Probability and Statistics

Transcription

1 Probability and Statistics

2 Probability and Counting Topics Methods of counting The Fundamental Counting Principle Permutations and Combinations Sample Spaces to compute probabilities

3 Systematic Counting - counting a set by listing its elements Example: How many ways can we do the following? 1. Flip a coin. 2. Roll a single die. A: Two we could flip a heads or tails 3. Pick a card from a standard deck of cards. A: Six we could roll 1, 2, 3, 4, 5, or 6 4. Choose a captain for a 9-player baseball team.

4 Example: The math department needs two students lead the volunteers at a recruiting event. One student will be in charge of the display table and the other student will lead tours of the department. There are four students available to help call them students A, B, C, and D. How many ways can the professor choose the two students to lead the volunteers? Let s choose a student to be in charge of the display table. How many choices are there? Once we ve chosen this student, how many choices are there for the tour guide? Say AB means student A does the display table and student B does the tour. (So the first student listed is always the one at the display table, and the second student does the tour.) This gives us AB, AC, and AD. If he choses student B to do the table, we get BA, BC, and BD. Similarly, we get CA, CB, CD and DA, DB, DC. Counting all the options, there are 12 ways to choose the two students to lead. What if there were 8 students available to help?

5 Example: You flip a coin three times and record the sequence of heads and tails. How many such sequences are there? A tree diagram displays all the possible outcomes of an event.

6 Example: If we roll two dice, how many different pairs of numbers can appear on the upturned faces? The question matters! If we roll two dice, how many different sums of the upturned faces can occur? In the game Settlers of Catan, players produce different items based on the total of two dice that are rolled. If wheat is produced when a total of 6 is rolled, how many different ways are there to roll a 6?

7 Careful: In some counting problems, objects can be repeated; in others, they can not. When choosing numbers for a garage keypad, we could repeat numbers with a combination like , for example. In our example with the students, we couldn t use the same student to lead both the display table and the tours. If objects are allowed to be repeated more than once, we use the phrase with repetition. If objects are not allowed to be repeated, we use the phrase without repetition.

8 Example: The Center for Academic Support needs to fill two time slots with math tutors. There are four student tutors: (K)ayley, (J)osh, (S)am, and (B)ecca. How many ways can they fill the two time slots if repetition is allowed? (So, KK is allowed, for example.) K J S B K J S B K J S B K J S B K J S B KK KJ KS KB JK JJ JS JB SK SJ SS SB BK BJ BS BB So there are 16 ways to fill the slots. In how many ways can the slots be filled if repetition is not allowed?

9 Example: A local burrito place offers a choice of 5 different meats, 2 different beans, and 4 different salsas. If you must choose one meat, one bean, and one salsa for your burrito, how many different burritos could you build?

10 The Fundamental Counting Principle (FCP) If we want to perform a series of tasks and the first task can be done in a ways, the second can be done in b ways, the third can be done in c ways, and so on, then all the tasks can be done in a x b x c x... ways. We already used this principle in the last example: Example: A local burrito place offers a choice of 5 different meats, 2 different beans, and 4 different salsas. If you must choose one meat, one bean, and one salsa for your burrito, how many different burritos could you build? You can choose 5 meats x 2 beans x 4 salsas = 40 burritos.

11 Example: An ice cream shop has 10 employees. If one person is to be in charge of opening every day and another is to be in charge of closing every day, in how many ways can these two positions be filled? We have two tasks: Assigning an opener 10 ways to assign the opener Assigning a closer 9 ways to assign the closer once the opener has been chosen. So, there are 10 x 9 = 90 ways to fill the two positions. If they also want to assign another employee to hold a sign outside to lure customers in, in how many ways can the three positions be filled?

12 When in doubt, drawing a picture always helps! Example: (a) In how many ways can three coins be flipped? (b) In how many ways can 9 four-sided dice (each of a different color) be rolled? (a) We ll draw a picture: 2 x 2 x 2 # of possibilities for first flip # of possibilities for second flip # of possibilities for third flip You try (b)!

13 Slot Diagrams We call a picture like the one we drew on the previous slide with blanks for the number of ways to do each task a slot diagram. 1st Task 2nd Task 3rd Task 4th Task 5th Task Number of Ways x Number of ways x Number of ways x Number of ways x

14 Example: A security keypad uses five digits (0 to 9) in a specific order. How many different keypad patterns are possible if any digit can be used in any position and repetition is allowed? 1st Task 2nd Task 3rd Task 4th Task 5th Task 10 x 10 x 10 x 10 x 10 any digit any digit any digit any digit any digit There are 10 5 = 100,000 possible keypad patterns. You try: How many different keypad patterns are possible if (b) the first and last digits can t be 0 and repetition is allowed? (c) The first two digits must be even and the last three must be odd with no repetition?

15 Handling Special Conditions - When counting, consider the special conditions first. Example: A choir director is assigning spots on the stage for choir members for an upcoming concert. The choir has 18 members. Sam and Julie have a duet during the concert, and must stand next one another in the front row. If there are eight spots in the front row, in how many ways can the director assign the singers to the front row? First, determine our tasks: Task 1 Assign Sam and Julie two spots next to one another. Task 2 Arrange Sam and Julie in the two spots. Task 3 Assign singers to the remaining six spots.

16 Example: The choir has 18 members. Sam and Julie have a duet during the concert, and must stand next one another in the front row. If there are eight spots in the front row, in how many ways can the director assign the singers to the front row? Task 1: Assign Sam and Julie two spots next to one another. If there are eight spots in a row, how many pairs of spots are there next to one another? Task 2: Arrange Sam and Julie in the two spots. Once two spots have been chosen, there are two ways to assign Sam and Julie to the spots.

17 Example: The choir has 18 members. Sam and Julie have a duet during the concert, and must stand next one another in the front row. If there are eight spots in the front row, in how many ways can the director assign the singers to the front row? Task 3: Assign singers to the remaining six spots. There are six remaining spots and 16 remaining singers from which we could fill them. Applying the fundamental counting principle, we have 16 x 15 x 14 x 13 x 12 x 11 = Apply the fundamental counting principle. She can seat the front row in 7 x 2 x = ways

18 Kristen is marrying Tim and they are working on seating arrangements for the head table. The bridal party consists of 4 bridesmaids (Lori, Jenn, Sarah, and Billy) and 3 groomsmen (Chris,Les, and Ben). The head table has nine seats in a row facing the wedding attendees. In how many ways can the bridal party be seated if (a) anyone can sit in any of the seats? (b) Kristen and Tim must sit together, but the rest of the bridal party can sit in any open seat? (c) Kristen and Tim must sit in the middle with the groomsmen next to Tim on the left and the bridesmaids next to Kristen on the right?

19 Permutations A permutation is an ordering of distinct objects in a straight line. If we select r different objects from a set of n objects and arrange them in a straight line, this is called a permutation of n objects taken r at a time. The number of ways to do this is denoted by P(n,r). P(n, r) Permutation total number of objects number of objects to be ordered

20 Permutations Example: Suppose you have 3 different colored marbles - (r)ed, (b)lue, and (y)ellow. How many permutations are there of the marbles? Write the answer using P(n,r) notation. Solution: Note, for example, rby and byr are two different permutations of the marbles. We can use a slot diagram to count the total number of permutations: 1 st marble 2 nd marble 3 rd marble 3 x 2 x 1 Use any color Can t repeat first color Can t repeat first two colors So, there are a total of 3 x 2 x 1 = 6 permutations of the marbles r,b,y. There are 3 marbles and we are taking 3 of them at a time, so we write P(3,3) = 6

21 Example: How many permutations of the letters a, b, c, d, e, f and g are there if we take 3 letters at a time? Solution: 1 st letter 2 nd letter 3 rd letter 7 x 6 x 5 Use any letter Can t repeat first letter P(7, 3) = 210 Can t repeat first two letters (7 total objects; permute 3 of them)

22 Factorial Notation If n is a counting number then the symbol n! is read n factorial and We define 0! = 1. Example: 4! = 4 x 3 x 2 x 1 = 24 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720

23 Factorials Compute each of the following:

24 Computing P(n,r) We can compute P(n,r) using the formula below:

25 Example: The 10-person party planning committee wishes to select one person to order food, a second to be in charge of decorations, and a third to handle music for their next party. In how many ways can the group fill these positions? Solution: This is a permutation of selecting 3 people from 10.

26 Example: The university French Club needs to select a President, Vice President, Treasurer, and Secretary from their 15 members. How many ways can they do this?

27 Combinations Suppose we don t care the order in which objects are selected. For example, suppose in the party planning example, we want to select three people to form a committee to plan the winter holiday party, but we won t assign specific jobs to each of those selected. Here the choice ABC would be equivalent to the choice BCA or CAB, etc. There are 6 = 3 x 2 x 1 ways of ordering A, B, and C. Thus the number 720 that we found in that example is too large we should divide this by 6. Therefore the number of three person committees we could choose is 720/6 = 120.

28 Combinations If we choose r objects from a set of n objects, we say that we are forming a combination of n objects taken r at a time. We say n choose r. The notation C(n,r) or denotes the number of such combinations. Combination number of objects to be chosen total number of objects

29 Combinations Example: (a) How many 4-element sets can be chosen from a set of 6 elements? (b) How many 3-person committees can be formed from a set of 12 people? Solution: (a) Because order is not important in choosing a 4-element set, we use the formula for C(n,r) with n = 6 and r = 4. You try (b).

30 When deciding if a problem requires a permutation or a combination, remember that in permutations the order matters, while in combinations order does not matter. (1) In a game of poker, five cards are drawn from a standard deck of 52 cards. How many different poker hands are possible? (2) In the game of Euchre, a hand consists of 5 cards drawn from a deck consisting of 9, 10, Jack, Queen, King, Ace of each suit. How many different Euchre hands are possible? (3) In a solitaire game, the player lays 13 cards selected from a standard deck out in a straight line on the table. How many different ways are there to lay the 13 cards out?

31 Combining Counting Methods Example: Three men and two women from our class will attend an extra study session. If there are 8 women and 10 men in the class, in how many ways can you choose the students to attend? Solution: The answer is not C(18, 5) since this includes options like five men and no woman being sent to the study session. Stage 1: Select the two women from the eight available. Stage 2: Select the two men from the ten available. Thus, choosing the women and then choosing the men can be done in 28 x 120 = 3360 ways.

32 Example: A charity organization has 20 members. They need to choose a president, vice-president, and secretary and they need to choose a four person executive board. Solution: We will count this using two tasks: (a) choosing the president, vice president, and secretary from the group, (b) choosing an executive board from the remaining members. Task 1: Choose the president, vice president, and secretary. Order matters, since we are assigning titles. This can be done in P(20, 3) ways. Stage 2: Select the executive board. Here, order doesn t matter, so this can be done in C(17, 4) ways, since after we choose the president, v.p., and secretary, there are 17 people from whom to choose the four member board. Total: P(20, 3) x C(17, 4) = 6840 x 2380 = 16,279,200 ways to assign the roles.

33 Example: The university will send 10 students to a conference on Journalism. They will send 5 of 10 Journalism majors, 2 of 8 English majors, and 3 of 9 Communications majors. In how many ways can they choose the 10 students to attend the conference?

34 The Basics of Probability Theory The probability of any outcome of a random phenomenon can be defined as the proportion of times the outcome would occur in a very long series of repetitions. For example, if I were to flip a fair coin 2 times, it wouldn t be too surprising if heads were turned up on all the flips. However, if I were to flip the coin hundreds of thousands of times, we would expect approximately half of the flips to be heads and half to be tails. The probability of heads is 0.5 and the probability of tails is 0.5. Coin Flip Example

35 In an experiment, the sample space is a list of all possible outcomes of the experiment. Example: Determine the sample space for the experiment of choosing a student from this class and determining the student s gender. Solution: The possible outcomes are male or female. The sample space is {male, female}. What if we randomly chose a student and determined their class rank? What would the sample space be then?

36 Sample space Important: It s the question that determines the sample space. A. A basketball player shoots three free throws. What are the possible sequences of hits (H) and misses (M)? H M H M H - M - H - M - HHH HHM HMH HMM S = {HHH, HHM, HMH, HMM, MHH, MHM, MMH, MMM } Note: 8 elements, 2 3 B. A basketball player shoots three free throws. What is the number of baskets made? S = {0, 1, 2, 3} C. A researcher designs a new maze for a young male white rat. What are the possible outcomes for the time to finish the maze (in minutes)? S = ( 0, ) = (all numbers > 0)

37 Sample Space Determine the sample space for the following experiments: (1)Three children are born to a family and we record the birth order with respect to gender. (2)We roll two six-sided dice and observe the pair of numbers showing on the faces. (Assume one die is red and one is green.) (3)We roll two six-sided dice and record the sum of the faces showing.

38 Events In probability theory, an event is a subset of the sample space. Example: In the example where we recorded the birth order according to gender, the sample space was S = {BBB, BBG, BGB, BGG, GGG, GGB, GBG, GBB}. The event that the family has exactly two girls is the subset E = {BGG, GBG, GGB} What is the event that the family has exactly one girl? What is the event that the family has at least one girl? Example: Write the event that a sum of five occurs on a pair of dice as a subset of the sample space for rolling two dice and recording their faces.

39 Probability The probability of an outcome in a sample space is a number between 0 and 1 inclusive. The sum of the probabilities of all the outcomes in a sample space must be 1. The probability of an event E, written P(E) is the sum of the probabilities of the outcomes in E. Example: A couple wants three children. What are the arrangements of boys (B) and girls (G)? Find the probability of an individual outcome in your sample space. Genetics tells us that the probability that a baby is a boy or a girl is the same, 0.5. Sample space: {BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG} All eight outcomes in the sample space are equally likely. The probability of each is thus 1/8.

40 Example: Give the sample space and probabilities of each individual outcome if a couple has 3 children and you record the number of girls they have. Solution: If they have 3 children, then they could have 0,1,2, or 3 girls, so the Sample Space = S = {0,1,2,3}. To find the probabilities for each of these outcomes, we will use the previous example: P(0) = P(BBB) = 1/8 P(1) = P({GBB, BGB, BBG}) = P(GBB) + P(BGB) + P(BBG) = 1/8 + 1/8 + 1/8 = 3/8 P(2) = P({GGB, GBG, BGG}) = P(GGB) + P(GBG) + P(BGG) = 3/8 P(3) = P(GGG) = 1/8 Note that P(S) = P(0)+P(1)+P(2)+P(3) = 1/8 + 3/8 + 3/8 + 1/8 = 1

41 Counting Probability when Outcomes are Equally Likely If E is an event in a sample space S with all equally likely outcomes, then the probability of E is given by the formula: where n(e) = the number of outcomes in E and n(s) = the number of outcomes in S Note: If all the outcomes in an event are not equally likely, you must add the probability of each outcome in the event to find P(E).

42 Example: (1) We flip three fair coins and record the sequence of faces up. What is the probability of each outcome in this sample space? (2) We draw a 5 card hand randomly from a standard deck of 52 cards. What is the probability that we draw a royal flush in hearts (A, K, Q, J, 10 all of hearts)?

43 Basic Properties of Probability The complement of an event E is the set of all outcomes not in E. It is denoted. Assume S is a sample space for some experiment and E is an event in S. Then

44 Example: Four friends belong to a 10-member club. Two members of the club will be chosen to attend a conference. What is the probability that two of the four friends will be selected? Solution: We can choose 2 of the 10 members in Event E, choosing 2 of the 4 friends, can be done in C(4, 2) = 6 ways. So,

45 Probability Trees We can use a tree to compute probabilities. We write the probability of the event on each branch and the outcome at the end of each branch. Example: Suppose you flip a coin three times. What is the probability of getting HHT? Begin

46 We multiply probabilities along the branches and add down the columns. So, P(HHT) = 0.5 x 0.5 x 0.5 = In fact, the probability of each outcome in the sample space for flipping a coin 3 times is If we add all the probabilities up, we get P(S) = P(HHH) + P(HHT) + P(HTH) + P(HTT) + P(THH) + P(THT) + P(TTH) + P(TTT) P(S) = 8 X = 1.0 Begin

47 Example: A box of 20 chocolates contains chocolates filled with caramel, fudge, and nougat. The side of the box says that it contains 10% caramel, 30% fudge, and 60% nougat. Suppose you select two chocolates. (a)draw a probability tree to display this experiment. (b)find the probability of drawing two fudge filled chocolates. (c)find the probability of drawing a caramel filled chocolate on the second draw. (d)if the first draw is caramel filled, find the probability that the second is nougat filled. (a) 2/20 6/20 C F 1/19 6/19 12/19 C F 2/19 5/19 12/19 N C F N 12/20 N 2/19 6/19 11/19 C F N 47

48 Example: A box of 20 chocolates contains chocolates filled with caramel, fudge, and nougat. The side of the box says that it contains 10% caramel, 30% fudge, and 60% nougat. Suppose you select two chocolates. (a)draw a probability tree to display this experiment. (b)find the probability of drawing two fudge filled chocolates. (c)find the probability of drawing a caramel filled chocolate on the second draw. (d)if the first draw is caramel filled, find the probability that the second is nougat filled. (a) 2/20 6/20 C F 1/19 6/19 12/19 C F 2/19 5/19 12/19 N C F N (b) (c) 12/20 N 2/19 6/19 11/19 C F N (d) This notation means the probability of getting N on the second draw, given C on the first draw. 48

49 Example: A box of 20 chocolates contains chocolates filled with caramel, fudge, and nougat. The side of the box says that it contains 10% caramel, 30% fudge, and 60% nougat. Suppose you select two chocolates. (e) What is the probability that you get at least one caramel in the two draws? 2/20 6/20 C F 1/19 6/19 12/19 C F 2/19 5/19 12/19 N C F N (e) We need to add up the probabilities of all possible combinations that include at least one C. Or, let E be the event that we draw at least one caramel. Then P(E) = 1 - P(not E). The outcomes in not E are FF, FN, NF, NN. 12/20 N 2/19 6/19 11/19 C F N P(not E) = P(FF) + P(FN) + P(NF) + P(NN) = 153/190 = So, P(E) = 1-153/190 = 37/190 =

50 Example: Annalise loves dessert. If Dad makes supper, there is a probability of 0.75 that she will; get dessert. If Mom makes supper, there is a 0.45 probability that she will get dessert. Mom cooks supper 4 nights a week and dad cooks supper 3 nights a week. Find the probability that Annalise will get dessert today. (Start by filling in the probabilities on the tree below.) dessert Dad no dessert dessert Mom no dessert 50

51 Example: A jar contains 8 red marbles, 10 blue marbles, and 4 green marbles. (a)what is the probability that you draw a red marble? (b)what is the probability that you draw a color other than red? (c)if you randomly draw 4 marbles, one at a time, what is the probability that all four of them are red? (Assume you don t replace the marbles in the jar.) (d)what is the probability of choosing a red marble and then a green marble? (Assume that you don t replace the marbles in the jar.)

52 Example Computer games in which the players take the roles of characters are very popular. These games use many different types of dice. A four sided-die has faces with 1,2,3, and 4 spots. (a)what is the sample space for rolling the die twice? (b)what is the assignment of probabilities to outcomes in this sample space? Assume that the die is perfectly balanced. (c)what is the probability of rolling a sum of 6? A sum of 4? Rolling anything but a sum of 6? (d)what is the probability of rolling two even numbers? (e)what is the probability the sum of the two dice is odd?

53 The Children Puzzle A couple has two children. One of them is a boy. What is the probability that the other one is also a boy? The Birthday Paradox How many people must you have in a room to ensure that the probability that two of them have the same birthday is at least 0.5? Birthday Paradox Simulation

54 Statistics Topics Organizing data population vs sample measures of center and spread normal distributions, z-scores confidence intervals

55 Organizing Data In statistics, we want to collect data (numerical information) to describe a population. It is often difficult to collect information from everyone in the population, so we use a subset of the population, called a sample. We want the sample to be as representative of the population as possible. We can display the numerical data we collect in a histogram.

56 State Percent Alabama 21.4 Alaska 26.0 Arizona 25.3 Arkansas 19.3 California 29.5 Colorado 35.0 Connecticut 34.7 Delaware 26.1 Florida 25.8 Georgia 27.1 Hawaii 29.2 Idaho 24.5 Illinois 29.5 Indiana 22.1 Iowa 24.3 Kansas 28.8 Kentucky 20.0 Louisiana 20.4 Maine 26.7 Maryland 35.2 Massachusetts 37.9 Michigan 24.7 Minnesota 31.0 Mississippi 18.9 Missouri 24.5 Montana 27.0 Nebraska 27.5 Nevada 21.8 New Hampshire 32.5 New Jersey 33.9 New Mexico 24.8 New York 31.7 North Carolina 25.6 North Dakota 25.7 Ohio 24.1 Oklahoma 22.8 Oregon 28.3 Pennsylvania 25.8 Rhode Island 29.8 South Carolina 23.5 South Dakota 25.0 Tennessee 21.8 Texas 25.2 Utah 28.7 Vermont 33.6 Virginia 33.6 Washington 30.3 West Virginia 17.3 Wisconsin 25.4 The range of values that a variable can take is divided into equal-size intervals. Histograms Number of States The histogram shows the number of individual data points that fall in each interval. Percent of adults aged 25 or older with a Bachelor s degree or higher, 2007 Percent The first column represents the number of states with % of adults with at least a bachelor s degree. There are three such states. You can see from the histogram that there are 5 states with % of adults with at least a bachelor s degree. Wyoming 23.4 Source:

57 Remarks You must specify whether your bins contain the right or left endpoints. There is no right choice of bins or classes in a histogram too few give a skyscraper graph; too many give a pancake shaped graph. The purpose of graphs is to help us understand the data. Things to Look For Overall pattern (shape, center, and spread) and deviations from that pattern. Outliers individual values that fall outside of the overall pattern.

58 Example: Suppose 25 students take a math exam and earn the following scores. Create a histogram to display the distribution of scores: 75, 62, 81, 88, 95, 97, 79, 72, 93, 67, 85, 77 91, 87, 75, 89, 83, 81, 90, 81, 72, 73, 68, 86, 76 Solution: First make a table to count the number of scores that go in each bin (range of scores). Range of scores Number of scores 60 to to to to 100

59 Example: To the right are the travel times to work for several Ohio counties. Describe the center and spread of this distribution. Mean Travel Time to Work of Workers 16 Years and Over Who Did Not Work at Home ACS Census 2000 County Name Estimate Estimate Allen County, Ohio Ashland County, Ohio Brown County, Ohio Butler County, Ohio Carroll County, Ohio Clinton County, Ohio Darke County, Ohio Erie County, Ohio Holmes County, Ohio Huron County, Ohio 20

60 Measuring Center: Mean The mean of a set of observations x 1,x 2,...,x n is or Mean Travel Time to Work ACS Census 2000 County Name Estimate Estimate Allen County Ashland County Brown County Butler County Carroll County Clinton County Darke County Erie County Holmes County Huron County 20 Example: The mean travel time to work in 2000 for Ohio counties was Find the mean travel time during

61 Measuring Center: the Median M The median is the midpoint of a distribution. When n is odd, the median is exactly the middle observation when the observations are ordered from smallest to largest ACS When n is even, the median is the mean of the two middle observations when the observations are ordered from smallest to largest Census

62 Measuring Spread: the Quartiles The quartiles Q 1 and Q 3 : 1. Arrange the observations in increasing order and find the median M. 2. The first quartile Q 1 is the median of the observations to the left of the overall median. 3. The third quartile Q 3 is the median of the observations to the right of the overall median ACS: Q 1 = (19+22)/2 = 20.5 Q 3 = (25+30)/2 = Census: You find Q 1 and Q 3 for the 2000 census.

63 The Five-Number Summary The five-number summary of a distribution is Min Q 1 M Q 3 Max ACS: Census: The five-number summary for the travel time to work in is What s the five-number summary for the travel time to work in 2000?

64 Example: Find the mean and the five number summary for the data sets below: (a) 8, 11, 23, 14, 15, 4, 3, 9, 10 (b) 24, 42, 45, 22, 18, 35, 33, 36

65 Measuring spread: The Standard Deviation The standard deviation is used to measure the spread of the data about the mean. If x 1, x 2,..., x n is the data you ve collected, the standard deviation is defined as follows: Example: We can find the standard deviation using our calculators by entering the data in a list and using the stat functions on the calculator. Find the standard deviation for the data set 8, 11, 23, 14, 15, 4, 3, 9, 10.

66 Choosing among summary statistics Because the mean is not resistant to outliers or skew, use it and the standard deviation to describe distributions that are fairly symmetrical and don t have outliers. Use the median and the five number summary otherwise.

67 Example: Here are the amounts of money (cents) in coins carried by 10 students in a stats class: The mean of the data is: (a) 37.2 (b) 42.5 (c) The median of the data is (a) 35 (b) 42.5 (c) The five number summary of the data is (a) 0, 0, 42.5, 76, 97 (b) 0, 29, 57.5, 81.5, 97 (c) 0, 29, 42.5, 75, 97

68 To the left is a histogram displaying the length of housefly wing spans for 100 flies, in millimeters. Note the shape of the histogram is bell-shaped. We say the data is approximately Normal. We can draw a Normal curve to approximate the shape of the data. What proportion of flies have a wing span greater than 48 mm? ( )/100 = 0.31 of the flies have wingspan greater than 48 mm. 68 source:

69 What proportion of flies have a wing span greater than 48 mm? ( )/100 = 0.31 of the flies have wingspan greater than 48 mm. The area under the curve to the right of 48 is

70 Features of Normal Curves Symmetric, single peaked, bell-shaped, with the highest point of the curve at the mean. Completely described by the mean µ and the standard deviation σ. we use µ to denote the mean and σ to denote the standard deviation because these numbers describe the population, not the sample the mean is equal to the median the total area under the curve is 1

71 The Rule 99.7% of scores 95% of scores 68% of scores µ 3σ µ 2σ µ σ µ µ + σ µ + 2σ µ + 3σ In a Normal distribution with mean µ and standard deviation σ, Approximately 68% of the data values fall within 1σ of µ. Approximately 95% of the data values fall within 2σ of µ. Approximately 99.7% of the data values fall within 3σ of µ.

72 Example: The distribution of scores on the ACT in a recent year were normally distributed with mean 20.8 and standard deviation 4.8. If 1000 incoming students took the test, how many students would we expect to have scores between 11.2 and 30.4? How many scores would we expect to be above 25.6? 95% of scores The Rule tells us that 95% of scores will be between 11.2 and 30.4, so we would expect 0.95 x 1000 = 950 scores to be in this range.

73 68% of scores We know that 68% of scores are between 16 and This means that = 32% of scores are below 16 or above By symmetry, we have that 32/2 = 16% of scores are above So we d expect 0.16 x 1000 = 160 scores to be above

74 Example: A machine fills 12 oz cans of soda. Suppose the fills are normally distributed with mean 12.1 and standard deviation 0.2. (a)what weight of soda cans covers 99.7% of the distribution? (b)what percent of soda cans weigh less than 11.9 ounces?

75 z-scores The standard normal distribution has a mean of 0 and a standard deviation of 1. There are tables that give the area under this curve between the mean and a number called a z-score. A z-score represents the number of standard deviations a data value is from the mean. For example, for a normal distribution with mean 20.8 and standard deviation 4.8, the value 25.6 is 1 standard deviation above the mean; that is, the value 25.6 corresponds to a z-score of 1. Table of z-scores

76 Example: Use a table to find the percentage of the data (area under the curve) that lie in the following regions for a standard normal distribution: a) between z = 0 and z = 1.25 b) between z = 1.52 and z = 2.15 c) between z = 0 and z = 1.83 Solution: (a) We first draw a picture: We want this area

77 (a) To find the area between z=0 and z=1.25, we find z=1.25 in the table. Note the first column gives the integer part and the first decimal of z and the top row gives the second decimal place for z. We find that the area between z = 1.25 and z=0 is So the percentage of z-scores between 0 and 1.3 is 39.44% is the area under the curve between z=0 and z= is the area under the curve between z=0 and z=0.36

78 (b) The area between z = 1.52 and z = 2.15 is the area between z = 2.15 and z = 0 minus the area between z = 1.52 and z = 0. = z=1.52z=2.15 z=2.15 = = z= So, about 4.85% of the data has a z-score between 1.52 and 2.15.

79 (b) By the symmetry of the normal distribution, the area between z = 0 and z = 1.83 is the same as the area between z = 0 and z = = z = 1.83 z = 1.83 =

80 Converting raw scores to z - scores Assume a normal distribution has a mean of µ and a standard deviation of σ. Then the z-score for a value x in the distribution is given by the equation

81 Example: Suppose the mean of a normal distribution is 30 and its standard deviation is 5. a) Find the z-score corresponding to the raw score 21. b) Find the z-score corresponding to the raw score

82 Example: Suppose you take a standardized test. Assume that the distribution of scores is normal and you received a score of 72 on the test. The distribution of all scores had a mean of 65 and a standard deviation of 4. What percentage of those who took this test had a score below yours? Solution: We first draw a picture of the shaded area we want. µ = 65 x =72

83 Next, we find the z-score for 72: Now, we use the z-score to find the area we want. We know that the area to the left of 0 in the standard normal distribution is 0.5 (since the curve is symmetric and the total area is 1). The area between z = 0 and z = 1.75 is So, the area to the left of 1.75 is = So there are 95.99% of the scores below a score of 72.

84 Example: The scores of high school seniors taking the SAT in 2015 are approximately Normal with mean Mathematics score 511 and standard deviation 120. The mean Critical Reading score was 495 with standard deviation 116. What proportion of students scored above 392 on the Mathematics portion of the SAT? 1 Draw a picture = + z = 0.99 z = Find the z-score: 3 Use the Table The area between z = 0.99 and 0 is the same as the area between z=0 and z=0.99, which is The area from z =0 to the right is 0.5, so the total area we want is = Conclude About 83.89% of seniors scored above 392. What percent of students scored between 392 and 650?

85 Finding a Value given a Percentage Example: SAT combined math and reading scores follow approximately the Normal distribution with mean 1006 and standard deviation 209. How high a score must a student to place in the top 10% of all students taking the SAT? 1 Draw a picture The area between 0 and z is = Use the table to find z We look in the table for the area closest to This is the entry corresponding to z = area = z area = Solve for x in the z-score equation 4 Conclude A student must score above 1273 to be in the top 10%.

86 Example: The thorax lengths in a population of male fruit flies follow a Normal distribution with mean mm and standard deviation mm. What are the median and the first and third quartiles of thorax length?

87 Confidence Intervals A level C confidence interval for the mean of a population has two parts An interval calculated from the data of the form estimate margin of error A confidence level C, which gives the probability that the interval will capture the true mean in repeated samples. That is, the confidence level is the success rate for the method.

88 Critical values and Confidence Intervals Numbers like z* that mark off specified areas are called critical values. area = C -z* 0 z* Most common confidence intervals: Confidence level C Critical value z* 90% 95% 99% Forming a Confidence Interval for the mean of a Normal Population: Draw a simple random sample of size n from a Normal population having unknown mean µ and known standard deviation σ. A level C confidence interval for µ is is called the margin of error.

89 Finding Critical Values z* To find the critical value z* for a level C confidence interval area = C -z* 0 z* 1. Write C as a decimal and find. This is the area between z = 0 and z* on the standard normal curve. 2. Find the z-score in your table with area closest to between it and z=0. This is your z*.

90 Example: Find the critical value z* for the following confidence levels: (a)a 82% confidence interval (b)a 67% confidence interval Solution: (a) First convert 82% to the decimal Then C/2 = We ll look for the area 0.41 in the table.

91 is closest to 0.41, so z* = is closest to =.67/2, so z* = 0.97 for a 67% C.I.

92 We say that we are 95% confident that the true mean score for the exam is in the interval to

93

94 The margin of error is the term the margin of error is, so

95 The Margin of Error The margin of error for a confidence interval is How could do we get a small margin of error? Make z* smaller. This makes the confidence level C. Make n larger. Larger sample sizes give more accurate estimates. Have a small standard deviation σ.

96 Changing the margin of error Example: The weight of raspberries from a U-pick farm is known to be normally distributed with a standard deviation of σ = 1.3 grams. A sample of 24 raspberries had a mean weight of 4.8 grams. Give a 90% and a 99% confidence interval for the mean weight of raspberries on the farm. How do the margins of error compare for your confidence intervals?

97 What does the margin of error include? - The margin of error in a confidence interval ignores everything except the sample-to-sample variation due to choosing the sample randomly. The margin of error doesn t cover all errors - The margin of error in a confidence interval covers only random sampling errors. - Practical difficulties like undercoverage and nonresponse are often more serious than random sampling error. The margin of error doesn t take these difficulties into account.

98 Sample size and experimental design You may need a certain margin of error (e.g., drug trial, manufacturing specs). In many cases, the population variability (σ) is fixed, but we can choose the number of measurements (n). So plan ahead what sample size to use to achieve that margin of error. Remember, though, that sample size is not always stretchable at will. There are typically costs and constraints associated with large samples. The best approach is to use the smallest sample size that can give you useful results.

99 Example: Suppose we are calculating a 95% confidence interval for the mean height of adult male residents of St Joseph. The mean height follows a Normal distribution with standard deviation σ = 3.94 in. (a) How large a sample would be needed to estimate the mean height µ in this population to within 1 with 95% confidence? (b) Suppose you want a margin of error no more than 0.5 in. How big a sample must we take to ensure the margin of error is this small?

100 z Z