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6.3b h.w : pg 404: 75, 77, 79 - 89

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. 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?

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6.3b h.w : pg 404: 75, 77, 79 - 89

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  1. Binomial FormulasTarget 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? SNNN NSNN NNSN NNNS • 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? SSNN SNSN SNNS NSSN NSNS NNSS • 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.

  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.

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

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

  10. Step 1: • Find the probability child #1 and #3 have type O blood. 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. How many arrangements? 5 choose 2 = 10 • So, P(X=2) = # arrangementsx prob. = 10(.25)2(.75)3 = .2637

  12. Recall: n! = n x (n-1) x (n-2) x … x 3 x 2 x 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) = pk(1-p)n-k (# arrangements xprob) 10(.25)2(.75)3

  13. Ex. Defective Switches (no calculators) • n = 10, p = 0.1, X is the # of switches that fail • Find the probability no more than 1 switch fails. • 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: Or, • (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). • 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. 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.

  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 n(1-p) ≥ 10 s/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.”

  21. What is the probability that 1520 or more of the sample agree? Check: • 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

  23. A normal distribution approximates our binomial distribution well. So do a normal calculation. • 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.

  25. What is the probability to 6 decimal places? • 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? • Standardize: = P(Z≤ -1.29) normcdf(-EE99,-1.29) = .0985

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