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Chapter 17: The Binomial Model. Part 4 When to substitute the normal model

Chapter 17: The Binomial Model. Part 4 When to substitute the normal model. AP Statistics B. Outline of today’s class. Problems/limitations with the binomial model Alternatives to the binomial model that are easier to calculate

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Chapter 17: The Binomial Model. Part 4 When to substitute the normal model

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  1. Chapter 17: The Binomial Model. Part 4When to substitute the normal model AP Statistics B

  2. Outline of today’s class • Problems/limitations with the binomial model • Alternatives to the binomial model that are easier to calculate • Applications of the alternatives when we have large populations (large n, that is)

  3. Warmup, Statistics • Write out, but DO NOT CALCULATE getting exactly 247 heads on 500 flips of an unbiased coin using the binomial model. Answer these among yourselves and agree on an answer. Worked out problem are on the next slides.

  4. Answer to warm-up • What’s the formula we use? • So for this case, the answer would be • Too much to calculate! (If you tried to calculate, please report immediately to the nurse to be examined…..)

  5. The binomial model:Limitations • Great for small numbers (less than 20) • Really REALLY bad for big numbers (500! has about 100,000 digits to it) • Complicated calculations • So, what’s a brother to do? • What LOOKS like a binomial distribution? • Well, let’s graph a few and see what we find….

  6. Binomial model: graphs (1) • The graph on the right is a typical example, here where n=21

  7. Binomial model: graphs (2) • Here’s a more typical example, one close to what we’ve worked with (n=15) • Here, p=1/3, which means what? • That we’re successful one-third of the time, or one time out of three. • Note that the curve is skewed—which way?

  8. What do these curves look like? • Chapter 6, maybe? • Et seq.? • They have means and standard deviations? • Yes, it’s our old friend (drum roll please……) • THE NORMAL DISTRIBUTION! • Let’s compare the two……

  9. The binomial model:Comparing binomial and normal distributions Binomial (discrete) Normal (continuous)

  10. OK, so they’re alike. How do we get from one to the other? • Well, what can you calculate using the binomial formula and what you know about the binomial distribution? • Mean? • Standard deviation? • So take a minute, review your notes, and have somebody write the mean and standard deviation for a binomial distribution!

  11. The binomial model: • That’s right; here they are: • Mean: • Standard deviation: • These are REALLY important for you to know cold (i.e., forget some of the lyrics to your favorite song from 7th grade and remember these guys instead)

  12. The binomial model:converting and to normal values • Compute and using the formulas just listed • For this case we have μ=n ×p=500 ×½=250 • σ is a bit more complex: • OK, so now what?????? • Now we have a mean and a standard deviation and a normal model:

  13. The binomial/normal model:what’s next? • Remember the formula? • Look back in your notes, and somebody write it on the board.

  14. The binomial/normal model:z-formula • That’s right: • For N(250, 11.18), what does this look like? • Y is 247 • Mu is 250 • Sigma is 11.18. Except what’s the problem with this analysis? Discuss among yourselves and come to a conclusion

  15. The binomial/normal model:z-formula • The problem is that this measures the areas to the LEFT of y, right? • So we don’t get an exact calculation. • How can we approximate?

  16. The binomial/normal model:z-formula • We approximate by calculating an interval. • Example: use y=246.5 for one calculation, and y= 247.5 for the second, and subtract the first from the second • Not precise, but very VERY close. • Most problems are easier. Let’s look at the one in the book.

  17. Practice (Red Cross example):The Binomial Horror P. 394 of the text: Red Cross looking for 0-negative donors. p=0.06 and n=32,000. Red Cross needs at least 1850 pints (each person donates 1 pint; any more and you get really dizzy, not to say ill) Using the binary distribution, you have to calculate each y from 1850 through 32,000. That is, you’re going to have 32,000-1850+1 calculations=30,151 calculations that look like this:

  18. Practice (Red Cross example):Normal curve to the rescue • Here, the normal curve really saves us • Mean is np, or 32,000×0.06=1920 • Standard deviation is • So now we can apply our z model and get • Our tables or calculators tell us that only about 0.05 lie below this point, and that’s your answer.

  19. Practice:How to apply this procedure • If n>9, you can use the normal model. • Calculate • Adopt normal model as • Apply normal formula (y-μ)/σ and solve as usual. • Warning: be sure you know whether you’re looking above or below y

  20. Practice:Chapter 17, problem 22 (p. 399) • Have someone read the problem • Work together at your tables to solve the problems. Everybody needs to be involved; no free riders (pace Hickman, although in my experience most libertarians don’t believe in free riders whereas lawyers always do) • Share answers among yourselves and reach a consensus • When you’re ready, move on to next slide for my analysis

  21. Practice:Problem 17(a) • This is a standard plug-and-chug problem • We know that • Here, n=200, p=0.8, and q= 0.2 • So μ=160 and σ=5.66

  22. Practice:Problem 17(b) • No problem with using the normal model here. Why? (you can find the answer in the text on p. 394. If nobody knows, find it and read it now!)

  23. Practice:Problem 17(c) • This is a common question you should expect on any statistics exam. It’s also easy to answer. • Take the mean, and add and subtract one standard deviation. Repeat for two- and three-standard deviations. • In notation, calculate μ±σ, μ±2σ, and μ±3σ • These correspond to the 68-95-99.7 model • Numerical answers: • 68% of the time she gets between 154.34 and 165.66 bull’s eyes • 95% of the time between 148.68 and 171.32 bull’s eyes, and • 99.7% of the time between 143.02 and 176.98 bull’s eyes

  24. Practice:Problem 17(d) • Use the answers from (c) to answer this question. • We know that 99.7% of the time, the archer will get between 143.02 and 176.98 bull’s eyes • If she gets 140 bull’s eyes or less, that’s only 0.0015%......VERY rare. • One more slide for today

  25. Homework: • Chapter 17, Exercises 24, 26, 27, 34 and 36. • Read Chapters 18 and 19….we will be going REALLY fast from now on. • (Yes, I know: “eez now time for fiendish plan!” which is from Boris Badenov of “Rocky and Bullwinkle”, which I expect you to become familiar with)

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