Which spinners?

6123 /www/nrich/html/content/id/6123/ 1 2011-08-16T14:28:42 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> We have two spinners, each numbered 1 to 4. We're going to spin them both and add the numbers. What possible totals might you get? What is the least likely/most likely total to occur? Use the interactivity below to test your hypotheses. <a href="/content/id/6123/experimenter.swf">Full Screen Version</a> (click 'back' on your internet browser to exit full screen mode) <mdo:flash height="400" id="/content/id/6123/experimenter.swf" width="550"><param name="allowfullscreen" value="true"></param><param name="allowfullscreen" value="true"></param> <param name="movie" value="/content/id/6123/experimenter.swf"></param> <param name="flashplayerversion" value="7"></param> </mdo:flash> Now make a new set of spinners. The interactivity can record either the sum of, or the difference between, the numbers on the spinners, and you can see the results displayed on the relative frequency bar chart. You can choose to run the interactivity lots of times so that the bar chart "settles down". Experiment with different pairs of spinners. What features do you notice on the bar charts that you produce? Can you come up with ways of predicting what a chart will look like before you produce it? The challenge The bar charts below were generated on the interactivity using different combinations of spinners. (You can download a <a href="/content/id/6123/Which%20Spinners%3F.pdf">pdf</a> with all eight bar charts here.) <table style="width: 500px;border-spacing:1px;" border="1"> <tbody> <tr> <td>A</td> <td><mdo:image src="Spinner%201a.png"></mdo:image></td> <td style="text-align: right;padding:1px;"> B</td> <td><mdo:image src="Spinner%202a.png"></mdo:image></td> </tr> <tr> <td>C</td> <td><mdo:image src="Spinner%203.png"></mdo:image></td> <td style="text-align: right;padding:1px;"> D</td> <td><mdo:image src="Spinner%204.png"></mdo:image></td> </tr> <tr> <td>E</td> <td><mdo:image src="Spinner%205.png"></mdo:image></td> <td style="text-align: right;padding:1px;"> F</td> <td><mdo:image src="Spinner%206.png"></mdo:image></td> </tr> <tr> <td>G</td> <td><mdo:image src="Spinner%207.png"></mdo:image></td> <td style="text-align: right;padding:1px;"> H</td> <td><mdo:image src="Spinner%208.png"></mdo:image></td> </tr> </tbody> </table> Can you deduce which spinners were used to create each bar chart? Can you explain how you used the information provided by the bar charts to work it out? <script type="text/javascript"> // // // // // // // // // // // // // // // // // $(document).ready(function(){ $("p#zoom a").fancybox();}); // // // // // // // // // // // // // // // // // </script>Final challenge Imagine you had 1-20 and 1-30 spinners. Describe in as much detail as you can what the relative frequency bar charts would look like for: <ul> <li>The sum of two 1-30 spinners</li> <li>The difference between two 1-20 spinners</li> <li>The sum of a 1-20 and a 1-30 spinner</li> <li>The difference between a 1-20 and a 1-30 spinner</li> </ul> Try to provide a good explanation to convince us that your descriptions of the bar charts are correct.</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Shaun from Wilson's School observed: Every graph of the sum of two spinners would bulge in the middle because there is more chance of making the middle number. There is little chance of getting the highest numbers so that would mean the higher and, equally, the lower numbers would come up less often. Thomas and Holly from Hymers College explained their thinking clearly: After figuring out the first few graphs and their corresponding spinners, we soon found out that the maximum sum of the spinners equalled the highest bar value for the sum. We also found that a sample space diagram could be applied to the graphs involving sums. For A and B, we saw a graph going up and down with equal steps, peaking at 8 and 12 for A and B respectively. Therefore, we deduced that the two spinner numbers had to be of equal value. As we had already figured out, the sum of A had to be 14, so we could see that the sum was 7 + 7. For B, the sum had to be 20, so the sum had to be 10 + 10. For C, we knew the sum was 14, and we counted 3 steps up to the peak, so we thought it was 3 + 11, still measuring the sum. It is very hard to get a graph exactly like shown in C, but 3 + 11 is the way to get it. For D, we followed the same method, but knew the sum was twenty with 6 steps, so after inputting 6 + 14; we got a graph similar to the one shown in D. For the remaining four, it is obvious from the shape of the graph that it must be on the difference setting, as it would be impossible to get 0 on sum setting. Since E has a maximum difference of 6, it has to be 7 + 7 (because the difference of 6 would be between 1 and 7 (as it is impossible to get 0 and 6)). We worked on a similar basis for the remaining 3. F was 11 + 11, G was 3 + 7 and H was 4 + 11. Jennifer from Taipei European School had a go at the final challenge: The difference between two 1-20 spinners. There will be 20 bars because the smallest bar is 1-1=0 and the biggest bar is 20-1=19 then 19+1=20 because the smallest number is 0 isn't 1. The second bar will be the longest because 1, the answer, appears most often when finding the difference between two numbers and the last bar will be the shortest because the only way to make 19 is 20-1. The sum of a 1-20 and a 1-30 spinner There will be 49 bars because the smallest is 1+1=2 and the biggest is 20+30=50 The first and the last bar will be the same and the two bars will be the shortest because only 1+1=2 and only 20+30=50. The difference between a 1-20 and a 1-30 spinner. There will be 30 bars because 1-1=0 and 30-1=29. The last bar (29) will be the shortest because the only way to make 29 is 30-1. The second bar will be the longest because 1, the answer, appears most often when finding the difference between two numbers. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3>Why do this problem?</h3> <a href="http://nrich.maths.org/public/viewer.php?obj_id=6123">This problem</a> provides a motivation to use sample space diagrams; it could be used to introduce or consolidate work on them. The interactivity offers an ideal context in which to observe the "messy" randomness of results after a small number of experiments, and the predictability of results after a large number of trials. The problem also offers a good starting point for considering different probability distributions and their features. <h3>Possible approach</h3> <a href="http://nrich.maths.org/public/viewer.php?obj_id=6033&part=">Interactive Spinners</a> provides a useful introduction to this problem, and the <a href="/6033/note">Teachers' Notes</a> suggest how it could be used in the classroom. Introduce the interactivity with a single spinner, one spin at a time. Clarify that it is not a frequency table, it's a relative frequency table. "What could happen to the chart after the next spin?" Collect together the different possibilities together with the students' explanations. "Let's see which one happens" and spin again. Repeat until students are secure in their understanding of the relationship between what is happening on the spinner and what appears on the relative frequency chart. Finally, "What would the chart look like after $50 000$ spins?" Allow time for them to discuss in pairs, then gather together suggestions and justifications before spinning to confirm their ideas. If computers are available, ask students to work in pairs and set them the following challenge: "I wonder whether you could predict what would happen if we had two spinners... In a while, I'm going to choose two spinners (which could be identical but may be different), and decide on either the sum or the difference, and you will need to be able to predict what the bar chart will look like, and explain how you know." "You have some time to work in pairs at the computer to prepare for this challenge by experimenting with different pairs of spinners and recording what you notice in order to help you to make predictions." Show how to set up the interactivity with two spinners, if necessary. While they are working, circulate and listen to students' noticings. Challenge them to explain what they have noticed, and be aware of any students who have useful recording methods, insights or explanations that could be shared with the rest of the class. Bring the class together to share their insights and explanations before handing out <a href="/content/id/6123/Which%20Spinners%3F.pdf">this worksheet</a>. "Can you work out how these graphs were created, WITHOUT using the interactivity?" Once pairs have had time to decide how the eight graphs were created, a nice way to finish off the task is to arrange the class into eight groups, give each group one of the graphs, and invite them to prepare a short explanation to present to the class. If computers for the students are not available... "What would happen if we had more than one spinner? I'm going to set the interactivity up with two spinners going from 1 to 3, and each time, it will record the sum of the two numbers. With your partner, talk about what you think the graph might look like after $50 000$ spins." Once they've had a chance to discuss, share ideas and ask for justifications before checking using the interactivity. If no-one has suggested using a sample space diagram, this would be a good opportunity to introduce the technique to explain the heights of the bars on the chart. Hand out the first page of <a href="/content/id/6123/Which%20Spinners%3F.pdf">this worksheet</a>. "These graphs were created using two spinners. Your challenge is to work out which two spinners were used in each case, and provide a convincing argument." Give students plenty of time to work on the challenge in pairs. As they are working, circulate and note which pairs have insights that are worth sharing. The second page of the worksheet could be handed out to pairs who have identified the spinners for the first page, for them to start thinking about. Bring the class together to discuss the first page, and invite those pairs with interesting ideas to share what they did, using the interactivity to check. Then hand out the second page to all and set them the same challenge as before. "These graphs look rather different - see what key features you notice and see if you can deduce how they were made and which spinners were used." Again, give students time to work on the challenge before bringing the class together to share what they found. If time allows, the first extension task below would provide a good final challenge. <a href="/6963">Here</a> is a version of the interactivity which allows you to choose the numbers on the spinners, rather than being restricted to consecutive numbers starting at $1$. <h3>Key questions</h3> What features of the chart do you think might be important? How could those features have been created? Can you deduce anything about the biggest numbers on the spinners? <h3>Possible extension</h3> Set students the final challenge from the problem: "Imagine you had 1-20 and 1-30 spinners. Describe in as much detail as you can what the relative frequency bar charts would look like for: <ul> <li>The sum of two 1-30 spinners</li> <li>The difference between two 1-20 spinners</li> <li>The sum of a 1-20 and a 1-30 spinner</li> <li>The difference between a 1-20 and a 1-30 spinner</li> </ul> Try to provide a good explanation to convince me that your descriptions of the bar charts are correct." A more challenging extension is to explore the charts produced by three spinners. <h3>Possible support</h3> <a href="http://nrich.maths.org/public/viewer.php?obj_id=6033&part=">Interactive Spinners</a> provides a good introduction to this problem. Begin by just exploring the first four graphs where only the sum is used rather than the difference. Then introduce the second page of the worksheet with an example of two 1 to 3 spinners finding the difference rather than the sum, and take time to show how a sample space diagram can explain the features of the graph. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> You may wish to look at the problem <a href="http://nrich.maths.org/public/viewer.php?obj_id=6033&part=">Interactive Spinners</a> before starting this problem. What similarities can you see between the bar charts? What differences can you see? For each bar chart, what does the minimum and maximum value tell you? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Charlie and Alison's version, Aug 2011 <ol> <li>a & b 1-7 twice, added</li> <li>a & b 1-10 twice, added</li> <li>1-3 and 1-11, added</li> <li>1-6 and 1-14 added</li> <li>1-7 twice, difference</li> <li>1-11 twice, difference</li> <li>1-3, 1-7, difference</li> <li>1-4, 1-11, difference</li> </ol> OLD VERSION?? a. 1-4, 1-4, 1-5 and 1-5 b. 1-9 and 1-9 (sum) c. 1-13 and 1-4 (difference) d. 1-13 and 1-13 (difference) e. 1-4, 1-4 and 1-4 (sum) f. 1-2, 1-4 and 1-6 (sum) Edgar from Dr Challoners Grammar School was the first to send in a detailed answer. He writes "For A, the first step I took was to note that the lowest value recorded was $2$. This told me that the sum of two spinners had been recorded; the lowest possible value each spinner can have is $1$, and $1+1=2$. The highest value recorded was $18$, and this told me that because there was a 'triangle' shape on the graph, the numbers were both $9$ because $9+9=18$. The 'triangle' shape I will now proceed to explain. "If there are two spinners of nine sides spun at once, there are $9\times 9=81$ possibilities for combinations they can produce. For $2$ or $18$ respectively, there is only one possibility in which they are produced - $1+1$ and $9+9$ respectively, giving them a probability of $\frac{1}{81}$. For $3$ and $17$, however, there are two possibilities - $1+2$, $2+1$ and $8+9$, $9+8$ respectively, giving them each a probability of $\frac{2}{81}$. This pattern continues all the way to $10$ which has $9$ possibilities - $1+9$, $2+8$, $3+7$, $4+6$, $5+5$, $6+4$, $7+3$, $8+2$ and $9+1$, and then the probability declines again down to eighteen. Providing each spin of the spinner is totally random, two spinners of the same number should always come up with this 'triangle'. "For B, the solution I came up with was the difference of two spinners of thirteen. The 'triangle' for differences is a right angled one, but it has an anomalous probability for zero, which can only be achieved when both numbers are the same. The reason for this is similar to the corresponding one for the sum, only reversed. Similarly, this 'triangle' shape can be achieved only when both the spinners must be the same. "For C, the spinners are a four and a thirteen, and the difference is taken. The graph is a sort of plateaued 'triangle' shape. This shape is created because certain numbers only be achieved in four ways (e.g. $8=12-4=11-3=10-2=9-1$, $7=11-4=10-3=9-2=8-1$) so their probabilities are not different, so the shape created is basically a 'triangle' with the middle stretched out in a sort of plateau shape. This 'plateau' can also be created when adding the spinners. "For D, the plateau shape appears again, only this time in the form of a symmetric sum graph. The two spinners involved are a fourteen and a four. I could tell that there were two spinners being added together because the lowest value is two ($1+1=2$), one of them was a four (because there are four steps up to the plateau) and then the other one must be the highest value eighteen minus four which gives fourteen. "For E, I first concluded that there must be three spinners because the lowest outcome was three ($1+1+1=3$). Because there was a bit of a jump up to the middle four on the graph, I concluded that the spinners might all be the same (when there are more than two spinners, intermediate results are made more likely). The highest outcome was twelve, and $\frac{12}{3}=4$, so therefore all the spinners must be four. "For F, there was no 'jump up' in the middle of the graph, and there were three spinners involved, and because of the absence of the 'jump up', I surmised that the numbers ought to be consecutive (this makes the 'triangle' shape) and the only three consecutive numbers I know which sum to $12$ are $3$, $4$ and $5$, so I concluded that these were the numbers of the spinners." An outstanding solution, Edgar, well done! Thomas from A Y Jackson also sent in a full, detailed solution to this problem. Interestingly, he had a different answer for F, suggesting that the spinners used were three, three and six. </mdoxml> 2 3 0 0 1 1 0 Which spinners? Can you work out which spinners were used to generate the frequency charts? 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