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Warm up

Warm up. How many ways are there to spell SUCCESS? (moving only along diagonals) S S S E E E C C C C C U U S. Warm up. How many ways are there to spell SUCCESS? S 18 S 9 S 9 E 3 E 6 E 3 C 3 C 3 C 1 C 2 C 1

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Warm up

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  1. Warm up • How many ways are there to spell SUCCESS? (moving only along diagonals) S S S E E E C C C C C U U S

  2. Warm up • How many ways are there to spell SUCCESS? S18 S9 S9 E3 E6 E3 C3 C3 C1 C2 C1 U1 U1 S1

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

  4. Our Problem… • A batter either gets a hit or he doesn’t • Suppose his lifetime batting average is 0.292 •  the probability that he will get a hit in any AB (at-bat) is 0.292 • Why is lifetime average more reliable than season average? • As the leadoff hitter he gets 5 AB in a game • What are the possible outcomes for the number of hits he gets in 5 AB? • Do they have different probabilities?

  5. 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 and failure)

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

  7. 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

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

  9. Example 1a • 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

  10. Another solution

  11. Example 1b • Suppose the game is changed so that you win if you get at least 3 heads • what is the probability of winning now?

  12. 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

  13. a. What is the probability that he will get no hits in the next 5 at bats?

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

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

  16. 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 is expected to get 3 hits in the next 10 at bats

  17. MSIP/ Homework • p. 299 #1, 3, 7, 8 – 12

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

  19. 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

  20. 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

  21. 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

  22. 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?

  23. 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

  24. 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

  25. 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

  26. 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

  27. 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

  28. 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

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

  30. 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

  31. 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

  32. 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%

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

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