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MM207 Unit 5 Sample Problems Similar Examples: 1. Use the Standard Normal Distribution table to find the indicated area under the standard normal curve. (References: example 3 - 6 pages 244 - 247, end of section exercises 21 – 40 page 249) a. Between z = 0 and z = 1.85 Solution: Refer to Table 4 – Standard Normal Distribution in the text. For z = 0 the area to the left is 0.5000 Or, using Excel, NORMSDIST(0) = 0.5000 For z = 1.85 the area to the left is 0.9678 Or, using Excel, NORMSDIST(1.85) = 0.9678 0.9678 – 0.5000 = 0.4678 b. To the left of z = -2.13 Solution: Refer to Table 4 – Standard Normal Distribution in the text. For z = -2.13 the area to the left is 0.0166 Or, using Excel, NORMSDIST(-2.13) = 0.0166 c. Between z = 0.43 and z = 1.68 Solution: Refer to Table 4 – Standard Normal Distribution in the text. For z = 0.43 the area to the left is 0.6664 For z = 1.68 the area to the left is 0.9535 0.9535 – 0.6664 = 0.2871 d. To the right of z = 2.07 Solution: Refer to Table 4 – Standard Normal Distribution in the text. For z = 2.07 the area to the left is 0.9808 1 – 0.9808 = 0.0192 Similar Example: 2. The thicknesses of widgets produced by a new machine are normally distributed with a mean of 0.72 cm and a standard deviation of 0.02 cm. What percent of the widgets will have a diameter less than 0.75 cm? Solution: P(x < 0.75) = x 0.75 0.72 P z P z P ( z 1.5) 0.02 Refer to Table 4 – Standard Normal Distribution in the text. For z = 1.5 the area to the left is 0.9332 or 93.32%. Or, using Excel, NORMDIST(x, mean, standard dev, cumulative) = NORMDIST(0.75,0.72,0.02,TRUE) = 0.9332 or 93.32%. Similar Example: 1 MM207 Unit 5 Sample Problems 3. For a certain IQ test, scores are normally distributed. The mean is 100 and the standard deviation is 15. Find the probability that an individual will have an IQ score less than 125. Solution: P(x < 125) = x 125 100 P z P z P ( z 1.67) 15 Refer to Table 4 – Standard Normal Distribution in the text. For z = 1.67 the area to the left is 0.9525. Or, using Excel, NORMDIST(x, mean, standard dev, cumulative) = NORMDIST(125,100,15,TRUE) = 0.9522. Similar Examples: 4. Answer the questions about the specified normal distribution. (References: example 4 and 5 page 264 - 265, end of section exercises 39 – 45 pages 267 - 268) a. Heights of young women are normally distributed with a mean of 64 inches and a standard deviation of 2.7 inches. Find the height that represents the 20th percentile. Solution: Refer to Table 4 – Standard Normal Distribution in the text. z for P20 is -0.84. To convert z to x use x = + z = 64 + (-0.84)(2.7) = 61.7 inches. Or, using Excel, NORMINV(probability, mean, standard dev) = NORMINV(.20,64,2.7) = 61.7 inches. b. Scores on the SAT test for one year were normally distributed with a mean of 1025 and standard deviation of 205. Find the score that marks the top 10%. Solution: For the top 0.10, the area to the left is 1.00 – 0.10 = 0.90. Refer to Table 4 – Standard Normal Distribution in the text. z = 1.28 To convert z to x use x = + z = 1025 + (1.28)(205) = 1287. Or, using Excel, NORMINV(probability, mean, standard dev) = NORMINV(.90,1025,205) = 1288. Similar Examples: 5. Find the probabilities. (References: example 5 and 6 page 276 - 277, end of section exercises 25 31 pages 280 - 281) a. Scores on the SAT test for one year were normally distributed with a mean of 1025 and standard deviation of 205. If the scores from 64 randomly selected individuals are chosen, what is the probability that the mean score would be greater than 1100? 2 MM207 Unit 5 Sample Problems Solution: 1100 1025 P( x > 1100) = P z P( z 2.93) 205 64 Refer to Table 4 – Standard Normal Distribution in the text. For z = 2.93 the area to the left is 0.9983. 1 - 0.9983 = 0.0017 Or, using Excel, NORMDIST(x, mean, standard dev, cumulative) = NORMDIST(1100,1025,205/SQRT(64),TRUE) = 0.9983. 1 - 0.9983 = 0.0017 b. Heights of young women are normally distributed with a mean of 64 inches and a standard deviation of 2.7 inches. If the heights from 25 randomly selected individuals are chosen, what is the probability that their mean height is less than 63 inches? Solution: 63 64 P( x < 63) = P z P( z 1.85) 2.7 25 Refer to Table 4 – Standard Normal Distribution in the text. For z = -1.85 the area to the left is 0.0322. Or, using Excel, NORMDIST(x, mean, standard dev, cumulative) = NORMDIST(63,64,2.7/SQRT(25),TRUE) = 0.0320. 3