Learning objectives (and summaries)Recognize the binomial setting and calculate probabilities of binomial events.- Understand that the binomial setting is the basis for the normal curve
- As n->infinity and p=0.5, the distribution becomes entirely normal. A large number of 50/50 choices that lead to an overall outcome is why the normal distribution is so common.
- Recognize the binomial setting from a non-binomial setting
- An event with a fixed number of trials and a fixed probability of success for each trial is binomial.
- Read P(X>4) notation for the probabilities of discrete events.
- P(X=4): the probability that you get exactly 4 successes
- P(X≥4): the probability that you get 4 or more successes
- P(X>4): the probability that you get more than 4 successes (5 or more)
- P(X≤4): the probability that you get 4 or fewer successes
- P(X<4): the probability that you get less than 4 successes (3 or fewer)
- Find the probability of a binomial event using binomcdf(n,p,x) or binompdf(n,p,x) on a TI-83.
- Exactly one value: binompdf(n=number of trials,p=probability of success,x=how many successes)
- Cumulative range of values: binomcdf(n=number of trials,p=probability of success,x=how many successes OR FEWER)
Assessment- Quiz (8 pts): 5 questions (2 short answer/MC, 3 numeric); 1 of these free response questions (2pts):
- Something about independence?
- How does Pascal's Triangle connect to the binomial distribution?
- Free throws as binomial trials project
- Complete worksheet at bottom of page (use a PDF app such as Notability)
InstructionIf you do not have any intention of doing Honors/AP, you can use this online calculator instead of the TI-83/4. PracticeDetermine if each of the following situations, determine if the count has a binomial distribution. If it does, explain why each condition is met. If it does not, say what condition was not met.- 1. I flip a coin and count how many times it takes to land a “head”.
- 2. I draw 10 cards from a deck, one at a time, without replacement, and see how many 7’s come up.
- 3. I roll a die 100 times and count how many of each number I get.
- 4. I spin a spinner 12 times and count how many times it lands on the first region.
- 5. I throw a bean bag at a target, receive coaching, and repeat 5 times. I count how many bags hit the target.
- 6. I use a random number generator to determine success or failure. I count how many times until I succeed.
In the following situations, it is common to use the binomial distribution as a model. What cautions should we consider when using it in these situations? - 7. A basketball player shoots 25 free throws and counts her made shots.
- 8. A baseball player throws 80 pitches and counts his strikes (two reasons).
Plug and chug: use your calculator to find the probability of a single value in the following scenarios.- 9. 5 trials, p=0.3, find P(X=4).
- 10. 8 trials, p=0.625, find P(X=2).
- 11. 15 trials, p=0.5, find P(X=6).
- 12. 100 trials, p=0.1, find P(X=10).
First, underline the values you are looking for in the scenario. Then, write a statement of how you would use your calculator to find the probability of the range of values. Finally, calculate.- 13. 8 trials, p=0.625, find P(X≤5).
- 14. 8 trials, p=0.625, find P(X>5).
- 15. 8 trials, p=0.625, find P(X=5).
- 16. 8 trials, p=0.625, find P(X≥5).
- 17. 8 trials, p=0.625, find P(X<5).
Write a calculator statement for each of the following scenarios. Then calculate the probability.Billy makes 77% of his free throws. His probability of making the shot is fairly constant across different games. If Billy shoots 48 free throws this season…- 18. How many free throws do you expect Billy to make? Round to the nearest integer.
- 19. Using your answer for for the expected number (the mean), how likely is Billy to make exactly this number of free throws?
- 20. How likely is he to make at least 34 free throws?
- 21. How likely is he to make more than 41 free throws?
- 22. How likely is he to make at most 34 free throws?
- 23. How likely is he to make fewer than half of his free throws?
Beth wants to do a sampling experiment on a city where 32% of the people prefer to not eat meat. Beth takes a SRS of 120 people.- 24. What is the probability that she finds less than 34 who don’t prefer meat?
- 25. What is the probability that she finds more than 43 who don’t prefer meat?
- 26. How confident are you (i.e. what is the probability) that Beth will find 34 to 43 people who don’t prefer meat?
EXTRA: First, underline the values you are looking for in the scenario. Then, write a statement of how you would use your calculator to find the probability of the range of values. Finally, calculate.- 1. 5 trials, p=0.3, find P(X≤3).
- 2. 5 trials, p=0.3, find P(X<3).
- 3. 5 trials, p=0.3, find P(X≥3).
- 4. 5 trials, p=0.3, find P(X>3).
- 5. 5 trials, p=0.3, find P(X=3).
EXTRA: Mary is a consistent, expert bean bag tosser who lands 61% of her throws in the hole. If Mary throws 100 bags…- 6. What is the mean of this binomial distribution? What does the “mean” mean in the context of this situation?
- 7. What is the standard deviation of this distribution?
- 8. What number of made throws is two standard deviations below the mean? Find the probability that Mary makes fewer than this number.
- 9. What number of made throws is two standard deviations greater than the mean? Find the probability that Mary makes more than this number.
- 10. How likely is Mary to make exactly 62 throws?
- 11. How likely is Mary to make at most 50 throws?
EXTRA: Jack wants to do a sampling experiment on a classroom where 57% of the people prefer Batman over Iron Man. Jack takes a random sample of 12 people.- 12. What is the probability that he finds 4 or fewer Batman fans?
- 13. What is the probability that he finds 10 or more Batman fans?
- 14. How confident are you (i.e. what is the probability) that Jack will find 5 to 9 Batman fans?
EXTRA: Practice crunching binomial probabilities.- 15. 9 trials, p=0.5, find P(X≤9).
- 16. 2 trials, p=0.3, find P(X<1).
- 17. 19 trials, p=0.25, find P(X≥5).
- 18. 100 trials, p=0.9, find P(X>90).
- 19. 51 trials, p=0.44, find P(X=40).
- 20. 12 trials, p=0.4, find P(X≤5).
- 21. 18 trials, p=0.8, find P(X>10).
- 22. 42 trials, p=0.23, find P(X=12).
- 23. 155 trials, p=0.52, find P(X≥70).
- 24. 805 trials, p=0.78, find P(X<5).
Practice solutionsDetermine if each of the following situations, determine if the count has a binomial distribution. If it does, explain why each condition is met. If it does not, say what condition was not met. - 1. No -- the number of trials is not predetermined.
- 2. No -- the probability changes with each draw since the cards are not put back into the deck.
- 3. No -- there are more than two possible outcomes (there are 6).
- 4. Yes -- 12 trials, success or failure, probability does not change between spins.
- 5. No -- the probability changes with each toss due to coaching.
- 6. No -- the number of trials is not fixed.
In the following situations, it is common to use the binomial distribution as a model. What cautions should we consider when using it in these situations? - 7. In any sport, the probability of success in each trial probably changes a little bit depending on the pressure of the situation or where the game is played.
- 8. In addition to the reason above, success is not as clearly defined when pitching since it is up to the judgement of the umpire.
Plug and chug: use your calculator to find the probability of a single value in the following scenarios.- 9. 0.028
- 10. 0.030
- 11. 0.157
- 12. 0.132
First, underline the values you are looking for in the scenario. Then, write a statement of how you would use your calculator to find the probability of the range of values. Finally, calculate.- 13.
__0 1 2 3 4 5__6 7 8, binomcdf(8, 0.625, 5) = 0.630 - 14. 0 1 2 3 4 5
, 1 - binomcdf(8, 0.625, 5) = 0.370**6 7 8** - 15. 0 1 2 3 4
6 7 8, binompdf(8, 0.625, 5) = 0.282**5** - 16. 0 1 2 3 4
, 1 - binomcdf(8, 0.625, 4) = 0.651__5 6 7 8__ - 17.
**0 1 2 3 4**5 6 7 8, binomcdf(8, 0.625, 4) = 0.349
Write a calculator statement for each of the following scenarios. Then calculate the probability.Billy makes 77% of his free throws. His probability of making the shot is fairly constant across different games. If Billy shoots 48 free throws this season…- 18. n*p = 48*0.77 = 36.96 ~ 37
- 19. binompdf(48, 0.77, 37) = 0.136
- 20. 0 ... 33
, 1 - binomcdf(48, 0.77, 33) = 0.880__34 35 ... 48__ - 21. 0 ... 33 41
, 1- binomcdf(48, 0.77, 41) = 0.053**42 ... 48** - 22.
**0 ... 33 34**35 ... 48, binomcdf(48, 0.77, 34) = 0.197 - 23.
**0 ... 23**24 25 ... 48, binomcdf(48, 0.77, 23) = 0.000 (0.0000112)
Beth wants to do a sampling experiment on a city where 32% of the people prefer to not eat meat. Beth takes a SRS of 120 people.- 24.
**0 ... 33**34 35 ... 120, binomcdf(120, .32, 33) = 0.169 - 25. 0 ... 42 43
, 1 - binomcdf(120, .32, 43) = 0.159**44 ... 120** - 26. Total area for any probability is 1, so take 1 - left side - right side = 1 - .169 - .159 = .672. Thus, you are 67.2% confident that you will find between 34 and 43 people who don't prefer meat.
EXTRA: First, underline the values you are looking for in the scenario. Then, write a statement of how you would use your calculator to find the probability of the range of values. Finally, calculate.- 1.
**0 1 2 3**4 5, binomcdf(5, 0.3, 3) = 0.969 - 2.
**0 1 2**3 4 5, binomcdf(5, 0.3, 2) = 0.837 - 3. 0 1 2
, 1 - binomcdf(5, 0.3, 2) = 0.163__3 4 5__ - 4. 0 1 2 3
, 1 - binomcdf(5, 0.3, 3) = 0.031__4 5__ - 5. 0 1 2
4 5, binompdf(5, 0.3, 3) = 0.132**3**
EXTRA: Mary is a consistent, expert bean bag tosser who lands 61% of her throws in the hole. If Mary throws 100 bags…- 6. n*p = 100*0.61 = 61. The mean on a binomial distribution is the expected value (the value with the highest individual probability).
- 7. √(n*p*(1-p)) = √(100*0.61*0.39) = 4.877
- 8. 61 - 2*4.877 = 51.245, P(X ≤ 51),
52 ... 100 since 51 < 51.245, binomcdf(100, 0.61, 51) = 0.027**0 ... 50 51** - 9. 61 + 2*4.877 = 70.755, P(X ≥ 71), 0 ... 70
since 70.755 < 71, 1 - binomcdf(100, 0.61, 70) = 0.024__71 72 ... 100__ - 10. binompdf(100, 0.61, 62) = 0.080
- 11.
**0 ... 49 50**51 ... 100, binomcdf(100, 0.61, 50) = 0.016
EXTRA: Jack wants to do a sampling experiment on a classroom where 57% of the people prefer Batman over Iron Man. Jack takes a random sample of 12 people.- 12.
**0 ...3 4**5 ... 12, binomcdf(12, .57, 4) = 0.087 - 13. 0 ... 9
, 1 - binomcdf(12, .57, 9) = .056**10 11 12** - 14. Total area for any probability is 1, so take 1 - left side - right side = 1 - .087 - .056 = .857. Thus, you are 85.7% confident that you will find between 5 and 9 people who prefer Batman.
EXTRA: Practice crunching binomial probabilities.- 15.
**0 ... 8 9**, use common sense -- every number is included so probability = 1 - 16.
**0**1 2, binomcdf(2, 0.3, 0) = 0.49 - 17. 0 ... 4
, 1 - binomcdf(19, 0.25, 4) = 0.535**5 6 ... 19** - 18. 0 ... 89 90
, 1 - binomcdf(100, 0.9, 90) = 0.451**91 ... 100** - 19. 0 ... 39
41 ... 51, binompdf(51, 0.44, 40) = 0.000 (4.4 x 10^-7)__40__ - 20.
**0 ... 4 5**6 ... 12, binomcdf(12, 0.4, 5) = 0.665 - 21. 0 ... 9 10
, 1 - binomcdf(18, 0.8, 10) = 0.984__11 ... 18__ - 22. 0 ... 11
13 ... 42, binompdf(42, 0.23, 12) = 0.095**12** - 23. 0 ... 69
, 1 - binomcdf(155, 0.52, 69) = 0.963__70 71 ... 155__ - 24.
**0 ... 4**5 6 ... 805, binomcdf(805, 0.78, 4) = 0.000 (the probability is VERY small)
AP Stats*AP Formula Sheet Here:*tinyurl.com/apstatssheet- AP FRQ 2004 #3 (solution)
- AP FRQ 2010B #3 (solution)
Notes |

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