CHAPTER I DEFINITION OF STATISTICS. Descriptive Statistics consists of the collection, organization, summation and presentation of data.

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1 CHAPTER I DEFINITION OF STATISTICS Statistics is the science of conducting studies to collect, organize, summarize, analyze and draw conclusions from data. 1.1 Descriptive and Inferential Statistics A variable is a characteristic or attribute that can assume different values. Data are the values (measurements or observations) that are variables can assume.variables whose values are determined by chance are called random variables. A collection of data values forms a data set. Each value in the data set is called a data value or a datum. Data can be used in different ways. Statistics can be divided into two main areas depending on how data are used. The two areas are Descriptive Statistics and Inferential Statistics. Descriptive Statistics consists of the collection, organization, summation and presentation of data. Example: 1) Nine out of ten on the job fatalities are men. ) Expenditure for the cable industry were 5.66 dollor billion in ) The median house hold income for people aged 54 is 5,888 dollar. 4) The national average annual medicine expenditure per person is 105 dollar. A population consists of all subjects (human or otherwise) that are being studied. A sample is a subgroup of the population. Inferential Statistics consists of generalizing from sample to populations, performing hypothesis testing, determining relationship among variables and making predictions. Example: 1) By 040atleast.5 billion people will run short of water. ) Experts say that mortgage rates may soon hit bottom. ) A diet high in fruits and vegetables will lower blood pressure. 4) In 00, the number of high school graduates will be. million students. 1. Variables and Types of Data Variables can be classified as quantitative or qualitative Qualitative variables are variables that can be placed into distinct categories,according to some characteristics or attribute. [Type text] Page 1

2 Example: 1) Marital status of nurses in a hospital. ) Colours of automobiles in a shopping centre parking lot. Quantitative variables are numerical in nature and can be ordered or ranked. Example: 1) Time it takes to run a marathon ) capacity of the NFL football stadium ) Ages of people living in a personal care home Discrete variables assume values that can be counted. Example: 1) Number of cups of coffee served in a restaurant. ) The number of ads on a one hour television show. ) Number of pizzas sold by Pizza Express each day. Continuous variables can assume all values between any two specific values.they are obtained by measuring. Example: 1) The time it takes a student to drive to school. )Lifetimes (in hours) of 15 ipod batteries. ) Blood pressure of runners in a marathon. 1. Data Collection and Sampling Techniques Random samples are selected by using chance methods or random numbers Systematic samples are obtained by numbering each vale in the population and then selecting the k th value. Stratified samples are selected by dividing the population into groups(strata) according to some characteristic and taking samples from each group. Cluster samples are selected by dividing the population into groups and then taking the samples of the groups. [Type text] Page

3 CHAPTER II FREQUENCY DISTRIBUTIONS AND GRAPHS Raw data: The original form of the data are called raw data. Class: Each raw data value is placed into a qualitative or quantitative category called a class. Frequency: The frequency of the class is the number of data values contained in the specific class. Frequency distribution : It is the organization of raw data in table form using classes and frequencies. The types of frequency distributions are Categorical frequency distribution and Grouped frequency distribution Example: Twenty five army inductees were given a blood test to determine their blood type. The data is A B B AB O O O B AB B B B O A O A O O O A AB A O B A Construct a frequency distribution for the data. Solution: Class Tally Frequency(f) Percent A 5 0 Percent= 100 n f B 7 8 O 9 6 AB 4 16 Total n=5 100 Example: The data shown here represent the amount of miles per gallon(mpg) that 0 selected four-wheel drive sports utility vehicles obtained in city driving. Construct a frequency distribution and a cumulative frequency distribution [Type text] Page

4 Solution: Highest value(h) = 19 ; Lowest value (L) = 1 Range = H L = 19-1 =7 Here the range of the data values is small, classes with single value data can be used. Class limits Class boundaries Tally Frequency Cumulative frequency Less than 11.5 Less than 1.5 Less than 1.5 Less than 14.5 Less than 15.5 Less than 16.5 Less than 17.5 Less than 18.5 Less than (0+6) 7 (6+1) 10(7+) 16 (10+6) 4 (16+8) 6 (4+) 9 (6+) 0 (9+1) Example: The number of total votes exercised by the past 0 Presidents is listed below.use the data to construct a grouped frequency distribution and a cumulative frequency distribution with 5 classes Solution: Highest value (H) = 65 ; Lowest value (L) = 6 Range = H L = 65 6 = 69 Range 69 Width = = = No.of classes 5 Class limits Class boundaries Tally Frequency Cumulative frequency Less than 5.5 Less than Less than 57.5 Less than Less than Less than Total (0+16) 19 (16+) 19(19+0) 19 (19+0) 0 (19+1) [Type text] Page 4

5 Example: The data represent the record high temperatures in degrees Fahrenheit ( o F) for each of the fifty states. Construct the frequency distribution for the data using 7 classes Solution: See the text book 1) The data shown are the number of grams per serving of 0 selected brands of cakes. Construct a grouped frequency distribution using 5 classes (Try by yourself) ) The number of stories in each of the world s 0 tallest buildings follow. Construct a grouped frequency distribution and a cumulative frequency distribution with 7 classes (Try by yourself) GRAPHS After organizing the data into frequency table, we can represent them in graphical form: 1) Histogram ) The frequency polygon ) The cumulative frequency graph (Ogive) Histogram: This graph displays the data by using contiguous vertical bars of various heights to represent the frequencies. The frequency polygon: This graph displays the data by using lines that connect points plotted for the frequency at the mid points of the classes. Ogive: This graph represents the cumulative frequencies for the classes in a frequency distribution. [Type text] Page 5

6 Example: The data show the number of rail road accidents for the 50 states of US for a specific year. Construct a histogram, frequency polygon and ogive for the data. Class limits Frequency Total 50 Solution: Histogram: Class limits Class boundaries Frequency Histogram frequency Frequency polygon: [Type text] Page 6

7 Class boundaries Mid point Frequency Frequency Polygon Mid points Ogive: Cumulative frequency 60 Cumulative frequency Less than 0.5 Less than 4.5 Less than 86.5 Less than 19.5 Less than 17.5 Less than 15.5 Less than 58.5 Less than 01.5 Less than Class boundaries Example: The data represent the record high temperatures in degrees Fahrenheit ( o F) for each of the fifty states. Construct a histogram, frequency polygon and ogive for the data. Class limits Frequency [Type text] Page 7

8 Solution: See the text book. Example : For 108 randomly selected college applicants, the following frequency table for entrance exam scores was obtained. Construct a histogram, frequency polygon and ogive for the data. Class limits Frequency (Try by yourself) The Pie Graph A Pie graph is a circle that is divided into sections according to the percentage of frequencies in each category of the distribution. Example: Construct a pie graph showing the blood types of the army inductees. Class A B O AB Frequency Solution: [Type text] Page 8

9 Class Frequency(f) Degree Percent A Degree = f 60 n B O Percent= 100 n f AB Total n= Example: Construct a pie graph for the following data; Class Frequency [Type text] Page 9

10 Solution: Class Frequency(f) Degree Percent Total n = Pie Chart [Type text] Page 10

11 CHAPTER III MEASURES OF CENTRAL TENDENCIES Statistical methods can be used to summarize the data. The most familiar of these methods is finding the averages. Example: The average American man is five feet, nine inches tall. On the average day,4 million people receive animal bites. Average means the center of the distribution. Measures of average is also called measures of central tendency. This include the mean, median,mode and midrange. The Measures that determine the spread of the data values are called measures of variation or measures of dispersion. These measures include the range, variance and standard deviation. A statistic is a characteristic or measure obtained by using the data values from the sample. A parameter is a characteristic or measure obtained by using all the data values from specific population. The Mean The mean (arithmetic average) is the sum of the values divided by the number of values. The sample mean is represented by X. X 1 + X + X X n X X = =, where n represents the total number of values in the sample. n n The population mean is represented by µ= X 1 + X + X X N = X, where N represents the total number of values in the population. N N [Type text] Page 11

12 Example: The data represents the the number of days off per year for a sample of individual selected from nine different countries. Find the mean. 0,6,40,6,,4,5,4,0 X Solution: X Hence, the mean of the number of days off is 0.7 days. Example: The data shows the number of patients in a sample of six hospitals who required an infection while hospitalized. Find the mean. 110,76,9,8,105,1. Solution: See the text book. Mean of grouped data f X m X =, where X m is the mid value of the class and n = f n Example: Thirty auto mobiles were tested for fuel efficiency (in miles per gallon).this frequency distribution was obtained. Find the mean. Class boundaries Frequency Solution: Class boundaries Frequency (f ) Mid point ( Xm ) f Xm [Type text] Page 1

13 n = f =0 f X m = 590 f X m 590 Mean= X = = = n 0 Example: The data represent the number of miles run during one week for a sample of 0 runners. Find the mean. Class Frequency Solution: See the text book A random sample of bonuses (in millions of dollars) paid by large companies to their executives is shown in the data. Find the mean. (Try by yourself) Class boundaries Frequency [Type text] Page 1

14 The Median The median is the midpoint of the data array (ordered data set).it id denoted by MD. n+1 th Note: If the number of data values (n) is odd, then the median (MD) is term. If the number of data values (n) is even, then the median (MD) is the average of n th n th and +1 term. Example: The number of rooms in the seven hotels in the downtown Pittsburgh is 71,00,618,595,11,401 and 9.Find the median. Solution: Arranging the data in the order, we get 9,00,11,401, 595,618,71. n+1 th 7+1 th th Here n =7, hence MD= = =4 term= 401. Example: The number of tornadoes that have occurred in United states over an 8 period follows. Find the median. 684,764,656,70,856,11,11,10. Solution: Arranging the data in the order, we get 656,684,70,764,856,11,11,10. n th n th (4 th +5 th )term Here n =8, hence MD is the average of and +1 = = = = 810 [Type text] Page 14

15 The Median for a grouped frequency distribution MD = n ( ) cf *w+ Lm where f n = f, cf - cumulative frequency of the class preceding the median class, w -width of the median class and L m - Lower boundary of the median class Example: The data represent the number of miles run during one week for a sample of 0 runners. Find the median. Class Frequency Solution: Class Frequency (f) ( f ) n = f =0 Cumulative frequency 1 6(cf ) n 0 Here = =10, f = 5,cf = 6, w = 5, L m = 0.5 n MD = f 10 ( ) 6 = *5+0.5= 4.5 ( ) cf *w+l m [Type text] Page 15

16 5 The data represent the ages of residents of a town. Find the median. (Try by yourself) Age Frequency The Mode The value that occurs most often in a data set is called the mode. Example:Find the mode of the signing bonuses of eight NFL players for a specific year.the bonuses in billions of dollars are 18.0,14.0,4.5,10,11.,10,1.4,10. Solution: Here $10 million occurs times. Hence the mode is 10. Example: The data show the number of nuclear reactors in the United States for the recent 15 year period. Find the mode Solution: Here the values104 and 109 occur 5 times. Hence the modes are 104 and 109.This data is said to be a bimodal. Example: Find the mode for the number of branches that six banks have. 401,44,09,01,7,5. Solution:Here each value occurs only once. Hence there is no mode. Modal class is the class with largest frequency Example: A random sample of bonuses (in millions of dollars) paid by large companies to their executives is shown in the data. Find the modal class. [Type text] Page 16

17 Class boundaries Frequency Solution: Here the highest frequency is 1.Hence the modal class is Example: A small company consists of the owner, the manager, the salesperson and two technicians all of whose salaries are listed here.(assume this is the entire population) Find the mean, median, mode. Staff Owner Manager Salesperson Technician Technician Salary $ X Solution: Mean of the population µ Median (MD) = $1000 Mode = $9000 The Midrange The midrange is the rough estimation of the middle. The midrange is the sum of the lowest and the highest values in the data set, divided by.it is denoted by MR Lowest value+ Highest value Midrange (MR) = = Example: Find the midrange of the signing bonuses of eight NFL players for a specific year. The bonuses in billions of dollars are 18.0,14.0,4.5,10,11.,10,1.4,10. [Type text] Page 17

18 Lowest value+ Highest value Solution: MR = = == == = $.5 million. The frequency distribution represents the data obtained from a sample of 75 copying machine service technicians. The values represent the days between service calls for various copying machines. Class boundaries Frequency Find the mean and the modal class. (Try by yourself) The Weighted Mean The weighted mean of a data is calculated by multiplying each variable X by its corresponding weights and dividing the sum of products by the sum of the weights. X = w 1 1 X + w X + w X w n X n = wx, where w 1,w,w,...,w n are weights and w 1 + w 1 + w +...w n w X 1, X,X,...,X n are the values. Example: The following table gives the credits of the courses and the grades obtained by a student. Find the student s grade point average. Course Credits(w) Grade(X) [Type text] Page 18

19 English Composition I Introduction to Psychology Biology I Physical Education 4 A(4 points) C( points) B( points) D(1 points) wx Solution: X = = = =.7 The grade point average is.7. MEASURES OF VARIATION w For the spread or variability of a data set three measures are used, namely, range, variance and standard deviation. The range is the highest value minus lowest value.it is denoted by R R = Highest,value Lowest value Example: The salaries for the staff of the XYZ Co. are shown here. Find the range. Staff Owner Manager Sales representative Workers Salary $100,000 40,000 0,000 5,000 15,000 18,000 Solution: The range is R = $100,000 - $15,000 = $85,000 [Type text] Page 19

20 VARIANCE AND STANDARD DEVIATION Population variance: (X µ) σ =, where X = individual value, µ= population mean, N = population size N Population standard deviation: The square root of variance is called the standard deviation. This is denoted by σ. = (X µ) σ N Sample variance: s sample size. Sample standard ( X X ) n 1 ( X X ) n 1 =, where X = individual value, X = sample mean, n = deviation: s = The shortcut formula for sample variance and standard deviation n X ( X) n X ( X) s = s = n n( 1) n n( 1) Example: A testing lab wishes to test the brand of outdoor paint to see how long it will last before fading. It makes 6 gallons of paints to test. Find the mean, variance and standard deviation. 10,60,50,0,40,0. X Solution: Mean =µ X (X µ) (X µ) [Type text] Page 0

21 Variance = Standard deviation = σ= 91.7 =17.1 (X µ) =1750 Example: The number of incidents in which police were needed for a sample of 10 schools in Allegheny country is 7,7,,8,48,11,6,0,10,. Find the variance and standard deviation. Solution: X X X (X X) (X X) (X X) 9 9 Variance =s = = = = 54.7 n Standard deviation = s = 54.7 = (X X) =9 [Type text] Page 1

22 Aliter: X X Variance n X ( X) 10*4061 (1) s = = = n n( 1) 10(10 1) 90 = = 54.7 Standard deviation = s = 54.7 = X =1 X =4061 Example: Find the sample variance and the standard deviation for the amount of European auto sales for a sample of 6 years shown. The data are in millions of dollars.11., 11.9, 1.0, 1.8,1.4,14. Solution: See the text book The normal daily high temperatures (in degrees Fahrenheit) in January for 10 selected cities are as follows: 50,7,9,54,0,61,47,8,4,61. Find the range, variance, standard deviation. (Try by yourself). VARIANCE AND STANDARD DEVIATION FOR GRUPED DATA n f X ( f X ) m m Variance = s = n n( 1) Example: A random sample of 0 states shows the number of low- power FM radio stations for each state. Find the mean and variance. Class limit Frequency [Type text] Page

23 Solution: Class limit Frequency f Midpoint X m f X m f Xm X m n = f = 0 f X m =67 f X m = 1990 m m (67) n f X ( f X ) 0(1990) Variance = s = = = = =167. n n( 1) 0(0 1) 0*9 870 Standard deviation = s = 167. =1.9 Example: The data represent the number of miles run during one week for a sample of 0 runners. Find the variance and standard deviation. [Type text] Page

24 Class Frequency Solution: See the text book This frequency distribution represents the data obtained from a sample of word processor repairers. The values are the days between service calls on 80 machines. Find the variance and the standard deviation. (Try by yourself) Coefficient of variation The coefficient of variation, denoted by CVar is the standard deviation divided by the mean. The result is expressed as a percentage. For samples s *100 X For populations σ *100 µ Example: The mean of the number of sales of cars over a -month period is 87 and the standard deviation is 5.The mean of the commissions is $55 and the standard deviation is $77.Compare the variations of the two. Solution: The coefficient of variations are s 5 CVar = *100 = *100 = 5.7% (sales) X 87 = *100 =14.8% (commissions) Since the coefficient of variation is large for commissions, the commissions are more variable than the sales. Example: The average age of the accountants at Three Rivers Corp.is 6 years with a standard deviation of 6 [Type text] Page 4

25 years: The average salary of the accountants is $1000 with a standard deviation of $4000. Compare the variations of age and income. (Try by yourself) CHAPTER IV PROBABILITY AND COUNTING RULES Probability can be defined as a chance of an event occurring. Sample spaces and probability A probability experiment is a chance process that leads to well defined results called outcomes. An outcome is a result of a single trial of a probability experiment. A sample space is the set of all possible outcomes of a probability experiment. Example: Find the sample space for rolling two dice. Solution: Die Die [Type text] Page 5

26 (1,1) (1,) (1,) (1,4) (1,5) (1,6) (,1) (,) (,) (,4) (,5) (,6) (,1) (,) (,) (,4) (,5) (,6) (4,1) (4,) (4,) (4,4) (4,5) (4,6) (5,1) (5,) (5,) (5,4) (5,5) (5,6) (6,1) (6,) (6,) (6,4) (6,5) (6,6) Example: Find the sample space of drawing a card from an ordinary deck of cards. Solution: There are four suits (hearts, diamonds, spades and clubs) Example: Find the sample space of the gender of the children if a family has three children. Solution: BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG An event consists of a set of outcomes of a probability experiment. Classical Probability assumes that all outcomes in the sample space are equally likely to occur. Equally likely events are events that have the same probability to occur. The probability of any event E, denoted by P(E) and is given by ( ) P(E) = = ( ) [Type text] Page 6

27 Examples: Tossing a coin, rolling a die, taking a card from a deck of cards. Rules of probability 1. If E is any event, then. If an event E cannot occur, then ( ) = 0. If an event E cannot certain, then ( ) = 1, that is sum of the probabilities of all the outcomes in the sample space is 1 Example: If a die is rolled one time, find these probabilities. (a) Getting a (b) Getting a number greater than 6 ( c) Getting an odd number (d) Getting a 4 or an odd number (e) Getting a number less than 7 (f) Getting a number greater than or equal to (g) Getting a number greater than and an even number. Solution: (a) P(Getting a ) = (b) P(Getting a number greater than 6) = 0 [Type text] (c) P(Getting an odd number) = = (d) P(Getting an 4 or an odd number) = 6 4 = % & (e) P(Getting a number less than 7) = = 1 (f) P(Getting a number greater than or equal to ) = = & (g) P(Getting a number greater than and an even number) = = Example: If two dice are rolled one time, find the probability of getting these results. (a) A sum of 9 (b) A sum of 7 or 11 (c) Doubles (d) A sum less than 9 (e) A sum greater than or equal to 10 Solution: [Type text] Page 7

28 (a) P(A sum of 9) = = Sample space = {(,6), (4,5),(5,4),(6,)} ) & (b) P(A sum of 7 or 11) = = Sample space = {(1,6),(,5),(,4), (4,),(5,),(6,1),(5,6),(6,5)} * (c) P(Doubles) = = Sample space = {(1,1),(,),(,), (4,4),(5,5),(6,6)} & (d) P(A sum less than 9) = = Sample space = {(1,1) (1,) (1,) (1,4) (1,5) (1,6) (,1) (,) ) (,) (,4) (,5) (,6) (,1) (,) (,) (,4) (,5) (4,1) (4,) (4,) (4,4) (5,1) (5,) (5,) (6,1) (6,) } (e) P(A sum greater than or equal to 10) = = Sample space = { (4,6) (5,5) (5,6) (6,4) (6,5) (6,6) } Example: If one card is drawn from a deck, find the probability of getting these results. (a) A queen (b) A club (c) A queen of clubs (d) A or an 8 (e) A 6 or a spade (f) A black king (g) A red card and a seven (h) A diamond or a heart (i) A black card. Solution: % (a) P(A queen) = = (There are 4 queen cards) +& (b) P(A club) = = (There are 1 club cards) +& % (c) P(A queen of clubs) = +& (There is 1 queen of club cards) ) & (d) P(A or an 8) = = (There are 4 cards with number and 4 cards with 8) +& (e) P(A 6 or a spade) = = % (There are cards with number 6 and 1 cards of spade) +& [Type text] Page 8

29 & (f) P(A black king) = = +& & (There are black king cards) & (g) P(A red card and a seven) = = (There are red cards with number 7) +& & & (h) P(A diamond or a heart) = = (There are 1 cards of diamond and 1 cards of heart) +& & & (i) P(A black card) = = (There are 6 black cards) +& & Example: If a family has three children, find the probability that two of the three children are girls. Solution: P(Two girls) = (There are 8 outcomes BBB,BBG,BGB,GBB, BGG,GBG,GGB and GGG and ) ways to have girls) Example: A couple has three children, find each probability. (a) All boys (b) All girls or all boys (c) Exactly two boys or two girls (d) At least one child of each gender 6 Solution: (a) P(All boys ) girls) = = 8 = (b) P(All boys or all girls) = (c) P(Exactly two boys or two 4 6 (d) P(At least one child of each gender) = = 8 4 Example: A card is drawn from an ordinary deck. Find the probabilities of (a) getting a jack (b) getting a 6 of clubs (c) getting a three or a diamond (d) getting a or a 6. Solution: See the text book. Example: When a single die is rolled, find the probabilities of [Type text] Page 9

30 (a) getting a 9 (b) getting a number less than 7 Solution: See the text book. The complement of an event is the set of outcomes in the sample space that are not included in the outcomes of event E. This is denoted by (,). Rules for complementary events [Type text] (,) = 1 ( ), ( ) = 1 (,), ( ) + (,) = 1 P(S) =1 Example: When a single die is rolled, find the probabilities of (a) getting an odd number (b) getting an even number. Solution: (a) P(Getting an odd number) = = (b) P(Getting an even number) = 1 = & Example: When two coins are tossed, find the probabilities of (a) getting all heads (b) getting at least one head. Solution: (a) P(Getting all heads) =% (b) P(E) = P(Getting at least one head) = = 4 Empirical Probability relies on actual experience to determine the likelihood of outcomes. [Type text] Page 0

31 Given a frequency distribution, the probability of an event being in a given class is / :1 5;<== 7 P(E) = = >89<; ?1=?4 4 This probability is called empirical probability and is based on observation Example: In a sample of 50 people, 1 had type O blood, had type A blood, 5 had type B blood and had type AB blood. Set up a frequency distribution and find the probabilities. (a) A person has type O blood (b) A person has type A or type B blood (c) A person has neither type A nor type O blood (d) A person does not have type O blood Type A B AB O Frequency 5 1 [Type text] Page 1

32 Solution: n = 50 (a) P(Type O blood) = (b) P(Type A or B blood) = (c) P(Neither type A nor O blood) = 48 (d) P(Does not have Type AB blood) =1 = Example: Hospital records indicated that knee replacement patients stayed in the hospital for the number of days shown in the distribution. Find the probabilities. No.of days stayed Frequency (a) A patient stayed exactly 5 days (b) A patient stayed at most 4 days (c) A patient stayed less than 6 days (d) A patient stayed at least 5 days. Solution: (a) P(Exactly 5 days) = (b) P(At most 4 days) = ( or 4) (c) P(Less than 6 days) = ( or 4 or 5) (d) P(At least 5 days) = (5 or 6 or 7) Example: Rural speed limit for all 50 states is indicated below. 60 mph 65 mph 70 mph 75 mph Dr. Yashpal Singh Raghav

33 Choose one state at random. Find the probability that its speed limit is (a) 60 or 70 miles per hour (b) greater than 65 km per hour (c) 70 miles per hour or less Solution: (a) P(60 or 70 mph) = = 0.8 (b) P(greater than 65 mph) = = 0.6 ( 70 or 75) (c) P(75 mph or less) = = 0.74 THE ADDITION RULES FOR PROBABILITY Two events are mutually exclusive events if they cannot occur at the same time (i.e., they have no outcomes in common) Examples: 1) When a single die is rolled, Getting an odd number and getting an even number (mutually exclusive) Getting a number less than 4 and getting a number greater than 4(mutually exclusive) Getting a and getting an odd number (not mutually exclusive) Getting an odd and getting a number less than 4(not mutually exclusive) ) When a single card is drawn from a deck, Getting a 7 or getting a jack (mutually exclusive) Getting a club or getting a king (not mutually exclusive) Getting a face card or getting an ace (mutually exclusive) Getting a face card or getting a spade (not mutually exclusive) Addition Rules 1.If two events A and B are mutually exclusive, the probability that A or B will occur is or B) = P(A) + P(B) P(A. If two events A and B are not mutually exclusive, the probability that A or B will occur is Dr. Yashpal Singh Raghav

34 P(A or B) = P(A) + P(B) P(A and B) Venn diagram for addition rules Example: A single card is drawn from a deck. Find the probability of selecting the following. (a) A 4 or a diamond (b) A club or a diamond (c) A jack or a black card Solution: (a) P ( Selecting a 4) =, P ( Selecting a diamond) =, P ( Selecting a 4 and a diamond) = P ( Selecting a 4 or a diamond) = + = = (b) P(A club) =, P(A diamond) =, P (A club and a diamond) = 0 ( mutually exclusive) P (A club or a diamond) = + = = (c) P(A jack) =, P(A black) =, P(A Jack and a black) = P(A Jack and a black) = + = = Example: A city has 9 coffee shops: Starbuck s, Caribou Coffees and 4 Crazy Mocho Coffees. If a person selects one shop at random to buy a cup of coffee, find the probability that it either a Starbuck s or Crazy Mocho Coffees. Dr. Yashpal Singh Raghav

35 Solution: P(Starbuck s or Crazy Mocho Coffees) = P(Starbuck s ) + P(Crazy Mocho Coffees) 4 7 = + = (Events are mutually exclusive) Example: A corporate research and development centres for three local companies have the following number of employees: Alcoa 750 U.S Steel 110 Bayer Material Science 50 If a research employee is selected at random, find the probability that the employee is employed by U.S Steel or Alcoa. Solution: See the text book. Example: At a particular school with 00 male students, 58 play foot ball, 40 play basket ball and 8 play both. What is the probability that the randomly selected male student plays neither sport? Solution: P(foot ball or basket ball) = P(foot ball ) P(basket ball) P(foot ball and basket ball) = + = = = P(Neither sport) = = 0.55 Example: In a statistics class there are 18 juniors and 10 seniors : 6 of the seniors are females and 1 of the juniors are males. If a student is selected at random. Find the probability of selecting the following. (a) A junior or a female (b) A senior or a female (c) A junior or a senior Solution: Male Female Total Senior Dr. Yashpal Singh Raghav

36 Junior (a) P(A junior or a female ) = P(A junior ) + P(A female ) - P(A junior and a female ) = + = = (b) P(A senior or a female ) = P(A senior ) + P(A female ) - P(A senior and a female ) = + = = (c) P(A junior or a senior ) = P(A junior ) + P(A senior ) = + = = Example: In a hospital unit there are 8 nurses are 5 physicians: 7 nurses and physicians are females. If a staff person is selected at random, find the probability that the subject is a nurse or female. Solution: See the text book MULTIPLICATION RULES AND CONDITIONAL PROBABILITY Two events A and B are independent events if the occurrence of A does not affect the the occurrence of B Example: Rolling two dice, drawing a card and replacing it and drawing a second card. Dr. Yashpal Singh Raghav

37 MULTIPLICATION RULES 1.When two events are independent, the probability of both occurring is P(A and B) = P(A). P(B).When two events are dependent, the probability of both occurring is P(A and B) = P(A). P(B/A) where P(B/A) is the conditional probability of B given that A has occurred and is given by P(A and B) P(A and B) P(B/A) =. Also P(A/B) =. P(A) P(B) Example: A coin is flipped and a die is rolled. Find the probability of getting a head on the coin and a 4 in the die. Solution: A: getting head B: getting 4 P(A and B) = P(A). P(B) = = (Events are independent) 6 1 Example: A card is drawn from a deck and replaced; then the second card is drawn. Find the probability of getting a queen and then an ace. Solution: A: Queen B: Ace P(A and B) = P(A). P(B) =. = (Events are independent) Example: An urn contains red balls, blue balls and 5 white balls.a ball is selected and its colour is noted. Then it is replaced. A second ball is selected and its colour is noted. Find the probability of each of these. (a) Selecting blue balls (b) Selecting 1 blue ball and then 1 white ball (c) Selecting 1 red ball and then 1 blue ball. Solution: (a) P( blue) =. = (b) P(1 blue and 1 white) =. = (c) P(1 red and 1blue) =. = Example: In a pizza restaurant, 95% of the customers order pizza. If 65% of the customers order pizza and a salad, find the probability that a customer who orders pizza will also order a salad. Dr. Yashpal Singh Raghav

38 Solution: P(order pizza) = 95% = 0.95, P(order pizza and salad) = 65% = 0.65 P(order pizza and salad) 0.65 P (order pizza also order a salad) = = = = 68.4% P(order pizza) 0.95 Example: If cards are selected from a standard deck of 5 cards without replacement, find the probabilities. (a) Both are spades (b) both are the same suit (c) Both are kings Solution: (a) P( Both are spades ) = = (Events are dependent) (b) P( Both are the same suit ) =. = (Events are dependent) (c) P( Both are kings ) =. = (Events are dependent) Example: A box contains black chips and white chips. A person selects two chips without replacement. If the probability of selecting a black chip and a white chip is and probability of selecting a black chip on the first draw is, find the probability of selecting a white chip on the second draw,given that the first chip selected was a black chip. Solution: A-Black chip B-White chip 15 P(B/A) = P(A and B) = 56 = = 5 P(A) Example: A recent survey asked 100 people if they thought women in the armed forces should be permitted to participate in combat.the result of the survey are shown. Gender Yes No Total Male Female Total Dr. Yashpal Singh Raghav

39 Find the probabilities. a. The respondent answered yes given that the respondent is a female b. the respondent is a male given that the respondent answered no Solution: (Respondent) M Male, F- Female, (Answer) Y-Yes, N- No 8 18 a. P(Y/F) = P(Y and F) = 100 = 8 = 4 b. P(M/N) = P(M and N) = 100 = 18 = P(F) P(N) PROBABILITIES FOR At least The multiplication rules can be used with the complementary rule to simplify solving probability problems involving at least. Example: A coin is tossed 5 times. Find the probability of getting at least 1 tail. Solution: P(E) = 1- P(),,, P(At least 1 tail) = 1- P(All heads) = 1 =1 = 1 Example: A game is played by drawing 4 cards from an ordinary deck and replacing each card after it is drawn. Find the probability that at least one ace is drawn. Solution: P(E) = 1- P(),,, P(At least one ace) = 1- P(No aces) = =1 1 4 = 0.7 Example: It is reported 7% working women use computers at work. Choose 5 working women at random. Find (i) the probability that at least one doesn t use a computer at work (ii) the probability that all the 5 use a computer in their jobs. Solution: P(E) = = 0.7 P(),,, = 1 - P(E) = 1 0.7= 0.8 (i) P(At least 1 doesn t use) = 1- P(All 5 use) = 1 (0.7) 5 = = Dr. Yashpal Singh Raghav

40 (ii) P(All 5 use) = (0.7) 5 = 0.19 Example: In 006, 86% of US households had cable TV. Choose households at random. Find the probability that a) None of the households had cable TV b) All households had Cable TV c) At least 1 of the households had Cable TV. Solution: P(E) = = 0.86 P(),,, = 1 - P(E) = = 0.14 P(None of the households had cable TV) = (0.14) = = 0.00 P(All households had cable TV) = (0.86) = 0.66 P(At least 1 of the households had Cable TV) = 1- P(None of the households had cable TV) = = Dr. Yashpal Singh Raghav