1 / 14

6.4 Normal Approximation to the Binomial Distribution

6.4 Normal Approximation to the Binomial Distribution. EQ: How can we use the normal curve as an approximation to the binomial distribution?. Activation:.

aristotle-alexander
Download Presentation

6.4 Normal Approximation to the Binomial Distribution

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. 6.4 Normal Approximation to the Binomial Distribution EQ: How can we use the normal curve as an approximation to the binomial distribution?

  2. Activation: • The probability that a new vaccine will protect adults from cholera is known to be .85. It is administered to 300 adults who must enter an area where the disease is present. What is the probability that more than 280 of these adults will be protected from cholera by the vaccine? • What type of problem from Chapter 5 is the above?

  3. BINOMIAL DISTRIBUTION • There are only two possible outcomes, success and failure. There is a sample size. You are looking for the probability of so many successes out of the number of trials.

  4. Summary of Binomial Distribution • n = number of trials • r = number of successes • p = probability of success • q = probability of failure = 1 – p

  5. What would you be able to pull out from the activation example? • You should have seen: • n = 300 • p = .85 • q = .15 • You would have been able to find

  6. How would we have done this example in Chapter 5? • We would have tried to use the Poisson distribution. • What criteria would we have had to meet? • Do we meet both of these criteria? • We do not so we need to find another way to do the problem.

  7. Binomial Distribution • We could fall back and use the formula and find • 1 – ( P(285) + P (286) + …+ P(300) ) • Do we really want to use this to find this probability? • Is there an easier way?

  8. Book Examples • Pages 324 and 325 • Figure 6-39 n = 3….not very normal looking • Figure 6-40 n = 10… we see the mound shape starting to emerge • Figure 6-41 n = 25…pretty darn symmetric with a mound shape but we aren’t quite there • Figure 6-42 n = 50…very symmetrical and very mound shaped • It gets closer and closer as n gets larger and larger.

  9. Conditions to Use the Normal Distribution as an Approximation of the Binomial Distribution

  10. Back to Activation Example • n = 300 • p = .85 • q = .15 • Now we normalize the data with z values • P( r > 280 ) becomes • P( x > 280 ) which we convert to z-values using • P ( z > 4.05 ) = • .5 – P ( 0 < z < 4.05) = • .5 - .4999 = .0001 • Is this a reasonable probability? • Think about p = .85 • sample size = 300 • and P( r > 280) • Think how far away from the mean 280 is….. • Yes it would be a small probability for that large of the population to be safe from the disease.

  11. Steps to Use • 1. Check does it meet the two requirements…is np>5 and nq>5? • 2. Find the mean and standard deviation • 3. Convert the x-values to z-values. • 4. Find the associated probability.

  12. Example 2 • The owner of a new apartment building must install 25 water heaters. From past experience in other buildings, she knows that Quick Hot is a good brand. It is guaranteed for 5 years only, but from her past experience she knows that the probability that it will last 10 years is .25.

  13. What is the probability that 8 or more of the 25 water heaters will last at least 10 years? • p = .25 • n = 25 • q = .75 • np = 6.25 • nq = 18.75 • We can use the normal distribution. • P ( r > 8 ) becomes • P ( x > 8 ) becomes • P ( z > .58 ) becomes

  14. How does this result compare with the result we can obtain using the binomcdf button on our calculator? • Pretty darn close for an approximation and if we don’t have the calculator to do the binomcdf….the normal approximation is much faster. • Assignment pg 329 1 – 15 odd

More Related