VIEWS: 7 PAGES: 16 POSTED ON: 4/30/2010 Public Domain
Achievement Standard 2.6 (90289) (Simulate probability situations, and apply the normal distribution) Level 2 Internal 2 credits Key words: simulation, probability, normal distribution, outcome, relative frequency, AND, OR, expected value, trial, mean, theta ( ), mu ( ), standard deviation, bell shape curve, symmetrical, certain, standard normal distribution, Z score, theoretical probability, tree diagram, conditional, independent, replacement, model, inverse normal, x bar, Achieve: (Design and use a simulation method to explore a situation involving probabilities AND Use the normal distribution model to find probabilities in straightforward problems.) Basic Probability Fraction, decimal or percentage A probability must only be expressed as a fraction, decimal or percentage One The greatest that a probability can be is 1 0 1 IMPOSSIBLE Certain Probability + NOT = 1 As 1 is the maximum probability, probabilities must add to 1 E.g. if P(Rain) = 0.23 the P(not Rain) = 1 – 0.23 = 0.77 Equally likely outcomes Look at shape of object to see that all events have an equal chance i.e. coin or dice. A' s P( A ) outcomes Long run relative frequency (exercise 26.1 is all practical) Similar to equally likely outcomes but based on a practical E.g. Drop toast experiment Buttered side up 10 Buttered side down 12 TOTAL 22 P(Up) = 10/22 = 5/11 This might be a good point to recap simplification of fractions 1 AND/OR AND = 1 1 1 1 1 E.g. if P(Ace) = and P(Tails) = then P(Ace AND Tails) = 6 2 6 2 12 This might be a good point to recap multiplication of fractions OR = + 4 4 4 4 8 2 E.g. if P(Jack) = and P(Queen) = then P(Jack OR Queen) = 52 52 52 52 52 13 This might be a good point to recap addition of fractions Page 274 Ex 26.2 Expected Value/Number (achieve and merit work) There are two types of questions, the first are obvious. i.e. if a coin is tossed 70 times, we would expect 35 heads and 35 tails. Expected number of = probability trials The second type of question involve tables Expected values from probability tables E.g. 1) We know that the probability of a fair die is 1/6 but what would be the average value of each roll be? Number 1 2 3 4 5 6 Probability 1/6 1/6 1/6 1/6 1/6 1/6 1/6 2/6 3/6 4/6 5/6 6/6 1 2 3 4 5 6 21 Expected value = 3.5 i.e. The mean average of all the numbers of a large 6 6 6 6 6 6 6 trial would be 3.5. E.g. 2) A biased dice is known to produce the following results Number 1 2 3 4 5 6 Probability 0.01 0.25 0.1 0.11 0.03 0.5 Expected value = sum of (number multiplied by probability) 0.01 0.5 0.3 0.44 0.15 3 Total = 4.4 In other words, the average number that we expect to obtain is 4.4 i.e. if we roll the dice 300 times adding the score each time then divide by 300 we would expect the result to be about 4.4 The following exercise needs knowledge of Venn Diagrams. This is not needed at level 2 but can do no harm being taught, as it is required at level 3 Page 292 Ex 26.6 2 Normal Distribution Do the following none calculator test with the students in silence 1) Solve (2x + 3)(x – 4) = 0 (Answer = –3/2 & 4) 2) Find distance between (2 , 3) and (5 , 7) (Answer = 5) 3) What is the product of the gradients of two perpendicular lines? (Answer = –1) 4) For the line y = 3x + 4, what is the gradient? (Answer = 3) 2 5) Factorise x – x – 6 (Answer = (x + 2)(x – 3)) 6) Expand (4x2)3 (Answer = 64x6) 7) Find mid point of (8 , 17) and (10 , 21) (Answer = (9 , 19)) 8) Gradient of line connecting (4 , 9) and (6 , 17) (Answer = 4) 1/2 9) Value of 9 (Answer = 3) A 10) Make r the subject of A = r2 Answer : r Now: 1) Find mean average ( ) 2) Find standard deviation ( ) (the calculator will do this for you) 3) Draw a histogram of class results with each bar being a mark out of 10 e.g. frequency 1 2 3 . . . etc. Marks out of 10 If there were enough people doing the test this should follow a normal distribution curve or bell shape. This is unlikely due to the small numbers involved. If not a perfect bell shape, more classes results are required. The more we collect, the closer it will be to a bell shape. What would a ‘cleverer’ class’s graph look like? What would a ‘less able’ class’s graph look like? What would the graph be like if the questions were harder? What would the graph look like if the questions were easier? 3 Casio Graphic Calculator: Averages from Discrete Data (This is not required for 2.6, but some students might be interested) To enter data 1) STAT mode (second in on left the top row) 2) Enter data into list 1 column i.e. 12, 13, 14 (press Exe after each entry) 3) F2 (CALC) 4) F6 (SET) make sure the first two lines of the menu is set up to 1Var XList :List1 1Var Freq :1 (The 1 indicates that there is one of each number in List1. If this is to be a frequency table this should read List1) 5) EXE To find various statistics 1) Once you have followed the instructions above and entered a set of data 2) EXIT or F2 (CALC) to obtain bottom menu with arrow above F6 and CALC above F2 3) F1 (1VAR) Using the arrow down key you can see various types of statistics including mean, mode, median, maximum, minimum, sum of data, number of pieces of data and quartiles. Take care with mode as if there is no mode it chooses the maximum value for the mode or if there is more than one mode it chooses the highest mode value. (If you are calculating averages and it has an error, check that you have followed instruction 4 from entering data correctly) To clear previously entered data 1) In STAT mode 2) EXIT or F2 (CALC) to obtain bottom menu with arrow above F6 and CALC above F2 3) F6 (arrow right) 4) To delete single piece of data, use arrow key to highlight piece of data and then press F3 (DEL) or 5) To delete all of row use arrow key to highlight any piece of data in row and then press F4 (DEL-A), F1 (YES) Graphically the mean and deviations look like: 0.5% 2% 13.5% 34% 34% 13.5% 2% 0.5% Xa likely/probable - 68% + very likely/very probable - 2 95% + 2 almost certain - 3 99% + 3 4 Notice that the normal distribution is symmetrical i.e. 50% either side of mean. Probability questions can now be asked using the bell shape curve i.e. on the above graph the probability that the result is bigger than Xa is 97.5% With the results that we have for our test result out of ten explain what the standard deviation and mean average mean i.e. if the mean is 6 and the standard deviation is 1.5 then we expect to find: 34% of the students with results between 6 and 7.5, 34% of the students with results between 6 and 4.5 47.5% (34 + 13.5) of the students with results between 7 and 10 etc. (with such a small amount of data we are unlikely to obtain realistic results) All normal distributions have different means ( ) and standard deviations ( ), but all have 68%, 95%, 99% pattern around mean. X Example: In a test, the mean is 70 and the standard deviation is 5. Assume normal distribution. Answer: From the above data the following graph can be drawn 0.5% 2% 13.5% 34% 34% 13.5% 2% 0.5% 55 60 65 70 75 80 85 From the above, certain probabilities can be stated as long as it involves the values that lie on multiples of standard deviations away from the mean. i.e. 0.5% of the students scored more than 85. 16% of the students scored less than 65. 50% of the students scored more than 70. 68% of the students scored between 65 and 75 i.e. it is likely that a student’s result is between these two values. 95% of the students scored between 60 and 80 i.e. it is very likely that a student’s result is between these two values. 99% of the students scored between 55 and 85 i.e. it is almost certain that a student’s result is between these two values. Alternatively, if I chose a student at random there is only a 1% chance that his score is outside of these two values. Page 350 Ex 30.1 (OK Q’s if time) Page 353 Ex 30.2 (good Q’s) The Standard Normal Distribution 5 The normal distribution graphs up to this point can only be analysed using values that are multiples of the standard deviation. This is rather limited. The standard normal distribution is a perfect normal distribution with a mean of 0 and a standard deviation of 1. Instead of using X for certain values, we use Z. The probability (area) of Z (+ve only) being between the mean and a certain value can be found on page 516. These values are as decimals i.e. they will not be bigger than 1 (0.5 as only half the graph is used in the table) e.g. P(0 < Z < 1.43) = 0.4236 or 42.36% P(0 < Z < 0.7) = 0.2580 P(–0.7 < Z < 0) = 0.2580 P(0 < Z < 2.163) = 0.4846 + 0.0001 = 0.4847 P(Z > 1.57) = 0.5 – 0.4418 = 0.0582 = 5.82% P(Z < 1.593) = 0.5 + 0.4445 = 0.9445 = 94.45% P(Z > 2.35) = 0.5 – 0.4906 = 0.0094 = 0.94% Page 358 Ex 30.3 Q1a–f, 2a–f Converting to the Standard Normal Distribution To be able to work with data we need to convert normal distribution to a standard normal distribution. This has the benefit that we know far more information about the standard normal distribution i.e. probability that data lies between mean and any point. This is given in table on page 516 All we need to do to convert a particular value (X) to see at what point it is on a standard normal X distribution we use the formula for Z value/score Z (on formula sheet). We will use this later. E.g.1) If year 11 boys had a mean height of 1.67m and a standard deviation of 9cm. Find the probability that a person chosen at random is bigger than 1.81m. Assume data is normally distributed 6 Question is asking i.e. P(X > 1.81) notice that this is not a Z. First using above formula convert X score to Z score 1.81 1.67 Z 1.5555555 use 1.556 0.09 Therefore the probability can now be written as P(Z>1.556) Draw diagram (stress importance of this step) 0.4401 0 1.556 From table: 1.556 gives 0.4401 (remember this value is the area between the mean and the stated value. We do not want this. We need the ‘other’ area) Therefore P(Z > 1.556) = 0.5 – 0.4401 = 0.0599 = 5.99% E.g. 2) Party bags of M and M‘s claim they have 255 in them. The company know that the mean average is 259 and has a standard of 2. What is the probability that a bag chosen at random has less than 255 M and M’s in them? Question is asking i.e. P(X < 255) notice that this is not a Z. First using above formula convert X score to Z score 255 259 Z 2 2 Therefore the probability can now be written as P(Z<-2) Draw diagram (stress importance of this step) 0.47725 Therefore P(Z > –2) = 0.5 – 0.47725 = 0.02275 = 2.275% Page 359 Ex 30.4 (using formula and practical problems, Achieve and Merit questions) 7 Casio Graphic Calculator: Practical problems using Normal Distribution Using the calculator at this stage will hinder your understanding of this topic, but if you are only after achieve …… 1) STAT mode 2) DIST (F5) 3) NORM (F1) 4) Ncd (F2) 5) Enter the lower value of X 6) Enter the Upper value of X 7) Enter the standard deviation 8) Enter mean 9) EXE E.g. 1) If year 11 boys had a mean height of 1.67m and a standard deviation of 9cm. Find the probability that a person chosen at random is between 1.7 and 1.81m. Assume data is normally distributed For step 5) above use 1.7 For step 6) above use 1.81 For step 7) above use 0.09 For step 8) above use 1.67 Answer = 0.30953 E.g. 2) If year 11 boys had a mean height of 1.67m and a standard deviation of 9cm. Find the probability that a person chosen at random is bigger than 1.81m. Assume data is normally distributed. Note that this is the same question in the above notes. For step 5) above use 1.81 For step 6) above use any large number i.e. 50 as this number is a lot bigger than the mean For step 7) above use 0.09 For step 8) above use 1.67 Answer = 0.059906 Compared to the answer obtained using tables, this is virtually the same. I would round the answer to 4 dp. Practicals Use of random numbers to:- 1) Choose a number that has a 60% chance. i.e. 1,2,3,4,5,6 is ok and 7,8,9,0 is not ok 2) Choose a number that has a 1 in 6 chance i.e. 1 is ok and 2,3,4,5,6. Take care of using Random6 to choose a number between 1 and 6. Monty Hall Problem Not a good example for working out statistics but good to use as a demonstration of how a practical activity can show if something works or not. You have three doors, you pick one, and Monte shows you which one of the other doors is the wrong choice. Do you keep your choice or do you swap. For a computer simulation go to http://www.cut-the-knot.com/hall.shtml There is much discussion as to which is the correct answer but the following convinces me: 8 The best explanation I can give is as follows. There are three possible places the prize can be, just for arguments sake lets say we always start by picking door number 1. There are two possible choices for the second question switch or do not switch. If we do not switch here are the outcomes: Prize is Monty I pick Result behind opens 1 1 2 or 3 Stay on 1, Win 2 1 3 Stay on 1, Lose 3 1 2 Stay on 1, Lose I will win 1/3rd of the time by sticking with my original choice. But if I do choose to switch: Prize is I Monty Result behind: pick opens 1 1 2 Switch to 3, Lose 2 1 3 Switch to 2, Win 3 1 2 Switch to 3, Win So, if I switch, I will win 2/3rds of the time. In the original problem, you have three doors, you pick one, and Monte shows you which of the other doors is the wrong choice. The key: he is showing you all but one of the remaining doors, and he will only show you ones which are wrong. The best bet is obfuscated by the fact that there're only two other doors - by showing you only one wrong answer, as that's all that are available without totally giving it away, the magnitude of the new information gets lost. Think of this instead: I have a standard deck of poker cards. The winner is the player with the ace of spades. You get to pick one, face down, and do not get to look at it yet. I have 51 cards left. I thumb through them and show you 50 of them that are not the AofS. Now, is the one you picked the AofS (you picked 1 card randomly out of 52), or is it the one I picked out of the remaining 51 cards (I get to look, and get to pick any of the 51 remaining cards)? I'd expect you to get the AofS correctly on a single pull, oh... about 1 out of 52 tries. I would expect me to have (and hence pick) the AofS about 51 out of 52 times. Put another way: if you managed to get the AofS on the first pull, I obviously have to go with something else; otherwise, I am gonna pick the AofS from what is left, and win. Now, let us get in line with the Monte Hall problem - after we both do our picks, you are betting on who has the AofS. You can either take the card you picked, or switch and take the one I picked. Go with my choice - it is the right bet. I have explained this to others before, and some have said, "It isn't the same problem." It IS the same problem, precisely. It just has the proportion set-up so the result is obvious. Reduce the number of cards from 52 to 3, and the correct choice is the same. Below is a convincing argument but not correct There are 24 possibilities: 12 possibilities of switching Prize is behind door Contestant chooses Monty Opens Switch to Result 1 1 2 3 L 1 1 3 2 L 2 1 3 2 W 3 1 2 3 W 2 2 1 3 L 2 2 3 2 L 9 1 2 3 1 W 3 2 1 3 W 3 3 1 2 L 3 3 2 1 L 2 3 1 2 W 1 3 2 1 W The above table shows that if you switch you will win 6 out of 12 times or 50% Prize is behind door Contestant chooses Monty Opens Result 1 1 2 W 1 1 3 W 2 1 3 L 3 1 2 L 2 2 1 W 2 2 3 W 1 2 3 L 3 2 1 L 3 3 1 W 3 3 2 W 2 3 1 L 1 3 2 L The above table shows that if you do not switch you will win 6 out of 12 times or 50% Greedy Pig Everyone stand up. Teacher rolls a dice. If the number is a two, you are out. Any other number you add it to your score. You can sit down at any point. Reaction time Work in pairs. One person drops the ruler and the other person catches it. Collect classes’ results. From the results, much statistics can be calculated. I.e. Probability person’s reaction time is bigger than…., smaller than… Excellence question could be what value are 60% of students better then. Weetbix packet In every Weetbix packet there is a playing card. Either a Jack, Queen, King or Ace of hearts. How many packets are needed to get a full set? A good use of random numbers here to simulate the choosing of cards. Mini Lotto Choose two numbers from 1 – 6. How many random numbers from 1 – 6 are required to win. Simulate with dice Theoretical probability can be found using trees Once mean and standard deviation are found, questions using normal distribution can be asked i.e. Find the probability that a student wins after 3 goes. Homework Sara has maths on 4 days each week and every day she has homework. Her teacher checks the class’s homework on only 1 of those days. This day is chosen at random by the teacher. Students who have not done their homework for that day get a detention. Sara says she did her homework on 32 out of the 40 days on which she had maths last term. Describe a simulation (probability experiment) that Sara could use to predict the probability that: i) she had not done her homework ii) the teacher checked 10 Casio Graphic Calculator: Generation of random numbers 1) Run mode 2) OPTN 3) F6 (arrow right) 4) PROB (F3) 5) Ran# (F4) 6) EXE Take care with trying to generate only certain random numbers. E.g. Pressing 6Ran# will only give numbers between 0 and 5.999999, not between 1 and 6. A way round this is choose a number one bigger then you require e.g. to generate numbers from 1 to 8 press 9Ran# and ignore any 0’s that appear a the start of the number. MERIT: (Use simulations and theoretical probability techniques to solve problems AND Solve problems using the normal distribution model) Probability Trees Probability trees are used to show with a picture what is going on mathematically. (I always use them!) Although both spinners are spun simultaneously, they can be treated as being spun one after the other Independent Green Green Yellow Blue Red Spinner A Spinner B P(Red) = ½ P(Yellow) = 2/3 P(Blue) = 1/3 P(Green) = 1/3 P(Green) = 1/6 11 P(Y) = 2/3 P(R) = ½ P(G) = 1/3 + P(Y) = 2/3 P(B) = 1/3 = P(G) = 1/3 1 P(G) = 1/6 P(Y) = 2/3 P(G) = 1/3 Using the above tree diagram, any question can now be easily asked. This is a good time to use AND = & OR = +. Describe the situations using the words AND & OR, this well indicate whether to multiply or add 1 2 2 1 P(Red and Yellow) = 2 3 6 3 1 1 1 17 P(not the same colour) = 1 - = 1 6 3 18 18 1 1 1 1 1 2 1 1 2 7 P(only one green) = P(RG or BG or GY) = = = 2 3 3 3 6 3 6 9 18 18 Page 281 Ex 26.3 Probability without Replacement (Conditional) Events cannot happen at the same time as the result of the first event affects the second event. 12 5 balls are placed in a bag (3 black & 2 white). A ball is chosen and put to one side. A second ball is chosen and placed to one side. This is the same as choosing two balls at once. P(B) = 2/4 P(B) = 3/5 P(W) = 2/4 P(W) = 2/5 P(B) = 3/4 P(W) = 1/4 3 2 6 3 P(B and then W) = 5 4 20 10 3 2 2 1 6 2 8 2 P(both the same colour) = P(BB or WW) = 5 4 5 4 20 20 20 5 3 2 6 14 7 P(not both Black) = 1 - 1 5 4 20 20 10 Page 286 Ex 26.4 More than one z-value Page 358 Ex 30.3 Q1ghi, 2ghi Excellence: (Interpret the outcome of modelling a probability situation) Inverse Normal Inverse normal problems: This is the reverse of the previous work and as such the tables are used in reverse i.e. probabilities are known and you are asked to find the Z value. E.g. 1) Find the value of Z if P(0 < Z < k) = 23.89% (this is an easy example compared to the exercises) You still need to draw diagrams. 0 k 23.89% = 0.2389 (from table this gives a Z score of 0.64). 13 If we had known mean and standard deviation of data we could work out X value i.e. using the X rearrangement of Z (on formula sheet) i.e. X Z (not on formula sheet) I.e. in the above example if the mean was 56 and the standard deviation was 3, X = 3(0.64) + 56 = 57.92 E.g. 2) Find k if P(Z > k) = 0.2576 Draw diagram 0.2576 0 k 0.5 – 0.2576 = 0.2424 Look up 0.2424 in body of table. 0.65 gives 0.2422 0.651 gives 0.2425 (this is closer) therefore answer = 0.651 Page 364 Ex 30.5 not Q8(finding Z score given probability from diagrams and values of probability) Casio Graphic Calculator: Inverse Normal Distribution Using the calculator at this stage will hinder your understanding of this topic and excellence exam questions will require more than just an answer to obtain full marks. 1) STAT mode (2) 2) DIST (F5) 3) NORM (F1) 4) InvN (F3) 5) Enter the area (this is the area from the LHS to the point in question) 6) Enter standard deviation 7) Enter mean 8) CALC (F1) E.g. Using a mean of 56 and a standard deviation of 3, find the value of Z if P(0 < Z < k) = 23.89% We need the value from the left so the area is 0.5 + 0.2389 = 0.7389 1) Enter area of 0.7389 2) Enter 3 for standard deviation 3) Enter 56 for mean 4) CALC (F1) The answer is 57.919 (same as example above) 14 Practical problems The Year 11 girls have a mean height of 1.59m and a standard deviation of 7 cm. Jo is 1.69m but is not sure of the minimum height that is required to join the basketball team. She knows to keep the numbers down the manager just lets the tallest 15% of students join. Can she join? 15% 1.59 Area between 1.59 and x is 0.35 (0.5 – 0.15) In body of table, area of 0.35 gives Z score of 1.036 or 1.037 Substitute either value of Z into X Z X = 0.07(1.036) + 1.59 This gives 1.66m so yes she can play. Page 364 Ex 30.5 Q8 Page 366 Ex 30.6 (practical inverse problems) For the exam students are required to know the difference between theoretical and experimental probability. Theoretical probability: use the shape or information you know about events to predict the probability of outcomes occurring. E.g. the probability of obtaining heads on a coin and 2 on a dice 1 1 1 is 2 6 12 Experimental probability: use the results of experiments to predict the probability of outcomes occurring. E.g. To find the experimental probability of obtaining heads on a coin we flick the coin numerous times and note the number of times the event happened. TEST 1) Go to the following website for past papers: http://www.nzqa.govt.nz/ncea/ and then type in the number 90289 2) Use the following link: www.studyit.org.nz 3) See Achieve mind map below 15 Expected Relative AND number frequency OR Basic On formula One value Not sheet. of z only. Probability X Fraction, Easy probability Standard normal z decimal or trees percentage 2.6 Achieve (Normal) Normal distribution (likely/probable = 68% 1 ), & Bell shape curve Design and use a simulation (very likely/very (almost probable = certain = Dice or 95% 2 ) 99% 3 ) random numbers etc 16