1. BCOR 1020 Business Statistics Lecture 14 – March 4, 2008

2. Overview Chapter 7 – Continuous Distributions –Exponential Distribution –Normal Approximation to the Binomial*

3. Chapter 7 – Exponential Distribution Characteristics of the Exponential Distribution: If events per unit of time follow a Poisson distribution, the waiting time until the next event follows the Exponential distribution. Waiting time until the next event is a continuous variable.

4. Chapter 7 – Exponential Distribution Characteristics of the Exponential Distribution: Probability of waiting more than x Probability of waiting less than x

5. Chapter 7 – Exponential Distribution Example: Customer Waiting Time Between 2 P.M. and 4 P.M. on Wednesday, patient insurance inquiries arrive at Blue Choice insurance at a mean rate of 2.2 calls per minute. What is the probability of waiting more than 30 seconds (i.e., 0.50 minutes) for the next call? Set = 2.2 events/min and x = 0.50 min P(X > 0.50) = e – x = e –(2.2)(0.5) =.3329 or 33.29% chance of waiting more than 30 seconds for the next call. The mean (expected) time between calls is 1/ =.45 minutes (or 27 seconds).

6. Chapter 7 – Exponential Distribution Example: Customer Waiting Time P(X > 0.50) = 0.3329

7. Clickers In Santa Theresa, false alarms are received at the downtown fire station at an average rate of 0.3 per day. If we are modeling the time between false alarms (in days), what it the appropriate distribution to use? A = Binomial B = Poisson C = Normal D = Exponential

8. Clickers In Santa Theresa, false alarms are received at the downtown fire station at an average rate of 0.3 per day. What is the mean time between false alarms (in days)? A = 0.3 B = 2.1 C = 3.3 D = 5.0

9. Clickers In Santa Theresa, false alarms are received at the downtown fire station at an average rate of 0.3 per day. What is the probability that more than a week will elapse between false alarms? A = 0.1225 B = 0.0001 C = 0.7408 D = 0.8775

10. Clickers In Santa Theresa, false alarms are received at the downtown fire station at an average rate of 0.3 per day. What is the probability that two consecutive false alarms will occur within 6 hours? A = 0.0750 B = 0.0723 C = 0.1653 D = 0.8347 E = 0.9277

11. Chapter 7 – Exponential Distribution Inverse Exponential: If the mean arrival rate is 2.2 calls per minute, we want the 90 th percentile for waiting time (the top 10% of waiting time). Find the x-value that defines the upper 10%.

12. Chapter 7 – Exponential Distribution Inverse Exponential: P(X x) =.10 So, e – x =.10 - x = ln(.10) = -2.302585 x = 2.302585/  = 2.302585/2.2 = 1.0466 min. 90% of the calls will arrive within 1.0466 minutes (62.8 seconds).

13. Clickers In Santa Theresa, false alarms are received at the downtown fire station at an average rate of 0.3 per day. What is the median time between false alarms? A = 0.30 B = 2.31 C = 3.33 D = 7.68

14. Chapter 7 – Exponential Distribution Mean Time Between Events: Exponential waiting times are described as Mean time between events (MTBE) = 1/ 1/MTBE = = mean events per unit of time In a hospital, if an event is patient arrivals in an ER, and the MTBE is 20 minutes, then = 1/20 = 0.05 arrivals per minute (or 3/hour).

15. Chapter 7 – Normal Approximation to the Binomial* When is Approximation Needed? Binomial probabilities are difficult to calculate when n is large. Use a normal approximation to the binomial. As n becomes large, the binomial bars become more continuous and smooth.

16. Chapter 7 – Normal Approximation to the Binomial* When is Approximation Needed? Rule of thumb: when n  > 10 and n(1-  ) > 10, then it is appropriate to use the normal approximation to the binomial. In this case, the binomial mean and standard deviation will be equal to the normal  and , respectively.  = n   = n  (1-  )

17. Chapter 7 – Normal Approximation to the Binomial* Example Coin Flips: If we were to flip a coin n = 32 times and  =.50, are the requirements for a normal approximation to the binomial met? Are n  > 5 and n(1-  ) > 10? n  = 32 x.50 = 16 n(1-  ) = 32 x (1 -.50) = 16 So, a normal approximation can be used with …