p. 403 – 404 #71 – 73, 75 – 78, 80, 82, 84 #72B: Binary? Yes – Success is a person is left-handed. I: Independent? Yes, since students are selected randomly, their handedness is independent. N: fixed number of observations? Yes; n = 15 S: Same probability of success for each student p = 0.10 Yes, this is a binomial setting. #74 (a) Yes, this is a binary setting. B: Binary? Yes – Success is reaching a live person I: Independent observations? It is reasonable to believe that each call is independent of the others. N: fixed number of observations? Yes, n = 15 S: Same probability of success? Yes, p = 0.2 (b) This is NOT a binomial setting because there are not a fixed number of attempts.

p. 403 – 404 #71 – 73, 75 – 78, 80, 82, 84 # There is about a 31.51% chance that exactly 1 of the 10 plants will die before producing any rhubarb # There is about a 1.15% chance that 3 or more of the plants will die before producing rhubarb. This would be surprising if it occurred. #80(a) (b) There is about a 6% chance of finding 4 or more lefties in a sample of 15. This would be moderately surprising but not completely unexpected. #82(a) μ x = 2.4. You would expect to find an average of 2.4 people that the machine finds to be deceptive when testing 12 people actually telling the truth. (b) σ x = In actual practice, you would expect the number of “deceivers” to vary from 2.4 by an average of #84(a) P(Y≥10) = (it is the same as P(X≤2) because if 2 or less are lying, then we are saying that 10 or more are telling the truth. (b) μ y = 9.6, which is 12 - μ x. The spread σ x and σ y are the same.

Warm-Up/Review Tastes as good as the Real Thing?

At the end of the lesson, students can: Sampling without replacement condition Determine when to use a normal approximation for a binomial distribution Graph binomial distributions on the calculator

Remember this example from last class? Deal 10 cards from a shuffled deck of 52 cards. X = # of red cards. We said this was NOT a binary setting because the cards were not being replaced, hence making it not independent. However, in almost all real-world sampling, such as taking an SRS from a population of interest, is done without replacement. When the population is much larger than the sample, a count of successes in an SRS of size n has approximately the binomial distribution ….but what counts as “MUCH LARGER”?

When taking a SRS of size n from a population of size N, we can use a binomial distribution to model the count of successes in the sample AS LONG AS.. n ≤ 1/10 (N) This is also known as the 10% rule – as long as the sample size n is less than or equal to 10% of the larger population N. This does NOT mean that we need small samples! If we have a sample that is larger than 10% of the population, it just means that we should NOT USE the binomial distribution.

Example: Almost everyone has one – a drawer that holds miscellaneous batteries of all sizes. Suppose that your drawer contains 8 AAA batteries but only 6 of them are good. You need to choose 4 for your graphing calculator. If you randomly select 4 batteries, what is the probability that all 4 of the batteries you choose will work? Problem: Explain why the answer isn’t: The actual probability is Solution: Since we are sampling without replacement, the selections of batteries aren’t independent. We can ignore this problem if the sample we are selecting is less than or equal to 10%. However, we are choosing 4 out of 8, which is 50%, so it is unreasonable to ignore the lack of independence.

The Normal Model to the Rescue!

As the number of observations ( n ) gets larger, the binomial distribution gets close to being a Normal distribution ! As n increases, our binomial formulas become LESS accurate and the Normal approximations become MORE accurate.

RULE OF THUMB: (remember this!) Success/Failure Condition: We will use the Normal approximation to the binomial distribution when n and p satisfy: 1.np ≥ 102. n(1 – p) ≥ 10 (# successes ≥10) (# failures≥10) Before you compute any probabilities, check to see if these conditions are satisfied. If they are, then use the Normal approximation; if not, use the binomial formula or calculator.

Suppose that X has the binomial distribution with n trials and probability of success p. When n is large (i.e. the Rule of Thumb is satisfied), then the distribution of X is approximately Normal with: Mean = μ x = np Standard Deviation = σ x = √np(1 – p) N (np, √np(1 – p) ) The accuracy of the Normal approximation improves as the sample size (n) increases. It is most accurate when p is close to 1/2 and is least accurate when p is close to 0 or 1..

Sample surveys show that fewer people enjoy shopping than in the past. A survey asked a nationwide random sample of 2500 adults if they agreed or disagreed that “I like buying new clothes, but shopping is often frustrating and time-consuming.” The population that the poll wants to draw conclusions about is all US residents aged 18 and over. Suppose that exactly 60% of all adult US residents would say “agree” if asked the same question. Let X = the number in the sample who agree. (a)Show that X is approximately a binomial random variable. (b)Check the conditions for using a Normal approximation to this setting. (c)Use a Normal distribution to estimate the probability that 1520 or more of the sample agree.

Sample surveys show that fewer people enjoy shopping than in the past. A survey asked a nationwide random sample of 2500 adults if they agreed or disagreed that “I like buying new clothes, but shopping is often frustrating and time-consuming.” The population that the poll wants to draw conclusions about is all US residents aged 18 and over. Suppose that exactly 60% of all adult US residents would say “agree” if asked the same question. Let X = the number in the sample who agree. (a) Show that X is approximately a binomial random variable. Binary? Yes  Success = people who agree Independent? Yes  We can assume that each adult’s response does not affect another’s response. Since we are sampling w/o replacement, check the 10% rule, too! It is reasonable to believe that there are at least 25,000 US adults in the population to draw the sample. Number? Yes  2500 adults Success? Yes  60%

Sample surveys show that fewer people enjoy shopping than in the past. A survey asked a nationwide random sample of 2500 adults if they agreed or disagreed that “I like buying new clothes, but shopping is often frustrating and time-consuming.” The population that the poll wants to draw conclusions about is all US residents aged 18 and over. Suppose that exactly 60% of all adult US residents would say “agree” if asked the same question. Let X = the number in the sample who agree. (b) Check the conditions for using a Normal approximation to this setting. 1.np ≥ 102. n(1 – p) ≥ 10 (2500)(.60) ≥ 10? (2500)(1 -.60) ≥ ≥ 10 ☺ 1000 ≥ 10 ☺

(c) Use a Normal distribution to estimate the probability that 1520 or more of the sample agree. Since this is a binomial setting that follows the rule of thumb for Normal approximation, we can use the Normal curve to find the P(X≥1520) with a mean of μ=np = 2500(.60) = 1500 and a standard deviation of √np(1 – p)= √2500(.60)(1-.60) = We want to find P(X≥1520), so to standardize the variable of X=1520, we will find the z-score = (1520 –1500)/ This will calculate z = Using Table A, we find that P(X<1520) = So P(X ≥1520) = 1 – = So there is about a 21% chance that the sample will agree that although they like to buy new clothes, but shopping is often frustrating and time-consuming.

(We answered this in the last problem, but if we were to do a full write up, this is what you would include): State: State what you are trying to find. Plan: First, check the BINS to decide if it’s a binomial distribution!! Then state why you can use a Normal approximation to the binomial. Do: Find the mean and standard deviation. Draw a Normal curve and standardize the variable. Use Table A, to find the probability. Conclude: Write your conclusion with the probability.

Read Textbook p. 393 – 397 Do exercises p. 404 – 405 #85, 87, 88, 91, and 92 Check answers to odd problems Happy Thanksgiving! I am very thankful for your continued efforts and good attitudes!

At the end of the lesson, students can: Sampling without replacement condition Determine when to use a normal approximation for a binomial distribution Graph binomial distributions on the calculator