Unformatted text preview: 71 = 0.9304 4. What is the probability that a standard normal variable Z is within 2 standard deviations of the mean? That is, find P(‐2 ≤ Z ≤ 2). 0.9772 – 0.0228 = 0.9544 or about 95% 5. What is P(Z ≤ 4.75)? 0.999999 from the “in the extreme” part of the table. 6. What is P(Z > 10.20)? Practically 0. 7. What is the 90th percentile of the standard normal N(0,1) distribution? Draw a picture of what you are trying to find first … Closest entry in the middle of the table to 0.90 is 0.8997 which corresponds to the z value of z = 1.28. 57 How to Solve General Normal Curve Problems To help us find areas and thus probabilities for any general N ( , ) distribution – we must use the idea of standardization. If the variable X has the N ( , ) distribution, then the standardized variable Z X will have the N (0,1) distribution. Let’s see how this standardization idea works through our next example. Try It! Scholastic Scores A proposal before the Board of Education has specified certain designations for elementary schools depending on how their students do on the Evaluation of Scholastic Scoring. This test is given to all students in all public schools in grades 2 through 4. Schools that score in the top 20% are labeled excellent. Schools in the bottom 25% are labeled "in danger" and schools in the bottom 5% are designated as failing. Previous data suggests that scores on this test are approximately normal with a mean of 75 and a standard deviation of 5. a. What is the probability that a randomly selected school will score below 70? P(X<70) = ? Find the z‐score: z = (70‐75)/5 = ‐1 So P(X<70) = P(Z< ‐1) = 0.1587. b. What is the score cut‐off required for schools to be labeled excellent? Show all work. Draw a picture: Want the upper 20th percentile (or 80th percentile) Closest z value is z = 0.84 so we need to unstandardize this z value. z = (x – 0.84 = (x – 75)/5 0.84(5) = x – 75 x = 0.84(5) + 75 = 79.2 Note: Once you find the percentile on the standard normal scale, say z, then you can convert it to the corresponding percentile on the X scale by: x = z Notes: (1) The normal distribution is a density for continuous random variables. (2) The normal distribution is not the only continuous distribution (recall you worked with a family of uniform continuous distributions a few pages back). (3) Computers and calculators often have the ability to find areas or percentiles for many density curves built right in to a function. You might be introduced to some of these in your lab sessions and are welcome to use them. But drawing a picture of what you are trying to find will be beneficial and serve as one way to show your work. 58 8.7 Approximating Binomial Distribution Probabilities Recall Our Left‐Handed Problem In an earlier problem (page 50) it was stated that about 10% of Americans are left‐handed. Let X = the number of left‐handed Americans in a random sample of 120 Americans (part c had us think about a sample size being 120 instead of 12). Then X has an exact Binomial distribution with n = 120 and p = 0.10. The mean or expected number of left‐handed Americans in the sample would be = np = 120(0.10) = 12. The standard deviation of X would be = np (1 p ) 120(0.10)(0.90) = 3.29. Suppose we want to find the probability that a random sample of 120 will contain 20 or fewer left‐handed Americans. Using the exact binomial distribution we would start with: P( X 20) P ( X 0) P ( X 1) P ( X 19) P ( X 20) This would not be too much fun to compute by hand since each of the probabilities for X = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 would be found using the n
k binomial probability formula: P( X k ) p (1 p ) k nk . But there is an easier way that will give us approximately the probability of having 20 or fewer. The easier way involves using a normal distribution. The normal distribution can be used to approximate probabilities for other types of random variables, one being binomial random variables when the sample size n is large. Normal Approximation to the Binomial Distribution If X is a binomial random variable based on n trials with success proba...
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 Summer '10
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