Introduction A family plans on having three children. Calculate the probability the family has ALL girls. P(3G) =.125 P(3G) =.125 (.5) 3 =.125 (.5) 3 =.125 Too Easy? Find the probability of having exactly 2 girls. Answer? Answer? Did you get that? How or Why NOT?? The above problems are examples of binomial probabilities. Let’s examine these a little closer!

What Gives Us a Binomial Setting? There are 4 key elements necessary to have a binomial setting. They are as follows: Each observation is either a “Success” or “Failure” Each observation is either a “Success” or “Failure” There are a fixed number n of observations. There are a fixed number n of observations. The n observations are all independent. The n observations are all independent. The probability of success, p, is the same for each observation The probability of success, p, is the same for each observation Let’s look at these elements in the context of a scenario.

What Gives Us a Binomial Setting? Scientists have randomly selected 10 rats known to have the flu and will inoculate them with a flu vaccine. The vaccine has been tested to have a 65% cure rate. Each observation is either a “Success” or “Failure” Each observation is either a “Success” or “Failure” In this case, the rats will either be Cured or Not Cured There are a fixed number n of observations. There are a fixed number n of observations. There are 10 observations or rats. The n observations are all independent. The n observations are all independent. The results of the vaccine in one rat do not effect the results for the next rat The results of the vaccine in one rat do not effect the results for the next rat The probability of success, p, is the same for each observation The probability of success, p, is the same for each observation The probability of success for each rat is.65

Examining the “Lingo” Binomial distribution Distribution of count of X successes in binomial setting Distribution of count of X successes in binomial setting n = Number of Observations p = Probability of Success on any ONE observation X = # of successes *Remember you must 1 st have a binomial setting (meet the 4 criteria) before you have a binomial distribution **One exception to independence is if you have a significantly large sample size **Look at Example 8.3 pg.440

Finding Binomial Probabilities We will see the formula later and won’t be required to use it, but may want to be familiar with it!! For now, let’s concentrate on using the calculator 2 nd VARS 2 nd VARS binompdf(n,p,x) binompdf(n,p,x) Remember: n = # of observations p – probability of success X = # of successes Don’t Forget!!! NPX… NPX… NPX

Let’s Practice Using the Calculator Allen Iverson is a career 73% free throw shooter. Against the Knicks, Allen went to the free throw line 12 times. 1. Find the probability that Allen makes all 12 free throws. 2. Find the probability that Allen makes exactly 10 free throws. 3. Find the probability that Allen makes only 1 free throw. 4. Find the probability that Allen makes less than two of his foul shots. binompdf(12,.73,12) =.0229 binompdf(12,.73,10) =.2068 binompdf(12,.73,1) = 0 (basically) binompdf(12,.73,1) + binompdf(12,.73,0)

Cumulative Function What about situations where we need the sum of more than one probability **like #4 The Cumulative Binomial Function in the calculator will find the probability from 0 successes to the # you desire The Cumulative Binomial Function in the calculator will find the probability from 0 successes to the # you desire Calculator Syntax Calculator Syntax 2nd Vars 2nd Vars binomcdf(n,p,x) binomcdf(n,p,x) This gives you the sum of the probabilities from 0 to X This gives you the sum of the probabilities from 0 to X Ex: binomcdf(19,.5,3) will find the probability of three or less successes out of 19 observations From the previous example, find the probability that Iverson hits 10 free throws or less.

Cumulative Function (Snag) What about the probabilities below: Probability Allen makes more than 5 shots Probability Allen makes more than 5 shots Probability Allen makes at least 8 shots Probability Allen makes at least 8 shots Probability Allen makes at least two thirds of his shots Probability Allen makes at least two thirds of his shots Let’s Think About It: What does the binomcdf(12,.73,5) give us? P(0) + P(1) + P(2) + P(3) + P(4) + P(5) There are 12 total probabilities in this distribution. P(6) + P(7) + P(8) + P(9) + P(10) + P(11) +P(12) This is what I need!!! Any Probability Distribution Adds up to? So… binomcdf(12,.73,5) P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + The Answer!!!

Get Some Answers!!! Find the probabilities below: Probability Allen makes more than 5 shots Probability Allen makes more than 5 shots Probability Allen makes at least 8 shots Probability Allen makes at least 8 shots Probability Allen makes at least one third of his shots Probability Allen makes at least one third of his shots P(X>5) = 1 – binomcdf(12,.73,5) 1 – binomcdf(12,.73,7) 1 – binomcdf(12,.73,3)

The Pesky Formula Below is the “by hand” formula used for binomial probabilities. Know it, but you will mostly use the calculator method. n = # of observationsx = # of successes p = probability of success nCx = combination – represents the number of ways of arranging the successes nCr Syntax 1)Type in the n 2)Math – Prb – nCr 3)Type in the r

Try Using the Formula Try these quickly 5C3 = 5C4 = Using the formula, calculate the following probability. Each child born to a particular set of parents has probability 0.25 of having blood type O. If these parents have 5 children, what is the probability that exactly 2 of them have type O blood? The Formula’s Big Drawback If the problem asks for a cumulative function, then you have to do this formula for each probability you need and then add all of them up. A LOT more time consuming!!! 105

Mean and Standard Deviation Remember, to find the mean of a probability distribution, you take the sum of the probabilities times their value **sum of xP(x) The binomial distribution has special formulas for the mean and standard deviation Mean(µ) = np Standard Deviation(σ) = Where: n = # of observationsp = probability of success Beware!!! These formulas are BINOMIAL specific and don’t work for other distributions.

Mean and Standard Deviation A university claims that 80% of its basketball players get degrees. An investigation examines the fate of all 20 players who entered the program over a period of several years that ended six years ago. Of these players, 11 graduated and the 9 remaining are no longer in school. What is the standard deviation for the group of 30? How many athletes can the university expect to graduate out of their next crop of 30 athletes? Answer 30(.55) = 16.5 If the same group had a probability of success of.8, what would the standard deviation be? … p =.9? Notice that our σ gets smaller as our p value gets closer to 1.

Normal Approximation As the number of trials n, gets larger, a binomial probability distribution gets close to becoming a Normal distribution What does that have to do with us? Statisticians often use a “normal” approximation to find binomial probabilities for high sample spaces. How do use a normal curve to find probabilities? Change your X to Z and find the appropriate area!! This method was used to shorten large cumulative binomial functions before we had cdf’s in the calculator. While the calculator is better, we still need to know how and when to use the normal approximation. Why would we use this?

The “When” of the Normal Approximation In order to use the normal approximation, the following 2 conditions must be met: np ≥ 10 np ≥ 10 n(1-p) ≥ 10 n(1-p) ≥ 10 **Be sure to check these 2 conditions before you use a normal approximation Check the conditions and decide if a normal approximation would appropriate. ≥ (.91) = 368 ≥ (.09) = 32 ≥ 10 Normal Approximation is appropriate. Many local polls of public opinion use samples of size 400 to 800. Consider a poll of 400 adults in Richmond that asks the question, “Do you approve of President George W. Bush’s response to the World Trade Center terrorist attacks in September 2001?” Suppose we know the President’s approval rating on this issue nationally is 92%. You are asked to calculate the probability that at most 358 of the 400 adults in the Richmond Poll answer “yes” to the question.

The “How” of the Normal Approximation Here are the steps to a normal approximation: Find the µ and σ of the distribution. Find the µ and σ of the distribution. Change the X to a Z score Change the X to a Z score Use the z chart or your calculator to find the appropriate area under the curve. Use the z chart or your calculator to find the appropriate area under the curve.

Let’s Do It!! Many local polls of public opinion use samples of size 400 to 800. Consider a poll of 400 adults in Richmond that asks the question, “Do you approve of President George W. Bush’s response to the World Trade Center terrorist attacks in September 2001?” Suppose we know the President’s approval rating on this issue nationally is 92%. Use the Normal Approximation to calculate the probability of at most 358 people in the sample approving. µ = 400(.92) = 368 σ = sqrt(400*.92*.08) = X -> Z (358 – 368) / (5.4259) = Z-area for =.0329 Try getting the answer using your calculator and the Binomial CDF function. binomcdf(400,.92,358 =..0441

Note about the Normal Approximation Notice the difference between our approximate answer and the exact answer of the pdf. (.0441) – (.0329) =.0112 difference (.0441) – (.0329) =.0112 difference Not large, but not really good either!!! Accuracy of the Normal Approximation The Normal Approximation is MORE accurate when p is closer to ½. The Normal Approximation is LESS accurate when p is closer to 0 or 1.

Homework Read Pages 457,58 on Simulating Binomial Experiments Do Problem #’s 25,27- 35, Worksheet

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