Registration link on the calendar page of the class blog FREE AP STAT MOCK EXAM SATURDAY, APRIL 28, 2018 Oconee Civic Center Registration link on the calendar page of the class blog

ACCURACY OF NORMAL APPROXIMATION: Increases when p is close to ½. Least accurate when p is close to 0 or 1. RULE OF THUMB: BINOMIAL OR NORMAL ? If n, p, and q satisfy the conditions np > 10 and nq > 10 you are EXPECTED to use a Normal Approximation instead of a Binomial Distribution.

Advanced Placement Statistics Section 8.2: Geometric Distributions   EQ: What are “wait-time” problems and how do you calculate the probability of these events?

Name Notation RECALL: Two Distributions Learned So Far: Normal N (µ,σ) Binomial B(n,p) What happens when n is NOT defined?

a) Define the random variable X. X = ___________________________ Ex. Suppose 40% of OCHS students have jumper cables in their car. Your car has a dead battery and you don’t have jumper cables. You decide to stop students who are headed to the parking lot and ask them if they have a pair of jumper cables. a) Define the random variable X.   X = ___________________________ number of students asked in the OCHS parking lot to locate jumper cables

Complete the following table: S FS FFS :   Sample Points x = # of students required for success S FS FFS : 1 2 3 Each possible outcome consists of 0 or more ________ followed by a single _________. failures success

Geometric Distributions --- “wait-time” P(X = x) = P(x trials to first success) P(FFFF…FS)  (1 – p)(1 – p) …(1 – p)(p)

P(X = n) = pqn - 1 Calculating Geometric Distributions Properties of a Geometric Distribution: two mutually exclusive outcomes __________ or ____________ success failure 2. trials are independent of each other 3. probability of success same for each trial

NOTE: What missing condition makes it different from a binomial distribution? *** X  0 WHY? You must have at least 1 attempt to have a success. Refer back to jumper cable problem:   d) What is the probability the first student you stop has jumper cable? P(X = 1) = ___________________ (0.4)1 = 0.4

means P(X = 1) + P(X = 2) + P(X = 3) e) What is the probability that three or fewer students must be stopped to get the jumper cables?   P(X < 3) = ___________________ (0.4)1 + (0.6)1 (0.4)1 +(0.6)2 (0.4)1 = .784 means P(X = 1) + P(X = 2) + P(X = 3)

Helpful Formulas for Geometric Distributions: Probability that it takes more than n trials to reach first success: P(X > n) = (1 – p)n What is the probability that more than 7 cars need to be stopped to get the jumper cables?

Helpful Formulas for Geometric Distributions: 2. Probability it takes at most n trials for the first success: P(X ≤ n) = 1 - qn What is the probability that at most 5 cars need to be stopped to get the jumper cables?

h) What is the mean number of cars you can expect to stop before getting a set of jumper cables?

Describe Histogram of Geometric Distribution: Why is it skewed right? Previous value multiplied by q until success is obtained.

In CLASS Assignment: p. 543 – 544 #41, 43 p. 550 #48 – 50 (a–g)