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Score on upper face123456frequency211924131726120 From these results they calculate the mean score :  EMBED Equation.3  = (1 x 21) + (2 x 19) +(3 x 24) + (4 x 13) + (5 x 17) + (6 x 26) = 424 = 3.533 120 120 Their teacher discusses with Laura and Bethany that what they are dealing with here is a random process, the outcome of which cannot be predicted in advance. The value that shows on the die is called a random variable, usually denoted by an uppercase X. Although each individual outcome of tossing a die cannot be predicted it is possible to assign a probability to each possible outcome. This is called the probability distribution function of X. For this situation the probability distribution function is x123456P(X = x)1/61/61/61/61/61/6 Students draw a graph for this probability distribution and comment on what they notice. Their teacher encourages them to think about what would have happened if their experiment had worked out perfectly. They realize that in that case they would have expected a roughly equal number of each score. Score on upper face123456Experimental frequency211924131726120Theoretical frequency202020202020120 Students draw a graph of these two distributions, comment on what they notice and visually predict the location of the mean. In this case the calculation of the mean score for each toss of the die is  EMBED Equation.3  = (1 x 20) + (2 x 20) +(3 x 20) + (4 x 20) + (5 x 20) + (6 x 20) = 420 = 3.5 120 120 This leads Laura and Bethany to conclude that to reach a total of 100 points a player would have to throw the die about 100 / 3.5 times, that is about 29 times. Laura and Bethanys teacher explains that this value, 3.5, is called the expected value of the score on the die, also called the mean, denoted by and explains that the formula to calculate it could be written like this:  EMBED Equation.3  =(1 x 20) + (2 x 20) +(3 x 20) + (4 x 20) + (5 x 20) + (6 x 20) = 420 = 3.5 120 120 To work towards E(X) formula Splitting this into separate fractions: (1 x 20) + (2 x 20) + (3 x 20) + (4 x 20) + (5 x 20) + (6 x 20) = 420 = 3.5 120 120 120 120 120 120 120 This can be simplified as follows: (1 x 20) + (2 x 20) + (3 x 20) + (4 x 20) + (5 x 20) + (6 x 20) = 420 = 3.5 120 120 120 120 120 120 120 which can be further simplified to (1 x 1) + (2 x 1) + (3 x 1) + (4 x 1) + (5 x 1) + (6 x 1) = = 3.5 6 6 6 6 6 6 Laura and Bethany notice that that this is equivalent to multiplying each of the possible outcomes 1 to 6 by the probability of that outcome occurring and finding the total, that is  EMBED Equation.3 for all values of x. Exemplar 2 Laura and Bethany play the game of Hog. They throw two dice, one red, one blue, taking it in turns. If a one shows they score zero, otherwise they score the total showing on the dice. They decide they will play until one of them reaches one hundred points. They wonder how many games they will have to play. They realize that the number of games they will need to play depends on the average score per turn. They play the game until one of them wins, keeping a tally of the scores they throw. This gives them a sample of one to estimate the number of games on average they will have to play for one of them to win. They realize that it will take a long time to do enough experiments to make a reliable estimate. They decide to use probability to help them work out their average score for each turn. They construct a table of possible scores: score on red die123456score on blue die100000020456783056789406789105078910116089101112Their teacher shows them how to construct the probability function for their scores, with all possible scores listed in the top row, with their respective probabilities underneath in the second row. Laura and Bethany realize that they can use the method developed above to calculate the mean (expected value) of the score for each turn. possible score, x0456789101112probability11/361/362/363/364/365/364/363/362/361/36probability0.28210.02560.05130.07690.10260.12820.10260.07690.05130.0256 Students draw a graph for this probability distribution, comment on what they notice, and visually predict the location of the expected value of a turn. Their calculation produces a value of 5.556 (3dp) for the expected value of a turn. From this Laura and Bethany are able to calculate that, on average they will need about 100 / 5.556 = 18 turns to get 100 points. Exemplar 3 Their teacher asks Laura and Bethany to consider the probability distribution of some unusual six-sided dice, labeled A to E. A has two faces labeled one, one face 3, one face labeled 4 and two faces labeled 6. B has three faces labeled 1 and 3 faces labeled 6. C has one face labeled 1, two faces labeled 3, two faces labeled 4 and one face labeled 6. D has two faces labeled 1, one face labeled 2, one face labeled 5 and two faces labeled 6. E is a normal six-sided die. Laura and Bethany draw dot plots to show the probabilities of the possible outcomes on each die. They notice that all five distributions have the same mean and range, but the shape of the distributions differs.  They notice that the spread of the possible outcomes varies even though the range is the same. The teacher discusses with them the possibility of having a measure of spread for the distribution which takes into account this spread of possible outcomes, based on the distance of each outcome from the mean, in the same way as a statistical distribution. Using technology, Laura and Bethany calculate the mean and standard deviation for each probability distribution. They mark the mean and standard deviation on their graphs. They make comments about what they notice (for example, from looking at the graphs of the distributions could they predict the location of the mean, and rank the distributions on standard deviation from smallest to largest?). ABCDEmean, E(X)3.53.53.53.53.5Range (X)1 to 61 to 61 to 61 to 61 to 6SD(X)2.06152.51.52.21741.7078 Laura and Bethany note that the higher the probability is for the extreme values of the random variable, the greater the value of the standard deviation. The teacher may take them through the theoretical calculation, showing that the standard deviation is the square root of the expected squared distance of X from its mean which can be regarded as a rough measure of the average distance from the mean. Expected squared distance of X from the mean, =  EMBED Equation.3  Standard deviation of X, SD(X) =  EMBED Equation.3  for all possible values of x. They then explore the mean and standard deviation for other situations, observing that the standard deviation averages out the distance from the mean of each possible value of the random variable. Their teacher leads them into real data that looks similar, for example, X = number of siblings each has; X= number of quoits you can get on a spike (given 6 quoits), X= number of kids blocks you can stack into a tower before it falls. Find , SD(X) for all. Teachers should note that the emphasis has shifted away from variance to standard deviation. 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