 = 33.33%.





Frequency table

p̂ Frequency Relative Frequency

0 6 6/151/2 8 8/15

1 1 1/15





Sampling distribution of p̂ when n = 2.

●

●

●

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.1

0.2

0.3

0.4

0.5

0.6

Rel

ativ

e Fr

eq.

p̂

Sampling Distribution when n=2

6/15

8/15

1/15





Predicting outcome of the election

Proportion of times we would declare Bert lost the election usingthis procedure= 6

15 = 40%.





Problem (revisited)

Next, we are going to explore what happens if we increase oursample size. Now, instead of taking samples of size 2 we are goingto draw samples of size 3.





List of all possible samples

{A,B,C} {A,C,E} {B,C,D} {B,E,F}{A,B,D} {A,C,F} {B,C,E} {C,D,E}{A,B,E} {A,D,E} {B,C,F} {C,D,F}{A,B,F} {A,D,F} {B,D,E} {C,E,F}{A,C,D} {A,E,F} {B,D,F} {D,E,F}





List of all possible estimates

p̂1 = 2/3 p̂6 = 1/3 p̂11 = 1/3 p̂16 = 1/3p̂2 = 2/3 p̂7 = 1/3 p̂12 = 1/3 p̂17 = 0p̂3 = 2/3 p̂8 = 1/3 p̂13 = 1/3 p̂18 = 0p̂4 = 2/3 p̂9 = 1/3 p̂14 = 1/3 p̂19 = 0p̂5 = 1/3 p̂10 = 1/3 p̂15 = 1/3 p̂20 = 0

mean of sample proportions = 0.3333 = 33.33%.standard deviation of sample proportions = 0.2163 = 21.63%.





Frequency table

p̂ Frequency Relative Frequency

0 4 4/201/3 12 12/202/3 4 4/20






●

●

●

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.1

0.2

0.3

0.4

0.5

0.6

Rel

ativ

e Fr

eq.

p̂


4/20

12/20

4/20





Prediction outcome of the election

Proportion of times we would declare Bert lost the election usingthis procedure= 16

20 = 80%.





More realistic example

Assume we have a population with a total of 1200 individuals. Allof them voted for one of two candidates: Bert or Ernie. Fourhundred of them voted for Bert and the remaining 800 peoplevoted for Ernie. Thus, the proportion of votes for Bert, which wewill denote with p, is p = 400

1200 = 33.33%. We are interested inestimating the proportion of people who voted for Bert, that is p,using information coming from an exit poll. Our ultimate goal is tosee if we could use this procedure to predict the outcome of thiselection.






●

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0.0 0.2 0.4 0.6 0.8 1.0

02

46

810

Rel

ativ

e Fr

eq.

p̂


p=0.3333






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46

810

Rel

ativ

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eq.

p̂


p=0.3333






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p=0.3333






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0.0 0.2 0.4 0.6 0.8 1.0

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810

Rel

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p=0.3333






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810

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p=0.3333






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p=0.3333






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p=0.3333






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0.0 0.2 0.4 0.6 0.8 1.0

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ativ

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eq.

p̂


p=0.3333





Observation

The larger the sample size, the more closely the distribution ofsample proportions approximates a Normal distribution.

The question is: Which Normal distribution?





Sampling Distribution of a sample proportion

Draw an SRS of size n from a large population that containsproportion p of ”successes”. Let p̂ be the sample proportion ofsuccesses,

p̂ =number of successes in the sample

n





Sampling Distribution of a sample proportion (cont.)

Then:

The mean of the sampling distribution of p̂ is p.

The standard deviation of the sampling distribution is√p(1− p)

n.

As the sample size increases, the sampling distribution of p̂becomes approximately Normal. That is, for large n, p̂ has

approximately the N

(p,√

p(1−p)n

)distribution.





Approximating Sampling Distribution of p̂

If the proportion of all voters that supports Bert isp = 1

3 = 33.33% and we are taking a random sample of size 120,the Normal distribution that approximates the samplingdistribution of p̂ is:

N

(p,

√p(1− p)

n

)that is N (µ = 0.3333, σ = 0.0430) (1)





Sampling Distribution of p̂ vs Normal Approximation

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0.0 0.2 0.4 0.6 0.8 1.0

02

46

810

Rel

ativ

e Fr

eq.

p̂

Normal approximation





Predicting outcome of the election with our approximation

Proportion of times we would declare Bert lost the election usingthis procedure = Proportion of samples that yield a p̂ < 0.50.Let Y = p̂, then Y has a Normal Distribution with µ = 0.3333and σ = 0.0430.Proportion of samples that yield a p̂ < 0.50=

P(Y < 0.50) = P(Y−µσ < 0.5−0.3333

0.0430

)= P(Z < 3.8767).





P(Z < 3.8767)

−4 −2 0 2 4

0.0

0.1

0.2

0.3

0.4

0.9999471





Predicting outcome of the election with our approximation

This implies that roughly 99.99% of the time taking a random exitpoll of size 120 from a population of size 1200 will predict theoutcome of the election correctly, when p = 33.33%.


Documents

STA 218: Statistics for Management and Economicsmath.unm.edu/~alvaro/sta218-p1.pdf · Picturing distributions with graphs. Describing distributions with numbers. Normal distributions