1 / 24

Review of Binomial Concepts

the discrete random variable. Review of Binomial Concepts. A single experiment, with probability of success p, is repeated n independent times. X = number of successes. Probability distribution for X when n=10 and p=0.5. P(X = k). Cumulative distribution for X when n=10 and p=0.5.

Download Presentation

Review of Binomial Concepts

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. the discrete random variable Review of Binomial Concepts A single experiment, with probability of success p, is repeated n independent times X = number of successes

  2. Probability distribution for X when n=10 and p=0.5 P(X = k) Cumulative distribution for X when n=10 and p=0.5 P(X  k)

  3. Excel Commands for Binomial Probabilities X is a binomial random variable with binomial parameters n and p P(X = k): BINOMDIST(k, n, p, 0) P(X  k): BINOMDIST(k, n, p, 1) Important Formulas for Binomial Random Variable E[X] = n * p X = sqrt [ n * p * (1 – p) ]

  4. Example for Binomial Distribution A study by one automobile manufacturer indicated that one out of every four new cars required repairs under the company’s new-car warranty, with an average cost of $50 per repair. • For 100 new cars, what is the expected cost of repairs? What is the standard deviation? • Provide a reasonable estimate of the most it might cost the manufacturer for a specified group of 100 cars. (Hint: Ensure a 95% probability of not exceeding amount.)

  5. Example for Binomial Distribution Suppose you are in charge of hiring new undergraduate accounting majors for Coopers and Young (C&Y), an accounting firm headquartered in Chicago. This year your goal is to hire 20 graduates. On the average, about 40% of the people you make offers to will accept. Unfortunately, offers have to go out simultaneously this year, so you plan to make more than 20 offers. • How many offers should you make so that your expected number of hires is 20? • How many offers should you make if you want to have an 80% chance of hiring at least 20 people? • Find a 95% probability interval if you make 50 offers.

  6. Normal Example Otis Elevator in Bloomington, Indiana, reported that the number of hours lost per week last year due to employees’ illnesses was approximately normally distributed, with a mean of 60 hours and a standard deviation of 15 hours. Determine, for a given week, the following probabilities: • The number of hours lost will exceed 85 hours. • The number of hours lost will be between 45 and 55 hours.

  7. Normal Example A mail-order company has estimated one-year orders for a popular item to be normally distributed with a mean of 180,000 units and a standard deviation of 15,000 units. • What is the probability of selling all of the stock on hand if inventory equals 200,000 units? • What inventory should the company have on hand if they want the probability of running out of stock to be 5%?

  8. The Normal Distribution (preview) • The normal distribution is related to a particular type of continuous random variable (as opposed to “discrete random variable”) • It is the “bell-shaped” curve that you may have heard about • It is used widely in statistics and appears in lots of practical problems • It has an expected value (or “mean”) and a standard deviation just like other random variables • We’ll discuss it in detail shortly, but a little background first….

  9. Continuous Random Variables a random variable takes a probability event and assigns a number, or value, to it Recall a discrete random variable deals with a finite number of events and numbers Recall A continuous random variable can take on any value within a specified interval of numbers There are an infinite number of events and an infinite number of values

  10. Examples of Continuous Random Variables X = the precise time that a train arrives, when it is scheduled to arrive at 8:00 PM X = the residual chemical levels in the blood 24 hours after taking a specific medication X = the precise amount of soda placed in a bottle by a soda filling machine (or should it be “pop”?) X = the amount of energy used in a home during one hour

  11. Continuous or Discrete? Or Both? Sometimes a discrete random variable has so many possible outcomes (still finite, however) that we consider it to be continuous Sometimes a continuous random variable is actually discrete when we measure it X = salary of an office manager X = price of a therm of natural gas in January X = grades of a student on the SAT X = length of a manufactured part (limits of our measurement techniques)

  12. How to Talk aboutContinuous Random Variables Since a continuous random variable has so many possible outcomes, we do not consider things like: P(X = 1.0) P(X = 0.3349) P(X = -0.5) Instead, we consider: P(X  1.0) P(X  0.3349) P(-0.5  X  0.3349) equalities (no), intervals (yes) ( In fact, P(X = k) = 0 for a continuous random variable. Again, this is “equal to.” )

  13. 0.50 0.25 8 9 10 11 12 Example: A Delivery Truck Suppose the warehouse says that the delivery truck will arrive sometime “between 8 AM and noon.” What is the probability that the truck will arrive between 9 AM and 10 AM? The total area of the red block equals 1.0, the total probability. So P(9  X  10) = 0.25 Probability density function (distribution); similar to probability mass function in discrete random variables

  14. The Normal Distribution • Bell-shaped • Horizontal axis represents values of the random variable • Expected value, or mean, is  • Symmetrical about  • Total blue area is 1.0 • Vertical axis is not very important

  15. The Normal Distribution (cont’d)  is the standard deviation of the normal random variable P(  -   X   +  ) = 0.6826 “Probability that X is within 1 std dev of its mean is 68.26%”

  16. The Normal Distribution (cont’d) P(  - 2  X   + 2 ) = 0.9544 “Probability that X is within 2 std dev of its mean is 95.44%”

  17. The Normal Distribution (cont’d) P(  - 3  X   + 3 ) = 0.9973 “Probability that X is within 3 std dev of its mean is 99.73%”

  18. The Normal Distribution (summary) A continuous random variable X is “normally distributed with expectation/mean  and standard deviation ” if its probability distribution is • Bell-shaped • Symmetrical about the value  • The following are true: • P(  -   X   +  ) = 0.6826 • P(  - 2  X   + 2 ) = 0.9544 • P(  - 3  X   + 3 ) = 0.9973 We say X is N(  , 2 )

  19. Excel’s Normal Functions If X is N(  , 2 ): P(X  k) = NORMDIST(k, , , 1) NORMINV( prob, ,  ) gives the value of k so that P(X  k) = prob “normal inverse”

  20. The Standardized Normal Distribution X is N(  , 2 ) Z is N( 0, 12 ) The standardized normal • For comparison of several different normal distributions • For calculations without a computer

  21. Formula relating standard Z and any given X: Z = ( X -  ) /  “Given the value for X, the corresponding value for Z is given by the above formula.” Compare your performance in all courses…

  22. The standardized normal is also useful for hand calculations of normal probabilities… Suppose X is normally distributed with mean 20 and standard deviation 4.3. What is P(X  23)? What is P(X  16.5)? (23 – 20) / 4.3 = 0.698 (16.5 – 20) / 4.3 = -0.814 Z = ( X -  ) /  P(Z  0.698) = ? P(Z  -0.814) = ? … = 0.7580 … = 0.2090 (See Excel)

  23. The Normal Approximationto the Binomial Both distributions have the same shape…

  24. Suppose X is a binomial variable with n = 900 and p = 0.35 E[X] =  = np = 315  = sqrt( np(1-p) ) = 14.31 Let Y be the normal variable with  = 315 and  = 14.31 P(X  300) = BINOMDIST(300, 900, 0.35, 1) = 0.155 P(X  300)  P(Y  300) = NORMDIST(300, 315, 14.31, 1) = 0.147 P(X = 300) = BINOMDIST(300, 900, 0.35, 0) = 0.0162 P(X = 300)  P(299.5  Y  300.5) = NORMDIST(300.5, 315, 14.31, 1) – NORMDIST(300.5, 315, 14.31, 1) = 0.0161 The second is called the “continuity correction”

More Related