1. Binomial Formulas Target Goal: I can calculate the mean and standard deviation of a binomial function. 6.3b h.w: pg 404: 75, 77, 79 - 89

2. 1.Suppose I have a group of 4 students and I want to choose 1 of them as a volunteer. In how many ways can I choose 1 out of 4 students? SNNNNSNNNNSNNNNS Call this “4 choose 1.” There are 4 ways. Call this “4 choose 1.” There are 4 ways.

3. 2. Suppose I have a group of 4 students and I want to choose 2 of them as volunteers. In how many ways can I choose 2 out of 4 students? SSNNSNSNSNNSNSSN NSNSNNSS Call this “4 choose 2.” There are 6 ways. Call this “4 choose 2.” There are 6 ways.

4. 3.Suppose I have a group of 5 students and I want to choose 1 of them as a volunteer. In how many ways can I choose 1 out of 5 students? SNNNN NSNNN NNSNN NNNSN NNNNS Call this “5 choose 1.” There are 5 ways. Call this “5 choose 1.” There are 5 ways.

5. Binomial Coefficient There is a mathematical way to count the total number of ways to arrange k out of n objects. This is called “n choose k” or the binomial coefficient. There is a mathematical way to count the total number of ways to arrange k out of n objects. This is called “n choose k” or the binomial coefficient.

6. Binomial Coefficient: the number of ways to arrange k successes in n observations. It is written and is called “n choose k.” It is written and is called “n choose k.” The value of “n choose k” is given by the formula: The value of “n choose k” is given by the formula:

7. Example: “5 choose 2”

8. So, there are 10 ways to arrange 2 out of 5 objects. Think of this as flipping a coin 5 times and getting 2 heads. Think of this as flipping a coin 5 times and getting 2 heads.

9. Ex. Inheriting Blood Type Recall: p = 0.25, probability child has type O blood p = 0.25, probability child has type O blood n = 5, family has 5 children n = 5, family has 5 children Find the probability exactly 2 children have type O. Which 2 children? Which 2 children?

10. Step 1: Find the probability child #1 and #3 have type O blood. Find the probability child #1 and #3 have type O blood. S F S F F S F S F F Prob:(.25)(.75)(.25)(.75)(.75) = (.25) 2 (.75) 3

11. Step 2: All possible arrangements of 2 successes and 3 failures will give (.25) 2 (.75) 3. All possible arrangements of 2 successes and 3 failures will give (.25) 2 (.75) 3. How many arrangements? 5 choose 2 = 10 So, P(X=2) = # arrangements x prob. So, P(X=2) = # arrangements x prob. = 10 (.25) 2 (.75) 3 = 10 (.25) 2 (.75) 3 =.2637 =.2637

12. Recall: n! = n x (n-1) x (n-2) x … x 3 x 2 x 1 0! = 1 0! = 1 Binomial Probability If X has the binomial distribution with n observations and probability p and k is one of the possible values then, P(X = k) = p k (1-p) n-k (# arrangements x prob) (# arrangements x prob) 10 (.25) 2 (.75) 3 10 (.25) 2 (.75) 3

13. Ex. Defective Switches (no calculators) n = 10, p = 0.1, X is the # of switches that fail n = 10, p = 0.1, X is the # of switches that fail Find the probability no more than 1 switch fails. Find the probability no more than 1 switch fails. P(X ≤ 1) = P(X = 0) + P(X = 1) P(X ≤ 1) = P(X = 0) + P(X = 1) = (.10) 0 (.9) 10 + (.10) 1 (.9) 9 = = 0.3487 + 0.3874 = 0.7361

14. Binomial Mean and Standard Deviation If X is a binomial random variable with parameters n and p, then the mean and standard deviation of X are: If X is a binomial random variable with parameters n and p, then the mean and standard deviation of X are: Or, (Only for binomial, not for discrete random variables.) (Only for binomial, not for discrete random variables.)

15. Ex: Bad Switches The count of bad switches with n = 10 and p = 0.1 (bad switches). The count of bad switches with n = 10 and p = 0.1 (bad switches). This is the sampling distribution an engineer would see if she drew all possible SRS’s of 10 switches from a shipment and recorded the value of X for each sample. This is the sampling distribution an engineer would see if she drew all possible SRS’s of 10 switches from a shipment and recorded the value of X for each sample.

16. Find the mean and standard deviation of the distribution. Find the mean and standard deviation of the distribution. σ = 0.9487

17. The Normal Approximation to Binomial Distributions As the number of trials n gets larger, the binomial distribution X gets close to a normal distribution. As the number of trials n gets larger, the binomial distribution X gets close to a normal distribution.

18. The probability histogram for the binomial distribution:

19. As a ”rule of thumb”, we use the normal approx.when n and p satisfy: np ≥ 10 np ≥ 10 n(1-p) ≥ 10s/a nq ≥ 10 n(1-p) ≥ 10s/a nq ≥ 10

20. Ex: Attitudes Toward Shopping A nationwide random sample surveyed 2500 adults about shopping. Suppose that in fact 60% of all adult U.S. residents would say they agree that “they like buying new cloths, but shopping is often frustrating and time consuming.” A nationwide random sample surveyed 2500 adults about shopping. Suppose that in fact 60% of all adult U.S. residents would say they agree that “they like buying new cloths, but shopping is often frustrating and time consuming.”

21. What is the probability that 1520 or more of the sample agree? Check: There are 195 million adults, so sample is independent. There are 195 million adults, so sample is independent. np = 2500(0.6) = 1500 ≥ 10 n(1-p) = 2500(0.4) = 1000 ≥ 10

22. So, we can assume N(μ,σ) μ = np = 1500 σ = = = 24.49 = 24.49

23. A normal distribution approximates our binomial distribution well. So do a normal calculation. A normal distribution approximates our binomial distribution well. So do a normal calculation. P(X≥1520) = P(X≥1520) = = P(Z≥0.82) = 1 – 0.7939 or normcdf(.82,EE99) = 0.2061 The probability that 1520 or more of the sample agree is approximately 20.61%.

24. Exercise 8.15: Attitudes on Shopping b. Use your calculator and the cumulative binomial function to verify the exact answer for the probability at least 1520 people in the sample find shopping frustrating is 0.2131. b. Use your calculator and the cumulative binomial function to verify the exact answer for the probability at least 1520 people in the sample find shopping frustrating is 0.2131.

25. What is the probability to 6 decimal places? P(X≥1520) = 1 – P(X≤1519) P(X≥1520) = 1 – P(X≤1519) = 1 - binomcdf(2500,0.6,1519) = 1 -.7868609113 =.213139

26. What is the probability that at most 1468 people in the sample would agree with the statement that shopping is frustrating? What is the probability that at most 1468 people in the sample would agree with the statement that shopping is frustrating? Standardize: Standardize: = P(Z≤ -1.29) normcdf(-EE99,-1.29) =.0985