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New Zealand Association of Mathematics Teachers 1 Level 1 Mathematics and Statistics 91037 (1.12): Demonstrate understanding of chance and data Example Task B 2011 Credits: Four Check that you have been supplied with the resource booklet for Mathematics and Statistics 91037 (1.12). You should answer ALL parts of ALL questions in this booklet. You should show ALL working. If you need more space for any answer, use the page(s) provided at the back of this booklet and clearly number the question. Check that this booklet has pages 2–7 in the correct order and that none of these pages is blank. YOU MUST HAND THIS BOOKLET TO YOUR TEACHER AT THE END OF THE ALLOTTED TIME. For Assessor’s use only Achievement Demonstrate understanding of chance and data. Achievement Criteria Achievement with Merit Demonstrate understanding of chance and data, justifying statements and findings. Achievement with Excellence Demonstrate understanding of chance and data, showing statistical insight. Overall Level of Performance NZAMT 91038 (1.12): Demonstrate understanding of chance and data Example Task B 2011 – page 1 of 12 You are advised to spend 60 minutes answering the questions in this booklet. QUESTION ONE Invest in Easy-to-win LOTTO Easy-to-win LOTTO has only four balls, painted with the numbers 1 to 4. To enter, pay $2 and choose two numbers. You win if we draw your two numbers (in any order) from our lotto bag. WIN $10 It’s so easy to win – half the numbers are winners - just like tossing a coin! Invest $2 and get 5 times that amount back! OR Try BIG-WIN LOTTO: 5 balls numbered 1 to 5 Pay $2 to enter, choose two balls, and win $20 if they are correct Evaluate the claims made in the advertisement for “Easy-to-win LOTTO”. In your answer you may find it helpful to comment on: What is the purpose of the advertisement? Are the claims made valid and sensible? Or can alternative/or additional claims be made? Would you recommend playing these games? NZAMT 91038 (1.12): Demonstrate understanding of chance and data Example Task B 2011 – page 2 of 12 NZAMT 91038 (1.12): Demonstrate understanding of chance and data Example Task B 2011 – page 3 of 12 QUESTION TWO Each school day you and your three friends have a treat from the canteen. Each time someone buys the treats for all four of you. Who is going to buy the four treats is decided by the following method: Everyone starts with a coin. (a) Each person tosses their coin. If everyone has the same (all heads, all tails) or if there are two heads and two tails, then everyone tosses again. If there is three heads and one tail, or if there is three tails and one head, then the person whose coin is different from everyone else’s is the “odd one out” and must they buy the treats for everyone that day. What is the probability that an “odd one out” will be found on the first coin toss? (b) What is the probability that you will have to buy the treats after the first coin toss? NZAMT 91038 (1.12): Demonstrate understanding of chance and data Example Task B 2011 – page 4 of 12 (d) One person leaves the group. You decide to change the game for 3 players, but you want it to last longer. OPTION 1 (3 coins) You still have a single coin each, but toss it twice. If there is a player who is “odd one out” in BOTH tosses, then they must buy the treats. If there is no one who is “odd one out” in the two consecutive throws the coins are tossed again. OPTION 2 (6 coins) You could have two coins each, toss them all at once. If there are 5 heads and 1 tail, or if there are 5 tails and 1 head, then the person with the “odd one out” is the one who buys the treats. If there is no “odd one out” all the coins are tossed again. Which game would be likely to last longer? Show all your calculations to justify your answer? NZAMT 91038 (1.12): Demonstrate understanding of chance and data Example Task B 2011 – page 5 of 12 QUESTION THREE Alcohol Heath Watch Director Rebecca Williams says that while New Zealand’s alcohol marketing in the past was predominantly focused on beer and aimed at men, it has switched to mixes and now young women are being targeted. The number of girls going to the Emergency Department as a result of binge drinking has increased 30% over the last four years. Rebecca suspects the problem is now as bad for females as it is for males. For the past two years Rebecca has been collecting data. Every month she records the number who come to the Emergency Department with problems associated with binge drinking. She calculated the mean number of girls presenting each month as 19.3 and the mean number of males was 21.9. A graph showing the data is given below. Monthly Number of arrivals at the Emergency Department as a result of Binge Drinking Rebecca concludes that males are still having more binge drinking problems than females. E valuate Rebecca’s conclusion. Make at least FOUR evaluative statements. You may comment on the data that Rebecca gathered the way she has analyzed it what can be concluded from her investigation what she could do next to further her investigation. NZAMT 91038 (1.12): Demonstrate understanding of chance and data Example Task B 2011 – page 6 of 12 NZAMT 91038 (1.12): Demonstrate understanding of chance and data Example Task B 2011 – page 7 of 12 Assessment Schedule Mathematics and Statistics 91037 (1.12): Demonstrate understanding of chance and data Assessment Criteria Achievement Demonstrate understanding of chance and data will involve using a range of appropriate concepts and terms to demonstrate an understanding of statistical literacy and probability. Merit Demonstrate understanding of chance and data, justifying statements and findings will involve: providing supporting evidence such as summary statistics, probabilities, data values, trends or features of visual displays reference to the context and the population. Excellence Demonstrate understanding of chance and data, showing statistical insight will involve: integrating the statistical and contextual information and knowledge to show a deeper understanding critical reflection on the validity of the processes and conclusions given in contexts involving probability or statistics. QUESTION ONE NO Purpose of article not identified. Claims not examined N1 Article is to sell Lotto game . Article implies you will always win N2 “toss a coin”, half the numbers are winners implies that you win ½ the time A3 Claim suggest better odds than there are. Recognition that there is less than a 50% chance of winning, partially complete list of possibilities made for either Easyto-Win or Big Win A4 List of all possible draws for Easyto-win (either with order:12,13,14,21,23,24,31,32,34, 41,42,43) (or without order: 12,13,14,23,24,34) M5 Calculation of probability of winning as 1/6 for Easy-to-win AND list of all possible draws for NZAMT 91038 (1.12): OR List of all possible draws for Big Win (either with order:12,13,14,15,21,23,24,25,31,32,34,35,41, 42,43,45,51,52,53,54,) (or without order: 12,13,14,15,23,24,25,34,35,45) Calculation of probability of winning as 1/10 for Big-win AND list of all possible draws for Easy Win Recommendation made without reference to probability Recommendation made without reference to probability Recommendation made without reference to probability or with incorrect reference to probability Recommendation made with reference to probability calculations (possibly with minor flaws) Recommendation made with reference to the probability Recommendation made with reference to the Demonstrate understanding of chance and data Example Task B 2011 – page 8 of 12 Big Win M6 Correct probability for both Easyto-win(1/6) and Big Win (1/10) E7 E(Easy-to-win)= $1.67 or some other equivalent calculation (eg if got back 6x investment it would be fair but you only get back 5x investment) OR E(Big) win calculated E8 Identification of conflicting claims within article (1/2 chance implied then “invest and get 5x amount”) Correct probability for both Easy-to-win(1/6) and Big Win (1/10) E(Big Win) = $2 or some other equivalent calculation (eg if I get back 10x the investment so it si fair) OR E(Easy-to-win) calculated All outcomes listed and probabilities calculated correctly for both Easy-to-Win and Big Win, Expected values correctly calculated for both Easy-to-Win and Big Win Expected coverage TWO 2(a) Pr(odd one out) = 8/16 = ½ or 0.5 TTTT,TTTH,TTHT,TTHH,THTT,THTH,THHT, THHH,HTTT,HTTH,HTHT,HTHH,HHTT,HHTH,HHHT ,HHHH 2(b) Pr(you) = 2/16 = 1/8 (bold in list above) 2(c) OPTION 1 Pr(you odd one out in first toss) = 2/8 = ¼ TTT, TTH, THT, THH,HTT,HTH,HHT,HHH So Pr(you in 2 consecutive)= ¼ x ¼ = 1/16 Three players, so Pr(winner in 2 consecutive tosses = 3/16) OPTION 2 Pr(one odd one out ) = 12/64 =3/16 Therefore both games would have same chance of repeating, but Option 2 requires two consecutive throws (which would take longer) so choose Option 2 NZAMT 91038 (1.12): Achieveme nt Demonstrate understandin g of chance by: Correctly deducing the probability in any form. Correctly deducing the probability in any form. Correct calculation of simple probability for one part of one of the games probability Recommendation refers to “fairness”or similar concept . Recommendation refers to expected return over a long-run situation, demonstrates understanding of expected value Recommendation refers to expected return over a long-run situation, demonstrates understanding of expected value Merit Excellence Demonstrate understandin g of chance, justifying statements by: Demonstrate understandin g of change, showing statistical insight by: Correctly calculating the overall probability for one of the games Correctly identifying probability for one of the games and minor error in the other game, with choice and justification based on (possibly only partially correct) result Correctly identifying probability for both Demonstrate understanding of chance and data Example Task B 2011 – page 9 of 12 games Scoring N1 – Incomplete list or flawed probability tree for calculations of probability in (a) and (b) and (d) N2 – One correct answer for either (a) or (b) NZAMT 91038 (1.12): A3 – 2a A4 – 3a OR 2a + m M5 – A4 + m OR A3+2m M6 – A4 +2m Replacement evidence for 2abc available. E7 – M + e with some valid reasoning (may be incomplete or communicate d clumsily) E8 – M + e with valicd clear and concise reasoning. Demonstrate understanding of chance and data Example Task B 2011 – page 10 of 12 Expected coverage Achievement Demonstrate understanding of chance by: THREE 3 Although the mode is the same for both male and female binge drinking admissions, median is lower for females (thereby suggesting females have “caught up to males” with binge drinking would not be justified). However female mean is strongly influenced by the two outliers of 8 and 9 presentations. As the data was collected over a two year period and Rebecca believes female binge drinking is increasing, it would be worth checking to see if the outliers were from the very start of the two year period. If so, it would be sensible to change the data to collection for both male and female to being over the last 22 months, and if this was done, the claim that now female binge drinking is as bad as male binge drinking would be justified. Variation in number of binge drinking presentations is much larger for females (larger range, IQ range, Std Dev). This could also be due to the binge drinking incidents changing over time for females, so lower values are more common in the first part of the two year period. If the two lowest values are excluded from the study then the measures of spread would be much closer. As Rebecca suspects the current situation is a result of a growing trend, it would have been more useful to plot the data as a time series (line graph). This would have more clearly shown whether females have “caught up” to males. Eg The problem of binge drinking could be an increasing one (if the highest 6 female values were all in the most recent months) Hospital presentations for binge drinking would generally be a good indicator of the incidence of binge drinking, but comparing populations when the data is gathered over time is problematic. Especially in this case when Rebecca knew that there has been an increase over time. Binge drinking presentations are also likely to peak during holiday periods. A line graph would have shown this. NZAMT 91038 (1.12): Using a measure of central tendency to make a valid conclusion Merit Excellence Demonstrate understanding of chance, justifying statements by: Demonstrate understanding of change, showing statistical insight by: Commenting that the mean will have been affected by outliers and adjusting conclusion. Justifying the possible exclusion of the outliers based on context Making an observation about a measure of spread Drawing a valid conclusion about spread Making a valid comment about the appropriateness of this data display for the question being asked. Making a valid comment about the relevance of the data to the question. Demonstrate understanding of chance and data Example Task B 2011 – page 11 of 12 Scoring N1- some aspect of graphs of summary statistics commented on, but with only partially correct interpretation N2 – some graphs comment on correctly, but interpretation is unclear or incomplete NZAMT 91038 (1.12): A3 – 1a A4 – 2a M5 – 1m OR 1e M6 – 3m OR 1m+1e E7- M + 1e E8 – M+ 2e Demonstrate understanding of chance and data Example Task B 2011 – page 12 of 12