I need help with a math project concerning probability distribution

Math | The Math Help Column | "Solutions for Your Everyday Math Dilemmas" The Math Help Column "Solutions for Your Everyday Math Dilemmas" By Student Name Dear Math Helper, I am in need of your help! I have a job where I design games. I am designing a game that involves a playing area with 12 equal sections. Players randomly end up in one of the sections. I labeled one section, "Gain 40 points," four sections, "Gain 10 points," and the rest, "Try Again." Also, each player automatically gives up 10 points with each move. After trying out the game on some volunteers, I noticed that players tend to lose points in the long run. Without changing the sizes of the sections, what adjustments can I make so that players are likely to earn at least one point, in the long run? Signed, Game Maker Dear Game Maker, It is easy to adjust your spinner so that players aren't so likely to lose points in the long run. First, you need to understand something called expected value. It's the amount you would expect to get, on average, after many spins. To find an expected value, multiply each outcome by its probability and find the sum of those products. As is, participants have a chance of landing on "Gain 40 points." Because they gave up 10 points at the beginning 1 12 of their turn, they earn 30 points each time this happens. They have a 4 12 chance of landing on "Gain 10 points." Because they gave up 10 points to begin with, they break even. Finally, they have a 7 12 chance of landing on "Try Again." Because they gave up 10 points, they end up losing the 10 points when this happens. I'll put all this, and the products of the outcomes and probabilities, in a table. Points earned Probabilities Product The expected value is 31 3 30 points 0 points 1 12 4 12 2.50 points 0 points 2.5 0 5.83 3.33 points 10 7 12 5.83 points . So, you can see that a player is likely to lose an average of about points in the long run. 2015 K12 Inc. All rights reserved. Copying or distributing without K12's written consent is prohibited. Page 1 of 7 Math | The Math Help Column | "Solutions for Your Everyday Math Dilemmas" Now, look what happens when you change the \"Gain 40 points\" section to \"Gain 100 points.\" Points earned Probabilities Product The expected value is average of 12 3 90 points 0 points 1 12 4 12 7.50 points 0 points 7.50 0 5.83 1.67 points 10 7 12 5.83 points . This is much better! In the long run, the players will earn an points per turn. This is just one option. You can play around with the outcomes until you achieve the desired results. Hope this helps! 2015 K12 Inc. All rights reserved. Copying or distributing without K12's written consent is prohibited. Page 2 of 7 Math | The Math Help Column | "Solutions for Your Everyday Math Dilemmas" Dear Math Helper, First, let me tell you how much I enjoy your column. It is the first thing I read in every issue. I run a carpet-cleaning company and need some advice. I want to offer steam-cleaner rentals because some people prefer to save money by cleaning their own carpets. But steam cleaners are expensive and they need repairs if not used properly, so I'm going to offer insurance policies. I'm not sure how much to charge. It will cost me $190 to replace one of my cleaners, and repairs average $100. I talked to some other people I know in the business and they said I should expect to replace a cleaner for about 1 of every 50 rentals and to repair a cleaner for about 1 of every 20 rentals. I'm not sure what to do with all this information. Please help me determine a fair price for my insurance policy. I want my rates to be low enough that customers buy the insurance, but high enough that I don't lose money by offering it. And please show me how you came up with your figure, so that I can make adjustments later if I need to. Your number one fan, Steaming in Seattle Dear Steaming in Seattle, Thank you for your kind words. The first thing you need to do is to find the expected cost for replacements and repairs, on average, per rental. Do this by multiplying the cost of each outcome (replacement, repair, neither) by its probability Then find the sum. I'll organize the information in a table for you, writing the probabilities you provided as decimals. Outcome Replacement Repair Neither Cost of Outcome $190 $100 $0 Probability 0.02 0.05 0.93 Product $3.80 $5.00 $0 The sum of the products is $8.80. In the long run, each rental will cost you $8.80 on average. Pricing the insurance policy at $8.80 offsets your costs. Any price over $8.80 gives you a profit. I would suggest charging about $10 for the insurance policy. This gives you a little wiggle room. 2015 K12 Inc. All rights reserved. Copying or distributing without K12's written consent is prohibited. Page 3 of 7 Math | The Math Help Column | "Solutions for Your Everyday Math Dilemmas" Dear Math Helper, I work in construction and am pretty good in math, especially with fractions, so I didn't think I would ever have to write to you. But this one has got me stumped. Last week I bought two boxes of nails, each from a different manufacturer. I poured the nails from one box into my left pocket and the other into my right. I started to use the ones in my left pocket, but they kept breaking and bending. I will never use them again. But here's my problem. I don't remember which box they came from. I contacted both manufacturers and "Manufacturer A" said its boxes have a mean of 205.4 nails and a standard deviation of 1.8 nails. "Manufacturer B" said its boxes have a mean of 190.6 nails and a standard deviation of 3.3 nails. Both said their number of nails is normally distributed. I counted the nails from my left pocket, including the used ones, and got 198. This is 7.4 nails less than Manufacturer A and 7.4 nails more than Manufacturer B. So I guess the standard deviation comes into play, but I don't know how. Please explain which manufacturer most likely made these defective nails. Sincerely, Stumped Builder Dear Stumped Builder, Yes, the standard deviation comes into play. It is an indication of how widely spread the number of nails in their boxes can be. I am going to show you a formula that converts the number 198 into a standard score for each manufacturer. Then, the standard score closer to 0, without regard to it being positive or negative, belongs to the more likely manufacturer. Standard Score for Manufacturer A 198 mean 198 205.4 4.1 standard deviation 1.8 B 198 mean 198 190.6 2.2 standard deviation 3.3 Standard Score for Manufacturer The score for Manufacturer B is closer to 0, so they are more likely to have made those defective nails. 2015 K12 Inc. All rights reserved. Copying or distributing without K12's written consent is prohibited. Page 4 of 7 Math | The Math Help Column | "Solutions for Your Everyday Math Dilemmas" Dear Math Helper, Please help settle this friendly dispute. My friend, let's call her Pam, took the same history course I did, but we had different professors. On the final exam, I scored 82 out of 100 points while she scored 47 out of 63 points. My percent score, 82%, is greater than her percent score, which is about 75%. So I think I did better than she did on the final exam. But she tells me that both sets of scores are normally distributed and that I have to look at the standard scores to find out who did better. Pam tells me that the mean and standard deviation for my test are 76.1 and 6.5, respectively, while hers are 40 and 4.2 respectively, so if I "do the math," then I should see why she performed better. Can you please explain what she means by standard scores and can you "do the math" that she mentions? Yours, Confused History Buff P.S. If she did perform better, then I have to buy her dinner! Dear Confused History Buff, It looks like you're buying your friend dinner. The percent's do not take into account the difficulty of the test. A better comparison looks at how each of you did in relation to the rest of the class who took the same test. You can do that with standard scores. Your standard score is what you get when you subtract the mean for your test from your score and then divide by the standard deviation for your test. Likewise, to find her standard score, subtract the mean for her test from her score and divide by the standard deviation for her test. Your standard score: score mean 82 76.1 0.91 standard deviation 6.5 Pam's standard score: score mean 47 40 1.67 standard deviation 4.2 Because her score is higher, she outperformed more people in her class than you did in yours. You did well, but not quite as well as she did. 2015 K12 Inc. All rights reserved. Copying or distributing without K12's written consent is prohibited. Page 5 of 7 Math | The Math Help Column | "Solutions for Your Everyday Math Dilemmas" Dear Math Helper, You write a great column, and I hope you stay around for years to come. I run a small boarding school. We are going to begin using test scores to help determine admissions eligibility. The scores for the test I have in mind are normally distributed with a mean of 300 and a standard deviation of 12.5. Can you show me how to determine the cutoff score that we should use to only accept applicants who score in the top 10% on the test? What cutoff score includes the top 15%? Thanks in advance. Regards, The Student Selector Dear Student Selector, Thank you, I hope to be here for years to come as well. To determine the top 10%, you have to first determine the standard score that has 90% of the distribution below it. You can find this number in a table in a statistics book. I will tell you that it is 1.28. Then use this formula: minimum score (standard score gstandard deviation) mean (1.28 g12.5) 300 16 300 316 If you want only the top 10%, they need a score of 316 or better. If you want the top 15%, you need the standard score for 85% of the distribution below it. This z-value or z-score is 1.04. Then use this formula: minimum score (standard score gstandard deviation) mean (1.04 g12.5) 300 13 300 313 Use a cutoff score of 313 if you decide to let in the top 15%. 2015 K12 Inc. All rights reserved. Copying or distributing without K12's written consent is prohibited. Page 6 of 7 Math | The Math Help Column | "Solutions for Your Everyday Math Dilemmas" Dear Math Helper, How's it going? I'm in the business of printing T-shirts and just got an interesting order. My sister is a pre-med student and wants to show off how well she performed on the Medical College Admission Test ( ). I told her I'd make her a T-shirt, but she's not making it easy for me. MCAT She wants the back of the shirt to read, "I scored at the nth percentile on the MCAT !" But instead of giving me the value of n, she told me her individual score was 38 and that all the test scores are normally distributed with a mean of 25 and a standard deviation of 6.4. I don't want to let her think she's smarter than me, even if she is going to be a doctor! Can you explain how I should use these numbers to find the value of n for her shirt? If so, maybe there's a free T-shirt in your future! Many thanks, T-shirt Guy Dear T-shirt Guy, It's going great and I'd love a new T-shirt. The percentile is the percent of scores below your sister's score. The first step is converting her score of 38 to a standard score. There is a formula for this: her score mean score 38 25 2.03 standard deviation 6.4 To determine the percent of scores below this standard score, you can look at a z-distribution table in a statistics book. The percent turns out to be about 97.88%. My guess is that she'll want you to round that to 98%. 2015 K12 Inc. All rights reserved. Copying or distributing without K12's written consent is prohibited. Page 7 of 7