1. Binomial Distributions Chapter 5.3 – Probability Distributions and Predictions Mathematics of Data Management (Nelson) MDM 4U

2. Our Problem… Students either like math or they don’t Suppose 5% of students like math 300 students, how likely is it that 20 like math? Can be modeled as a binomial distribution How likely it is that a situation has occurred randomly? More significant if it is unlikely to have occurred randomly

3. Binomial Experiments a binomial experiment is any experiment that has the following properties:  n identical trials  two possible outcomes for each trial, termed success and failure  the probability of success is p  the probability of failure is 1-p  the probabilities remain constant  the trials are independent Bernoulli Trials - repeated independent trials with 2 possible outcomes (success/failure)

4. Bernoulli? Jakob Bernoulli (Basel, December 27, 1654 - August 16, 1705)Basel December 271654 August 161705 Swiss Mathematician one of the great names in probability theory

5. Binomial Distributions in a binomial experiment the number of successes in n repeated Bernoulli Trials is a discrete random variable (usually called X) X is termed a binomial random variable and its probability distribution is called a binomial distribution the following formula provides a method of solving highly complex situations involving probability

6. Binomial Probability Distribution consider a binomial experiment in which there are n Bernoulli trials, each with a probability of success of p the probability of k successes in the n trials is given by:

7. Example 1 Consider a game where a coin is flipped 5 times. You win the game if you get exactly 3 heads. What is the probability of winning? we will let heads be a success n = 5 p = ½ k = 3

8. Example 1 continued suppose the game is changed so that you win if you get at least 3 heads what is the probability of winning now?

9. The Batting Example the Expected Value of a binomial experiment that consists of n Bernoulli trials with a probability of success, p, on each trial is  E(X) = n(p) Example: Consider a baseball player who has a lifetime batting average of 0.292 this means that his probability of getting a hit each time he is at bat is 0.292 let a hit be a success where p = 0.292

10. a. What is the probability of no hits in the next 5 at bats?

11. b. What is the probability of 2 hits in the next 8 at bats?

12. c. What is the probability of at least 1 hit in the next 10 at bats?

13. d. What is the expected number of hits in the next 10 at bats? E(X) = np E(X) = (10)(0.292) = 2.92 → 3 therefore the player can expect to get 3 hits in the next 10 at bats

14. MSIP/ Homework p. 299 #1, 3, 7, 8, 9, 10, 11, 12

15. Normal Approximation of the Binomial Distribution Chapter 5.4 – Probability Distributions and Predictions Mathematics of Data Management (Nelson) MDM 4U

16. Recall… the probability of k successes in n trials (where p is the probability of success) is this formula can only be used if we have a binomial distribution:  each trial is identical  the outcomes are either success or failure

17. This calculation is easy in simple cases… Find the probability of 30 heads in 50 trials So there is about a 4.2% chance However, if we wanted to find the probability of tossing between 20 and 30 heads in 50 trials, we would need to perform 11 of these calculations But…there is an easier way

18. Graphing the Binomial Distribution If the distribution is normal, we can solve complex problems in the same way we did in the last chapter the question is: is the binomial distribution a normal one? it turns out that if the number of trials is relatively large, the binomial distribution approximates a normal curve

19. What does it look like? when graphed the distribution of probabilities of heads looks like this what will the mean be? what will the standard deviation be?

20. So how do we work with all this it turns out that a binomial distribution can be approximated by a normal distribution if:  n x p > 5 and n x (1 – p) > 5 if this is the case, the distribution is approximated by the normal distribution

21. But doesn’t a normal curve represent continuous data and a binomial distribution represent discrete data? Yes! so to use a normal approximation we have to consider a range of values rather than specific discrete values The interval for a value is from 0.5 below to 0.5 above, i.e., the interval for 10 goes from 9.5 to 10.5

22. Example 1 Tossing a coin 50 times, what is the probability that you will get tails less than 20 times let success be tails, so n = 50 and p = 0.5 n x p = 50(0.5) = 25 > 5 n x (1 - p) = 50(1 - 0.5) = 25 > 5 now we can find the mean and the standard deviation

23. Example 1 continued we will consider 0-19.5 times (values below 20 - the interval from 19.5-20.5), and use it to calculate a z-score z = 19.5 – 25 = -1.55 3.54 therefore P(X < 19.5) = P(z < -1.55) = 0.0606 there is a 6% chance of less than 20 tails in 50 attempts

24. In terms of the normal curve, it looks like this all the values less than 19.5 are found in the shaded area 19.5 25.0

25. Example 2 Two dice are rolled and the sum recorded 40 times. What is the probability that a sum greater than 6 occurs in over half of the trials? let p be the probability of getting a sum greater than 6 p = 6/36 + 5/36 + 4/36 + 3/36 + 2/36 + 1/36 p = 7/12 now we can do some calculations

26. Example 2 continued the probability of getting a sum greater than 6 on at least half of the trials is 100 – 18 = 82%

27. Example 3 you have a drawer with one blue mitten, one red mitten, one pink mitten and one green mitten if you closed your eyes and picked a mitten at random 200 times (with replacement) what is the probability of choosing the pink mitten between 50 and 60 times (inclusive)? so, success is considered to be drawing a pink mitten, with n = 200 and p = 0.25

28. Example 3 Continued check to see whether the normal approximation can be used np = 200(0.25) = 50 n(1 – p) = 200(0.75) = 150 since both of these are greater than 5 the binomial distribution can be approximated by the normal curve now find the mean and standard deviation

29. Example 3 Continued the probability of having between 50 and 60 pink mittens (inclusive) drawn is 0.9564 – 0.4681 = 0.4883 or about 49%

30. MSIP/ Homework Read the example on page 310 do p. 311 # 4-10