Chapter 4. iclicker Question 4.4 Prelecture. Part 2. Binomial Distribution. J.C. Wang. iclicker Question 4.4 Prelecture


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1 Chapter 4 Part 2. Binomial Distribution J.C. Wang iclicker Question 4.4 Prelecture iclicker Question 4.4 Prelecture
2 Outline Computing Binomial Probabilities Properties of a Binomial Distribution Computing Binomial Probabilities by TI Calculators Binomial Random Variables Normal Approximation Example When Approximation is Good Examples Binomial Example Normal Example Properties of a binomial distribution 1. Only two possible outcomes for each observation 2. Probability of success is p probability of failure is q = 1 p 3. Observations are independent Note: Sample size is fixed Number of successes, X, is of interest
3 Example of binomial distribution Stat 2160 multiple choice quiz has 5 questions with 4 choices for each question success def. = answering a question correctly probability of success =? probability of failure =? sample size =? What is probability that a student will answer exactly 3 questions correctly, in other words, P(X = 3 n = 5, p = 0.25, q = 0.75) =? Computing Binomial Probabilities P(X = 3 n = 5, p = 0.25, q = 0.75) =? Formula 5! 3!(5 3)! p3 q 5 3 = Note: 0! = 1 and p 0 = 1 and n! = 1 2 n Using TI Calculator <2nd><DISTR> binomialpdf(5,.25, 3)<ENTER> and get
4 Computing Binomial Probabilities using TI calculator keyword: exactly 20% of TV buyers at TV+More purchase the store s extended warranty. Say that 10 TV sets were sold in one day. What is the probabilities that exactly 3 extended warranties were sold, in other words, P(X = 3 n = 10, p =.2) =? <2nd><DISTR> binompdf(10,.2,3)<enter> and get Cumulative Probabilities cumulative distribution function keyword: at most (i.e, lefttail probability) P(X j n, p) = P(X = 0 or 1 or or j n, p) = P(X = 0) + P(X = 1) + + P(X = j) 20% of TV buyers at TV+More purchase the store s extended warranty. Say that 10 TV sets were sold in one day. What is the probability that at most 3 extended warranties were sold, in other words, P(X 3 n = 10, p =.2) =? <2nd><DISTR> binomcdf(10,.2,3)<enter> and get
5 iclicker Question 4.5 iclicker Question 4.5 Cumulative Probabilities (right) tail probability keyword: at least (i.e, righttail probability) P(X j n, p) = P(X = j or j + 1 or or n n, p) = P(X = j) + P(X = j + 1) + + P(X = n) = 1 P(X j 1 n, p) 20% of TV buyers at TV+More purchase the stores extended warranty. Say that 10 TV sets were sold in one day. What is the probability that at least 3 extended warranties were sold, i.e., P(X 3 n = 10, p =.2) = 1 P(X (3 1) 10,.2) = 1 P(X 2 10,.2) =? 1 <2nd><DISTR> binomcdf(10,.2,2)=
6 Example of binomial distribution An important part of the customer service responsibilities of a telephone company relates to the speed with which troubles in residential service can be repaired. Suppose past data indicate that the likelihood is 0.7 that troubles in residential service can be repaired on the same day. For the first five troubles reported on a given day, what is the probability that at most three will be repaired on the same day? Given: n = 5, p = 0.7, j = 3, P(X 3) =? P(X 3) = binomcdf(5, 0.7, 3) = Example of binomial distribution, continued For the first five troubles reported on a given day, what is the probability that at least three will be repaired on the same day? Given: n = 5, p = 0.7, j = 3, P(X 3) =? P(X 3) = 1 P(X 2) = 1 binomcdf(5, 0.7, 2) =
7 iclicker Question 4.6 iclicker Question 4.6 Binomial Random Variables Expected value = E[X] = np, i.e. expected number of successes Standard deviation = sd(x) = np(1 p) = npq
8 Binomial Random Variables TV+More Example Suppose that 20% of TV buyers at TV+More purchase the store s extended warranty. If 26 TVs were sold last week, the (expected) number of extended warranties should be around np = = 5.2 give or take np(1 p) = (1 0.2) = Say that the extended warranty cost is $100, how much revenue will be generated? ($ ) ± ($ ) = $520 ± $ Note: Multiplication rule of location and spread. Graphical Representation of binomial probabilities bar graph and probability histogram Binomial(n=15,p=0.5) Binomial(n=14,p=0.5) 0.20 P(X=x) = bar height 0.20 probability P(X=x) P(X=x) = bar height = bar area x P(X=x) 0.10 Note: if probability of a success is 0.5, the shape is symmetric about n/ x
9 Normal Approximation of binomial probabilities X~Binomial(n=26,p=0.2) np = 5.2, npq = Binomial Probability Y~Normal(µ = 5.2, σ = ) Probability actual probability approximate probability P(X=x) P(x 0.5<Y<x+0.5) x P(X x) 0.20 P(X x) 0.15 P(Y<x+0.5) 0.15 P(Y>x 0.5) x x Normal Approximation TV+more example Say n = 26 and p =.2. Using the Normal Curve; make sure np > 5 and nq > 5. Here 26.2 = 5.2 > 5 and 26 (1.2) = 20.8 > 5. Therefore, conditions hold and the standard deviation is npq = Let s say X 5, P(X 5) P(Y < 5.5) = normalcdf( 9999,5.5,5.2,2.0396) =
10 Normal Approximation TV+more example, continued P(X 5) P(Y < ) = P(Y<5.5) Normal Approximation TV+more example comparison with exact binomial probability Say n = 26 and p =.2 and X 5 Using the Binomial Probability P(X 5 n = 26, p =.2) = binomcdf(26,.2, 5) = Note: Normal approximation gives a value close to the precise binomial method.
11 Conditions for Good Approximation Normal curve gives a reasonable approximation for the binomial probabilities whenever both np > 5 and nq > 5. Note: np = expected number of successes nq = expected number of failures iclicker Question 4.4 Postlecture iclicker Question 4.4 Postlecture
12 Binomial Example travel agents example The rate of commission that commercial airlines pay travel agents has been declining for several years. In an attempt by travel agents to raise revenue, many agents are now charging their customers a ticket fee, typically between 10 and 15 dollars. According to the ASTA, about 90 percent of travel agents charge customer fees when purchasing an airline ticket. Travel Agents Example continued Suppose that a random sample of 55 travel agents is selected. Assume that the number of the 55 travel agents charging a ticket is distributed as a binomial random variable. What are the mean and sd of this distribution? What assumptions are necessary? What is the probability that none of the travel agents will charge a fee, in other words, P(X = 0 n = 55, p =.9) =?
13 Travel Agents Example continued What is the probability that at least 46 of our sample of travel agents will charge a fee, in other words, P(X 46 n = 55, p =.9) =? What is the probability that at most 50 of our sample of travel agents will charge a fee, in other words, P(X 50 n = 55, p =.9) =? Normal Example packaging bags example Plastic bags used for packaging produce are manufactured so that the breaking strength of the bag is normally distributed with a mean of 5 pounds per square inch and a sd of 1.5 pounds per square inch. Given: µ = 5, σ = 1.5 What is the proportion of bags produced having a breaking strength of less than 3.17 pounds per square inch, in other words, P(X < 3.17) =? What is the proportion of bags produced having a breaking strength between 3.2 and 4.2 pounds per square inch, in other words, P(3.2 X 4.2) =?
14 Packaging Bags Example conitinued What is the proportion of bags produced having a breaking strength of at least 3.6 pounds per square inch, in other words, P(X 3.6) =? Between what two values symmetrically distributed around about the mean, will 95% of the breaking strength fall, in other words, P(a X b) =?
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