0

0

2881 days ago,
914 views

PowerPoint PPT Presentation
Insights Activities for a High School Mathematics Class Room ... Not important to have an in number details foundation to be a secondary school math educator

Measurements Activities for a High School Mathematics Class Room Megan McLennan May 2, 2005

Outline Introduction Activities "The Price is Right" Master Key, Plinko "Likelihood of a Kiss" Activity 1, Activity 2, Activity 3 Hypothesis Testing versus Jury Trial Conclusion

Introduction Why I picked this subject High teacher Have a few thoughts for how to fuse details in a math classroom before I go out into the field Personal encounters with insights in secondary school Not important to have a solid details foundation to be a secondary school math instructor

"The Price Is Right" The amusement demonstrate The Price is Right comprises of competitors playing item evaluating diversions keeping in mind the end goal to win prizes. Require the learning of costs, however there is additionally a component of possibility. There are approximately 70 diversions on TPIR, I will experience two of the amusements that can be made into a classroom movement for secondary school understudies Master Key, Plinko

Master Key-about the diversion There are three prizes the competitor can win, a little prize, and medium prize and a substantial prize There are five keys arbitrarily put before the challenger, one for the little prize, one for the medium prize, one for the huge prize, one for every one of the prizes, and one that is a "failure" The contender has an opportunity to get to two keys. He is demonstrated an item at which two costs are given, in the event that he surmises the right value, he gets a key. This rehashed The contender utilizes the keys he earned to attempt to open the three bolts and wins whatever he opens

Master Key-Goals Computing Probabilities Conditional Probabilities Bayes' Rule Counting Combinations Ex. In the event that you have 10 questions and pick 3 of them, what number of mixes are conceivable?

Master Key-Questions Assume that the challenger has no evaluating information of the two items, and accordingly his/her choices of picking the right cost for every item are autonomous and each has a half shot of being correct. Figure the accompanying probabilities:

Master Key-Questions A. What is the likelihood that the contender wins no prizes? B. What is the likelihood that the competitor wins a prize, however not the huge prize? C. What is the likelihood that the contender wins the huge prize? D. Given that a candidate has won the auto, what is the likelihood that he had earned stand out key?

Master Key-Solutions Consider the appropriation for the quantity of keys earned, X, and the restrictive probabilities for what prizes can be won given X keys were earned. Let: A = win no prizes, B= win the little as well as the medium prize however not the vast prize, C= wins the extensive prize

Master Key-Solutions The circulation of X and the contingent probabilities of A, B, and C given X can be shown in a tree graph:

Master Key-Solutions Breaking down the tree: P(A | X=0) = 1 P(A | X=1) = 1/5 P(B | X=1) = 2/5 P(C | X=1) = 2/5 P(A | X=2) = 0 P(B | X=2) = 3/10 5 C 2 = 10, approaches to pick the two keys 3 C 2 = 3, ways can win the little and additionally medium prize, yet no auto P(C | X=2) = 7/10

Master Key-Solutions Using the tree (and p1 = p2 = 0.5) A. P(A) = (1-p1)(1-p2)(1) + [p1(1-p2) + p2(1-p1)](1/5) P(A) = 14/40 B. P(B) = [p1(1-p2) + p2(1-p1)](2/5) + p1p2(3/10) P(B) = 11/40 C. P(C) = [p1(1-p2) + p2(1-p1)](2/5) + p1p2(7/10) P(C) = 15/40

Master Key-Solutions D. Given that a competitor has won the extensive prize, what is the likelihood that he had earned stand out key? P(X=1 | C) can be figured utilizing Bayes Rule (on side): P(X=1 | C) = 8/15

Plinko-about the amusement The hopeful is given one chip and has the chance to win 4 more. To gain the other 4 the hopeful is given items that are given a cost and should figure if the genuine cost is higher or lower. For each right reaction, a chip is remunerated.

Plinko-about the amusement After the challenger has their chips, they should drop the chips from any of the nine openings at the highest point of the Plinko board. On its way down, the chip will experience 12 pegs. On the off chance that the chip hits a peg beside the divider it will fall into the main open space, else it will tumble to one side or the privilege of the peg with equivalent likelihood. The most elevated prize for one chip is $10,000, so up to $50,000 can be won.

Plinko-Goals Binomial Experiments comprising of the perception of a succession of indistinguishable and free trials, each of which can bring about one of two results. Expected Value of a Random Variable Counting Combinations of n items taken r at once

Plinko-Question Contestants with different chips for the most part fluctuate the spaces from which they discharge the chips. Does the underlying arrangement of the chip matter? To choose, answer the accompanying inquiries:

Plinko-Question/Answer Question 1: For each of the three center openings at the highest point of the Plinko board (spaces 4,5, and 6), discover the likelihood that a chip beginning in every space brings about winning $10,000.

Plinko-Question/Answer 1: Let Y = # of pegs out of 12 that outcome in the chip tumbling to one side. (Y is not a binomial arbitrary variable in view of the limitations forced by the dividers of the load up) For a chip dropped in opening 5 to win $10,000, the chip must tumble to one side precisely 6 times (Y=6). (On the off chance that the chip hits a mass of the load up it has moved to one side or comfortable 8 times, and couldn't wind up in the $10,000 container. In this way, we may utilize the binomial conveyance with n=12 and p=0.5) P( win $10,000 beginning from opening 5) = P(Y=6) = ( 12 C 6 )(1/2)^12 = .2256 For a chip dropped in space 4 it will win $10,000 just if the chip tumbles to one side precisely 5 times (and to the right 7 times). Binomial conveyance applies here as above. P( win $10,000 beginning from opening 4) = ( 12 C 5 )(1/2)^12 = .1934 Same response for space 6 as opening 4.

Plinko-Question/Answer Question 2: If a chip is dropped in the center space of the Plinko board (opening 5), the sum won, U, has the accompanying circulation: If a chip is dropped in both of the openings nearby the center opening (space 4 or 6), the sum won, V, has the accompanying dissemination: Compute the normal rewards for a chip dropped in opening 5 and the normal rewards for a chip dropped in space 4 or 6.

Plinko-Question/Answer 2: Using opening 5: Using space 4 or 6:

Price is Right-Conclusion I like this venture on the grounds that a great many people like TPIR. Heaps of approaches to present this movement For instance, can begin by watching clasps of the recreations to get the understudies energized.

Probability of a Kiss-Activities Activity 1 Collecting and breaking down information Activity 2 Make expectations and showing information Activity 3 Properties of the appropriation of a specimen extent

Probability of a Kiss-Activity 1 Materials 10 plain Hershey's Kisses 16-oz plastic glass Students ought to be in gatherings of 3 or 4

Probability of a Kiss-Activity 1 Procedure Students talk about and appraise the likelihood a Kiss will arrive on its construct when it is hurled in light of the work area Leads to exchange of three sorts of probabilities, exact, subjective, and hypothetical. Observational likelihood can be considered as the most precise logical "guess" in view of the aftereffects of analyses to gather information around an occasion. Subjective likelihood portrays an individual's close to home judgment about how likely a specific occasion is to happen. Hypothetical likelihood is the proportion of the quantity of ways the occasion can jump out at the aggregate number of potential outcomes in the specimen space. For this situation, subjective probabilities are being doled out.

Probability of a Kiss-Activity 1 Now gathers put 10 Kisses into the glass and spill them onto the work area and record the quantity of Kisses that have arrived on their base in a table, including a line for the aggregate number of base arrivals. (Rehash 10 times). After the gatherings are done they are solicited to refine their past estimates from the likelihood the Kiss will arrive on its base. The understudies take part in a class dialog where they are asked how they could be increasingly sure of the probabilities. (More hurls fundamental)

Probability of a Kiss-Activity 2 Materials notwithstanding the 10 plain Kisses, likewise 10 almond Kisses. 16-oz plastic glass Students in gatherings of 3 or 4 once more

Probability of a Kiss-Activity 2 Procedure Students look at the two sorts of Kisses and examine which would have the higher likelihood of arriving on its base. Understudies then put every one of the 20 Kisses into the glass and spill them on the table and record the quantity of Kisses that have arrived on their base for every sort. Rehash 10 times. Understudies figure out how to manage "muddled information"

Probability of a Kiss-Activity 2 Displaying Data Stem and Leaf Plots Ex: Plain Almond | 1 | 8 9 8 4 | 2 | 0 2 4 6 7 8 7 6 5 4 2 1 0 | 3 | 0 1 3 6 3 0 | 4 |

Probability of a Kiss-Activity 2 Displaying information Boxplots Ex: Reviews discovering 1 st and 3 rd quartiles, medians, max and mins Students ought to discover exceptions for both information sets Calculate means and standard deviations

Probability of a Kiss-Activity 3 Materials: 30 plain Kisses, 30 almond Kisses, 16-oz container Procedure: Groups spill 10 plain Kisses onto the work area and record the number that arrive on its base (rehash 5 times) Repeat utilizing 20 plain Kisses Repeat utilizing 30 Kisses Do the same for the almond Kisses

Probability of a Kiss-Activity 3 Procedure (cont.) Groups join comes about with an accomplice gathering to get five hurls for n=60 and n=90 for every sort of Kiss. Record extents from both plain and almond Kisses in

SPONSORS

No comments found.

SPONSORS

SPONSORS