1. Continuous Distributions Chapter 6 MSIS 111 Prof. Nick Dedeke

2. Learning Objectives Appreciate the importance of the normal distribution. Recognize normal distribution problems, and know how to solve them. Decide when to use the normal distribution to approximate binomial distribution problems, and know how to work them. Decide when to use the exponential distribution to solve problems in business, and know how to work them.

3. Probabilities If a man had six children and he wanted to go a Red Sox game with one of them, what is the likelihood that he will choose Buba, his first son? A.1 B.6 C.1/6 D. I have no idea!

4. Probability Example 1 If a man had six children and he wanted to go a Red Sox game with one of them, what is the likelihood that he will choose Buba? Probability of event A = Number of ways event A could occur Number of events in the sample space Events that could occur (EVENTS SAMPLE SPACE)  He takes Buba  He takes Maria  He takes Bakuba  He takes Melinda  He takes Susan  He takes Elisha Probability (A= He takes Buba) = 1/6

5. Probability Example 2 If a man had ten bills in his pocket. Two $20 bills, three $100 bills and five $ 50 bills. He wanted to buy a baseball for $20. What is the probability that if he drew a bill out of his pocket it would be a $50 bill? Probability of event A = Number of ways event A could occur Number of events in the sample space Events that could occur (EVENTS SAMPLE SPACE)  He draws out first $20 bill  He draws out second $20 bill  He draws out a $50 bill (five possibilities)  He draws out a $100 bill (three possibilities) Probability (A= It is a $50 ) = 5/(5+3+2) = 0.5

6. Probability Example 3 If a man had ten bills in his pocket. Two $20 bills, three $100 bills and five $ 50 bills. He wanted to buy a baseball for $20. What is the probability that if he drew a bill out of his pocket it would be a $20 bill? Probability of event A = Number of ways event A could occur Number of events in the sample space Events that could occur (EVENTS SAMPLE SPACE)  He draws out a $20 bill (two possibilities)  He draws out a $50 bill (five possibilities)  He draws out a $100 bill (three possibilities) Probability (A= It is a $20 ) = 2/(5+3+2) = 0.2

7. Probability (Freq. Table) Example 3 If a man had ten bills in his pocket. Two $20 bills, three $100 bills and five $ 50 bills. He wanted to buy a baseball for $20. What is the probability that if he drew a bill out of his pocket it would be a $50 bill? Probability of event A = Number of ways event A could occur Number of events in the sample space Xi Fi Rel. Freq or P(x) $20 2 2/10 = 0.2 $50 5 5/10 = 0.5 $100 3 3/10 = 0.3 N 10 1 Probability (x = Bill is a $20 ) = 0.2 Probability (x = Bill is $50 or less) = 0.2 + 0.5 = 0.7 Probability (x = Bill is > $50) = 1 - Probability (x = Bill is $50 or less) = 1 – 0.7 = 0.3

8. Probability Distribution and Areas & Curve Example 5 Xi FiRel. Freq or P(x) $2022/10 = 0.2 $5055/10 = 0.5 $10033/10 = 0.3 10 1.0 Probability (B= Bill is $50 or less) = 0.2 + 0.5 = 0.7 P(x) 1.0 $100 $50 $20 0.5 Probability distribution of an event X is the listing of all possible events and their probabilities (in table of graphical form). If we put the results in form of an equation, we will have a probability density function.

9. Continuous Variable Problem Xi (Prices)Fi $2057 $25Not measured $35Not measured $40234 $60123 Relative frequency approach not advisable

10. Normal Distribution Probably the most widely known and used of all distributions is the normal distribution. It fits many human characteristics, such as height, weight, length, speed, IQ scores, scholastic achievements, and years of life expectancy, among others. Many things in nature such as trees, animals, insects, and others have many characteristics that are normally distributed. It is used for measured data not counted ones

11. Normal Distribution Many variables in business and industry are also normally distributed. For example variables such as the annual cost of household insurance, the cost per square foot of renting warehouse space, and managers’ satisfaction with support from ownership on a five-point scale, amount of fill in soda cans, etc. Because of the many applications, the normal distribution is an extremely important distribution.

12. Normal Distribution Discovery of the normal curve of errors is generally credited to mathematician and astronomer Karl Gauss (1777 – 1855), who recognized that the errors of repeated measurement of objects are often normally distributed. Thus the normal distribution is sometimes referred to as the Gaussian distribution or the normal curve of errors.

13. Properties of the Normal Distribution The normal distribution exhibits the following characteristics: It is a continuous distribution. It is symmetric about the mean. It is asymptotic to the horizontal axis. It is unimodal. It is a family of curves. Area under the curve is 1. It is bell-shaped.

14. Graphic Representation of the Normal Distribution

15. Probability Density of the Normal Distribution Note: The probability distribution changes if either mean or SD changes.

16. Family of Normal Curves

17. Standardized Normal Distribution Since there is an infinite number of combinations for  and , then we can generate an infinite family of curves. Because of this, it would be impractical to deal with all of these normal distributions. z distribution Fortunately, a mechanism was developed by which all normal distributions can be converted into a single distribution called the z distribution. standardized normal distribution This process yields the standardized normal distribution (or curve).

18. Standardized Normal Distribution The conversion formula for any x value of a given normal distribution is given below. It is called the z-score. A z-score gives the number of standard deviations that a value x, is above or below the mean.

19. Standardized Normal Distribution If x is normally distributed with a mean of  and a standard deviation of , then the z- score will also be normally distributed with a mean of 0 and a standard deviation of 1. Since we can covert to this standard normal distribution, tables have been generated for this standard normal distribution which will enable us to determine probabilities for normal variables. The tables in the text are set up to give the probabilities between z = 0 and some other z value, z 0 say, which is depicted on the next slide.

20. Standardized Normal Distribution

21. Z Table Second Decimal Place in Z Z0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.000.00000.00400.00800.01200.01600.01990.02390.02790.03190.0359 0.100.03980.04380.04780.05170.05570.05960.06360.06750.07140.0753 0.200.07930.08320.08710.09100.09480.09870.10260.10640.11030.1141 0.300.11790.12170.12550.12930.13310.13680.14060.14430.14800.1517 0.900.31590.31860.32120.32380.32640.32890.33150.33400.33650.3389 1.000.34130.34380.34610.34850.35080.35310.35540.35770.35990.3621 1.100.36430.36650.36860.37080.37290.37490.37700.37900.38100.3830 1.200.38490.38690.38880.39070.39250.39440.39620.39800.39970.4015 2.000.47720.47780.47830.47880.47930.47980.48030.48080.48120.4817 3.000.49870.49870.49870.49880.49880.49890.49890.49890.49900.4990 3.400.49970.49970.49970.49970.49970.49970.49970.49970.49970.4998 3.500.49980.49980.49980.49980.49980.49980.49980.49980.49980.4998

22. Applying the Z Formula Z0.00 0.01 0.02 0.000.00000.00400.0080 0.100.03980.04380.0478 1.000.34130.34380.3461 1.100.36430.36650.3686 1.200.38490.38690.3888 What is the probability that a measured value would be less than or equal to 600 but greater than or equal to 485?

23. Applying the Z Formula 0.5 + 0.2123 = 0.7123 What is the probability that a measured value would be less than or equal to 550?

24. Exercise: Applying the Z Formula Use chart on page 788 to find Probability

25. Applying the Z Formula 0.4738+ 0.3554 = 0.8292 Use chart on page 788 to find Probability

26. Exercise: Applying the Z Formula Use chart on page 788 to find probability

27. Applying the Z Formula 0.5 – 0.4803 = 0.0197

28. Exercise: Applying the Z Formula Use chart on page 788 to find Probability

29. Applying the Z Formula 0.4738+ 0.3554 = 0.8292

30. Exercise: Applying the Z Formula

31. Exercise 1 The weekly demand for bicycles has a mean of 5,000 and a standard deviation of 600. Assume normal distribution. What is the probability that the firm will sell between 3,000 and 6,000 (>= 3,000 and <=6,000) in a week?

32. Exercise 2 The distribution of customer spending has a mean of $25 and standard deviation of $8.  What is the probability that a randomly selected customer spends less than $35 at the store?  What is the probability that a randomly selected customer spends between $15 and $35?  What is the probability that a randomly selected customer spends more than $10 at the store?  What is the $ amount such that 75% of all customers spend no more than this amount?  What is the $ amount such that 80% of all customers spend at least this amount?

33. Exercise 2 Response The distribution of customer spending has a mean of $25 and standard deviation of $8.  What is the probability that a randomly selected customer spends less than $35 at the store?  What is the probability that a randomly selected customer spends between $15 and $35?  What is the probability that a randomly selected customer spends more than $10 at the store?

34. Exercise 2 Response What is the $ amount such that 75% of all customers spend no more than this amount? What is the $ amount such that 80% of all customers spend at least this amount? 0.67 is positive because its above mean

35. Exercise 2 Response What is the $ amount such that 80% of all customers spend at least this amount? 0.84 is negative because its below mean

36. Exercise 3 A furniture store has to decide if the current range of prices of its goods is appropriate for its strategy. The mean amount that customers spend in the store is $ 3,500/sale. The standard deviation is $ 550. Assuming that the distribution is normal, the firm wants its prices to cover persons about 45% of the customers around the mean. The current price range of goods is $4200 and $ 2500. Does the price range match the expectation?

37. Exercise 3 Response A furniture store has to decide if the current range of prices of its goods is appropriate for its strategy. The mean amount that customers spend in the store is $ 3,500. The standard deviation is $ 550. Assuming that the distribution is normal, the firm wants its prices to cover persons about 45% of the customers around the mean. The current price range of goods is $4200 and $ 2500. Does the price range match the expectation? = 0.4656 + 0.3980 = 0.8636 (86.36%). Scope too wide!

38. Exercise 4 A furniture store decided to change its price range of its goods. The mean amount that customers spend in the store is $ 3,500. The standard deviation is $ 550. Assuming that the distribution is normal, the firm wants its prices to cover persons about 45% of the customers around the mean. The new price range of goods is $3800 and $ 2800. Does the price range match the expectation?

39. Exercise 4 Response A furniture store decided to change its price range of its goods. The mean amount that customers spend in the store is $ 3,500. The standard deviation is $ 550. Assuming that the distribution is normal, the firm wants its prices to cover persons about 45% of the customers around the mean. The current price range of goods is $3800 and $ 2800. Does the price range match the expectation? = 0.2054 + 0.2054 = 0.4108 (41.08%). Scope is fine!

40. Exercises