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Normal Distribution Exercises Turn in your answers on the Normal Distribution Exercises Answer Form. This is covered in B&M Chapter 11. For use of the JMP calculator to compute the standard normal distribution and standard normal quantiles, read Sall and Lehman (2001) Chapter 3. In JMP, see Help, Contents, User's Guide, Using the Formula Editor. But use of the JMP calculator is not required or necessary. All of this can be done using a standard normal CDF table. 0.0013 1. Standard Normal Distribution, Use a standard normal CDF table or the JMP calculator PROBABILITY function NORMAL DISTRIBUTION to "compute" the probability that a standard normal random variable is a. less than −3, b. less than −2, c. between −3 and −2, d. less than −1, e. between −2 and −1, f. less than 0, g. between −1 and 0, h. between 0 and 1, i. between 1 and 2, j. between 2 and 3, k. greater than 3. 2. If you haven't already done so, draw a standard normal curve and divide the area under the curve into eight (8) regions with vertical lines through the Z axis at −3, −2, ..., 3. In each region write the probability of each interval: {Z < −3}(done as an example), {−3 < Z < −2}, {−2 < Z < −1}, . . ., {2 < Z < 3}, {Z > 3}, which you've already computed. This can be done in JMP calculator by using the PROBABILITY function NORMAL DENSITY. But it's best to just draw the curve by hand. 3. Now use the picture you've just drawn to compute the probability that a standard normal random variable is a. between −1 and +1, b. within two (2) standard deviations of 0, c. within three (3) standard deviations of the mean, d. not between −1 and +1, e. not within two (2) standard deviations of 0, f. not within three (3) standard deviations of the mean. 4. Suppose that Z is a standard normal random variable. a. Using a standard normal table, or the JMP PROBABILITY function NORMAL QUANTILE to find the 1-st percentile, i.e., the number z0.01, that satisfies the equation P{Z < z 0.01} = 0.01 (Answer: z 0.01 = −2.326 or −2.33, done as follows, as shown in the graph: Look for the probability 0.0100 in the standard normal CDF table. The closest to 0.0100 in the standard normal CDF table is 0.0099. This is in the row for z = −2.3 and in the column for 0.03, so the Z value corresponding to 0.0099 is z = −2.33, and that’s a valid answer. We can also write z 0.01 = −2.33 to denote the 0.01 quantile, i.e., the 1.0 percentile. And we say, “−2.33 is the 0.01 quantile, i.e., the 1st percentile, of the standard normal distribution.” The more precise value of −2.326 can be obtained using the JMP calculator.) b. Find the 5-th percentile, z0.05. c. Find the 10-th percentile, z0.10. d. Find the 90-th percentile, z0.90. e. Find the 95-th percentile, z0.95. f. Find the 99-th percentile, z0.99. 5. Assume that Z is standard normal. a. Find a number z such that P{−z < Z < z} = 0.90. Answer: z = 1.645 because P{−1.645 < Z < +1.645} = 0.90 b. For that number z compute: P{Z < −z} = c. For that number z compute: P{Z < z} = d. For that number z compute: P{Z > z} = e. The number z is which percentile of the standard normal distribution? 6. Assume that Z is standard normal. a. Find a number z such that P{−z < Z < z} = 0.95. b. Compute: P{Z < −z} = c. Compute: P{Z < z} = d. Compute: P{Z > z} = e. The number z is which percentile of the standard normal distributio n? 7. The Normal Family. The yearly growth of dwarf-apple-tree seedlings can be measured by the increase in the length of the central leader. Suppose that the second-year growth of such trees is normally distributed with a mean of 20 cm and a standard deviation of 6 cm. a. Compute the probability that the second-year growth of a randomly selected twoyear-old dwarf-apple-tree seedling is less than 15 cm. b. Compute the percentage of such dwarf-apple-tree seedlings that would grow more than 25 cm. c. Compute the fraction of such dwarf-apple-tree seedlings that would be expected to have a second-year growth of between 10 and 30 cm. d. Find a number x such that the second-year growth of 90% of the seedlings is more than x. e. Find two numbers a and b such that the second-year growth of 90% of the seedlings is between a and b and such that the second-year growth of 5% is less than a and the second-year growth of 5% is more than b. 8. A veterinarian found that the average time it takes residents to perform a certain procedure is 12 minutes. Assume that the time it takes residents to perform the procedure is normally distributed with a mean of 12 minutes and a standard deviation of 2 minutes. a. Compute the fraction of residents that you would expect to perform the procedure within two minutes of the expected time of 12 minutes. b. Compute the proportion of residents that you would expect to take less than 10 minutes or more than 14 minutes to perform the procedure. c. Compute the percentage of residents that you would expect to take between 8 and 10 minutes to perform the procedure. d. Compute the probability that a randomly selected resident would take between 9 and 11 minutes to perform the procedure. e. Compute the proportion of residents that you would expect to take between 10 and 12 minutes to perform the procedure. f. Compute the probability that a randomly selected resident would take more than 15 minutes to complete the procedure. g. For purposes of planning and scheduling, find the time before which 99% of residents would be expected to complete the procedure. h. If 50 residents were randomly selected, how many would you expect to be able to perform the procedure in 8 minutes or less? (Hint: The answer need not be an integer) Golde I. Holtzman, Department of Statistics, College of Arts and Sciences, Virginia Tech (VPI) Last updated: September 6, 2011 © Golde I. Holtzman, all rights reserved.