FINAL REVIEW W/ANSWERS

FINAL REVIEW W/ANSWERS ( 03/15/08 - Sharon Coates) Concepts to review before answering the questions: A population consists of the entire group of people or objects of interest to an investigator, while a sample refers to the part of the population that the investigator actually studies. In certain contexts, a population can also refer to a process (such as flipping a coin or manufacturing a candy bar) that in principle can be repeated indefinitely. With this interpretation of population, a sample is a specific collection of process outcomes; e.g., the number of heads in n tosses or the weights of n candy bars. A parameter is numerical characteristic of a population, while a statistic is a numerical characteristic of a sample. Examples: (population) parameter (sample) statistic mean (average) μ x standard deviation σ s proportion (of successes ) p A statistic from a random sample or a randomized experiment is a random variable. The probability distribution of the statistic is its sampling distribution. The binomial distributions are an important class of discrete probability distributions. The important skill for using binomial distributions is the ability to recognize situations to which they do and don t apply. In the binomial setting, the count X has a binomial distribution. The count takes whole numbers between 0 and n. If both np and n(1 p) are at least 10, probabilities about the count X can be approximated by the appropriate normal distribution. The proportion is always a number between 0 and 1. The proportion does not have a binomial distribution, but we can do probability calculations about by restating them in terms of the count X and using binomial methods. The sampling distribution of is approximately normal if np and n(1 p) are both at least 10. Question 1. Consider a game in which a red die and a blue die (assumed fair dice) are rolled. Let the random variable X represent the value showing on the up face of the red die. Since X is a discrete random variable, its probability distribution is tabulated: X 1 2 3 4 5 6 P(X) 1/6 1/6 1/6 1/6 1/6 1/6

Let the random variable Y represent the value showing on the up face of the blue die. In parts a through f below, give values. a. The mean of the probability distribution of X is. b. The standard deviation of the probability distribution of X is. c. Shortened version of question) The mean of Y is. d. The standard deviation of Y is. Suppose that you are offered a choice of the following two games: Game 1: costs $7 to play. You win the sum of the values showing on the up faces of the red and blue die. Game 2: doesn t cost anything to play initially, but you win 4 times the difference of the up faces; i.e., 4(X Y). If 4(X Y) is negative, you must pay that amount. If 4(X Y) is positive, you receive that amount. Net amount = total winnings cost to play. e. For Game 1, the net amount won in a game is the random variable X + Y 7. (1 pt) The mean of (the probability distribution of) X + Y 7 is $. (1 pt) The standard deviation of X + Y 7 is $. f. For Game 2, the net amount won in a game is 4(X Y) or 4X 4Y. (1 pt) The mean of (the probability distribution of) 4(X Y) is $. (1 pt) The standard deviation of 4(X Y) is $. g. (2 pts) Based on your responses to parts e and f above, if you had to play and you like excitement, which game should you choose and why? Question 2. Girls Get Higher Grades was a headline from USA Today, Aug 12, 1998. Eighty percent of the girls in grade 12 surveyed said it was important to them to do their best in all classes, whereas only 65% of the boys in grade 12 surveyed responded this way. Suppose you take a SRS of 20 female students and another sample of 20 male students in your region of the country. 2

a. To find the probability that more than half of the girls in the sample want to do their best in every class, organize your response: In a complete, English sentence, define the random variable in this context. What is an appropriate symbol for the random variable? Complete the mathematical sentence to find the probability by replacing event with an inequality involving the random variable, connecting (with an equals sign) the appropriate distribution function, and rounding the resulting probability to the nearest ten-thousandth: P(event) = b. What conditions would have to be satisfied in order to use your approach? Question 3. Airlines routinely overbook popular flights because they know that not all ticket holders show up. If more passengers show up than there are seats available, an airline offers passengers $100 and a seat on the next flight. A particular 120-seat commuter flight has a 10% no-show rate. a. Suppose the airline sells 130 tickets for the flight. If a ticketed passenger is a trial, how many trials are there? A ticketed passenger either shows up or doesn t show up for the flight. Assume that there are no groups traveling together. What is the probability that an individual ticketed passenger will show up for the flight? On average, how many ticketed passengers will show up for a 120-seat commuter flight? When does the airline have to make a payout on a 120-seat commuter flight? 3

Let the random variable Y represent the amount of the payout if the airline sells 130 tickets for the flight. In the space provided, create the probability distribution of the discrete random variable Y where the values of Y are in the first row of the table and the corresponding probabilities P(Y) are in the second row. A table is appropriate because Y is discrete. Make the table horizontal to save space. How much money does the airline expect to have to pay out per flight? Hint: the expected value of the payout is the mean of the probability distribution of payout amounts based on the sale of 130 tickets. b. To calculate how many tickets the airline should sell per flight if they want their chance of giving any $100 amounts to be about 0.05, organize your response: Let the random variable X count the number n of ticketed passengers that will show up for the 120-seat commuter flight. The range of values of X is 0 to n. In terms of X, the airline has to make a pay out on a 120-seat commuter flight when. If the airline sells n = 135 tickets, will the probability that the airline will have to make a pay out be only about 0.05? To find out, complete the mathematical sentence by replacing event with an inequality involving the random variable, connecting the appropriate distribution function (with an equals sign), and rounding the resulting probability to the nearest ten-thousandth: P(event) = What conditions would have to be satisfied in order to use the distribution function that you selected? 4

You want to find out what n should be so that the chance the airline will have to make a pay out is only about 5%. Use trial and error to obtain n, using your result above to give you some direction. Then complete the sentence below by replacing event with an inequality that involves the random variable. After the equals sign, give the appropriate distribution function and let the result verify that the probability of a pay out is about 5%: The airline can sell tickets for the 120-seat commuter flight because P(event) = Question 4. The process for manufacturing a ball bearing results in weights that have an approximately normal distribution with mean 0.15 g and standard deviation 0.003 g. Define the random variable in each case. Start the mathematical sentence P(event) where the event involves the random variable. Include the appropriate calculator distribution function in the sentence. a. If you select 1 ball bearing at random, what is the probability that it weighs less than 0.148 g? b. If you select 4 ball bearings at random, what is the probability that their average (mean) weight is less than 0.148g? c. If you select 10 ball bearings at random, what is the probability that their average (mean) weight is less than 0.148 g? 5

Question 5. Suppose that a particular candidate for public office is in fact favored by 48% of all registered voters in the district. A polling organization will take a random sample of 500 voters and will use the sample proportion probability that result of the upcoming election? to estimate the true proportion p. What is the approximate will be greater than 0.5, causing the polling organization to incorrectly predict the Question 6. Two business associates have opened a wine bottling factory. The machine that fills the bottles is fairly precise. The distribution of the number of ounces of wine in the bottles is normal. Although σ, the true population standard deviation, is not usually known, suppose for illustrative purposes that σ = 0.06 oz. The mean μ is supposed to be 16 oz, but the machine slips away from that occasionally and has to be readjusted. The associates take a random sample of 10 bottles from today s production run and weigh the wine in each. The weights are: 15.91 16.08 16.08 15.94 16.02 15.94 15.96 16.03 15.82 15.96 Should the machine be adjusted? Use a 95% confidence interval to estimate μ in order to support your advice. Question 7. Heights of women in the U.S. are normally distributed and have a standard deviation of about 2 inches. Suppose you want to estimate the mean height μ of all the women in your city from a random sample of size n at the 90% confidence level. What minimum sample size would you need in order to estimate the mean height μ to within: a. two inches? b. one-half inch? 6

Answers: Question 1. a. The mean of the probability distribution of X is 3.5. b. The standard deviation of the probability distribution of X is 1.70783. c. (Shortened version of question) The mean of Y is 3.5. d. The standard deviation of Y is 1.70783. e. The mean of (the probability distribution of) X + Y 7 is: $0.00. Fast way : μx + Y 7 = μx + μy 7 = 3.5 + 3.5 7 = 0 The standard deviation of X + Y 7 is: $2.415 approx. 2 2 2 2 Fast way: σ X + Y 7 = σ X + σy = 1.70783 + 1.70783 2.4152 The fast way is used to calculate the standard deviation of X + Y 7 because X and Y are independent random variables: the outcome of one die does not influence the outcome of the other die. Slow way: Give the probability distribution of X + Y. X + Y 2 3 4 5 6 7 8 9 10 11 12 P(X + Y) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 Then give the probability distribution of the net winnings $(X + Y -7): X + Y - 7 -$5 -$4 -$3 -$2 -$1 $0 $1 $2 $3 $4 $5 P(X + Y - 7) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 Enter the first row values of the distribution of net winnings in List 1 (L1). Enter the second row values into List 2 (L2). Calculate 1-Var Stats on L1, L2 or List 1 with Freq: List 2, depending on your Ti. The mean μ and the standard deviation σ will match the results above. (The symbol x is replaced with the appropriate symbol μ.) f. The mean of (the probability distribution of) 4(X Y) is: $0.00. Fast way : μ4x 4Y = μ4x μ4y = 4 μx 4 μy = 4(3.5) 4(3.5) = 0 The standard deviation of 4(X Y) is: $9.661 approx. ( ) ( ) 2 2 2 2 2 2 2 2 Fast way: σ4x 4Y = 4 σx + 4 σy = 4 1.70783 + 4 1.70783 9.66095 7

The fast way is used to calculate the standard deviation of 4(X Y) because X and Y are independent random variables: the outcome of one die does not influence the outcome of the other die. Slow way: Give the probability distribution of X - Y. X - Y 0 1 2 3 4 5-1 -2-3 -4-5 P(X - Y) 6/36 5/36 4/36 3/36 2/36 1/36 5/36 4/36 3/36 2/36 1/36 Then give the probability distribution of the net winnings $(4X 4Y): 4(X - Y) $0 $4 $8 $12 $16 $20 -$4 -$8 -$12 -$16 -$20 P(4X 4Y) 6/36 5/36 4/36 3/36 2/36 1/36 5/36 4/36 3/36 2/36 1/36 Then find μ and σ with your calculator. g. A gambler should choose Game 2 because although both games winnings in the long run average out to $0, there is more variability in the distribution of the net winnings in Game 2 because the standard deviation exceeds the standard deviation of the distribution of net winnings in Game 1. Variability indicates that the gambler could win a lot of money, but he could also lose a lot of money too. There is greater risk involved in Game 2 than in Game 1 and that is the exciting part for a gambler. Question 2. a. The random variable is the count in the sample of 20 girls in grade 12 who would say they want to do their best in every class. Note: The conditions for approximating the distribution of the sample proportion who want to do their best are not met because n(1 p) is only 4, not at least 10. So use the count X, not the proportion as the random variable in this scenario. The random variable could be symbolized by X (the count of those with the trait of interest in the sample). Ti 89/Voy 200: ( ) The math sentence is: P(X 11) = binomialcdf 20, 0.8, 11, 20 0.9974. Ti 83 family/ti 86: ( ) The math sentence is: P(X 11) = 1 - binomialcdf 20, 0.8, 10 0.9974. b. The Binomial Setting is satisfied: There is a fixed number of trials (attitudes): n = 20. 8

Each trial results in one of two outcomes: a success (the girl says that she wants to do her best in all her classes) or a failure (the girl doesn t say that she wants to do her best in all her classes). The sample must be collected so that one girl in the sample couldn t influence another girl in the sample, so that the trials are independent or the population size > 20 times n. The probability of a success is constant from trial to trial: p = 0.8. The count X has a binomial distribution with n = 20 and p = 0.8. The shape of this binomial distribution of successes is not roughly normal because the sample size n fails to meet the size requirement: if both np and n(1 p) are at least 10, then the distribution of the count X is approximated by a normal curve. However: np = 20(0.8) = 16, but n(1 p) = 20(0.2) = 4, therefore the distribution function normalcdf( ) to estimate the probability should not be employed. Question 3. a. n = 130 trials p = 1 0.10 = 0.90 On average, np = 130(0.90) = 117 ticketed passengers will show up for the flight. The airline makes a pay out when the number of ticketed passengers is at least 121. The distribution of the amount of payout Y is: Payout Y $0 $100 $200 $300 $400 $500 $600 $700 $800 $900 $1000 P(Y).8479.064.0425.0249.0126.0055.0019.00055.000117.000016.000001 If X represents the number of ticketed passengers out of 130 that will show up for the 120-seat commuter flight, then X and Y are related: Ti 89: P(Y = 0) = P(X < 120) = binomialcdf(130, 0.9, 0, 120) 0.8479 Ti 83/Ti 86: P(Y = 0) = P(X < 120) = binomialcdf(130, 0.9, 120) 0.8479 All Ti calculators use binomialpdf( ) to complete the table: P(Y = 100) = P(X = 121) = binomialpdf(130, 0.9, 121) 0.064 to P(Y = 1000) = P(X = 130) = binomialpdf(130, 0.9, 130) 0.000001. Enter the values of Y into List 1 and the corresponding probabilities in List 2. Calculate 1-Var Stats on List 1, List 2: The airline expects to pay out about $31.79 per flight, on average. The airline has to make a pay out on a 120-seat flight when X > 121. 9

If n = 135 tickets are sold, the probability that the airline will have to make a payout is better than 62%, an unwise policy: Ti 89/ Ti 92Plus/ Voy 200: P(X 121) = binomialcdf(135, 0.90, 121, 135) 0.6262 Ti 83 family/ Ti 86: P(X 121) = 1 - binomialcdf(135, 0.90, 120) 0.6262 The Binomial Setting is satisfied: i. Fixed number of trials (ticketed passengers): n = 135. ii. Each trial results in one of two outcomes: a success (a ticketed passenger shows up) or a failure (a ticketed passenger does not show up). iii. The trials are independent if there are no family groups; e.g., no couples traveling together. iv. The probability of a success (a ticketed passenger shows up for the flight) remains constant from trial to trial where p = 0.90. v. The random variable X that counts the number of ticketed passengers that show up for the flight has a binomial distribution with n = 135 and p = 0.90. vi. The normal approximation to the binomial distribution probably can be used because np = 135(0.90) = 121.5 and n(1 p) = 135(0.10) = 13.5. But using the normal approximation to the binomial distribution gives a result that underestimates the probability: P(X 121) normalcdf ( 121,, 121.5, 135(0.90)(0.10) ) 0.5570 The airline can sell 128 tickets for the 120-seat commuter flight and the chance they will have to make any payout is about 5% because: Ti 89/Ti 92Plus/ Voy 200: P(X 121) = binomialcdf(128, 0.90, 121, 128) 0.05. Ti 83 family/ti 86: P(X 121) = 1 - binomialcdf(128, 0.90, 120) 0.05. Question 4. a. The random variable X is the weight of an individual ball bearing. X is N(0.15, 0.003). 0.148-0.15 z = - 0.667 0.003 P( Z < - 0.667) = normalcdf(-, - 0.667, 0, 1) = 0.2524 approx. Faster way. You can find the probability without standardizing first: P(X < 0.148) = normalcdf(-, 0.148, 0.15, 0.003) 0.2525 10

Interpretation. If you select an individual ball bearing at random, the chance it will weigh less than 0.148 g is about 25%. b. The random variable x is the mean weight of a sample of n = 4 ball bearings. The statistic x is N(0.15, 0.003 ). 4 z = 0.148-0.15 0.003 4-1.333 P( Z < - 1.333) = normalcdf(-, - 1.333, 0, 1) = 0.0913 Faster way. You can find the probability without standardizing first: P( x < 0.148) = normalcdf(-, 0.148, 0.15, ) 0.0912 0.003 4 Interpretation. If you average the weight of 4 ball bearings, the chance that the average will be less than 0.148 g is about 9%. c. The random variable x is the mean weight of a sample of 10 ball bearings. The random variable x is N(0.15, 0.003 ). 10 Fast way. P( x < 0.148) = normalcdf(-, 0.148, 0.15, ) 0.0175 0.003 10 Interpretation. There is less than a 2% chance that the weight averaged over 10 ball bearings will be less than 0.148 g if the true mean weight is 0.15g. Sample averages are more stable (tend to be closer to the population average) than individual measurements. The larger the sample size n, the more stable the sample mean: the Law of Large Numbers states that the average value observed in many trials must approach μ. Question 5. The value of the parameter p = 0.48. Since np = 500(0.48) > 10 and n(1 p) > 10, the distribution of the random variable, the statistic, is approximately ( ) 0.48(0.52) N 0.48,. 500 0.48(0.52) ( ˆ ) ( 500 ) P p 0.5 = normalcdf 0.5,, 0.48, 0.1854 Interpretation. If in fact 48% of all registered voters in the district actually favor the candidate, there is about a 19% chance that a random sample of 500 will incorrectly predict the result of the upcoming election by concluding that better than 50% of all registered voters favor the candidate. 11

Question 6. The mean of the sample of size n = 10 is x = 15.974 oz. A 95% confidence interval to estimate the true mean weight of the wine is: 0.06 x ± z* σ x 15.974 ± 1.96 15.94 oz, 16.01 oz 10 ( ) Interpretation. With 95% confidence the true mean weight μ of the wine from today s production run lies between 15.94 oz and 16.01 oz. Since 16 oz falls within the interval, the machine doesn t need adjusting yet. Does the true mean weight of the wine μ actually lie between these endpoints? We have no way of knowing. The confidence we have is in the method we used, not the results. If we took 100 random samples and constructed a 95% confidence interval for each sample s mean x, then we could expect that 95 of the intervals would capture the true mean and 5 would not, but we would not know specifically which ones did and which ones did not. Question 7. a. To estimate the true mean height μ of the women in your city to within 2 inches with 90% confidence, the smallest random sample that will accomplish this goal is : n 2 2 z * σ 1.645(2) = = desired margin of error 2 3 women b. To estimate the true mean height μ of the women in your city to within one-half inch with 90% confidence, the smallest random sample that will accomplish this goal is: n 2 2 z * σ 1.645(2) = = desired margin of error 0.5 44 women 12