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nat is the probability that sample will be larger than 670 mg? 2. In a study of the life expectancy of 400 people in a certain geographic region, the mean age at death was 70 years, 'and the standard deviation was 5.1 years. If a sample of 50 people from this region is selected, what is the probability that the mean life expectancy will be less than 68 years?

nat is the probability that sample will be larger than 670 mg? 2. In a study of the life expectancy of 400 people in a certain geographic region, the mean age at death was 70 years, 'and the standard deviation was 5.1 years. If a sample of 50 people from this region is selected, what is the probability that the mean life expectancy will be less than 68 years?

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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is the probability that ble will be larger than 670 mg? 2. In a study of the life expectancy of 400 people in a certain geographic region, the mean age at death was 70 years, and the standard deviation was 5.1 years. If a sample of 50 people from this region is selected, what is the probability that the mean life expectancy will be less than 68 years? 3. The average cholesterol content of a certain canned goods is 215 milligrams, and the standard deviation is 15 milligrams. Assume that the variable is normollh distributod If
Transcribed Image Text:is the probability that ble will be larger than 670 mg? 2. In a study of the life expectancy of 400 people in a certain geographic region, the mean age at death was 70 years, and the standard deviation was 5.1 years. If a sample of 50 people from this region is selected, what is the probability that the mean life expectancy will be less than 68 years? 3. The average cholesterol content of a certain canned goods is 215 milligrams, and the standard deviation is 15 milligrams. Assume that the variable is normollh distributod If
Solution for #1.a: Step1: Identify the parts of the problem. Given: u= 46.2 minutes; o= 8 minutes; X= 43 minutes; Find: P(X < 43) n = 50 students %3D Step 2: Use the formula to find the z-score. Solution: 43 - 46.2 8 Given: u= 1200 hours; o = 250 hours; Vn V50 X = 1 150 & 1 250 hours n = 100 bulbs Unknown: P(1150 < X < 1250) z = -2.83 Step 2: Use the formula to find the z-score. Step 3: Use the z-table to look up the z-score you calculated in step 2. z = -2.83 has a corresponding area of 0.4977 1150 1200 1250 1200 250 Vn V100 250 Step 4: Draw a graph and plot the z-score and its corresponding area. Then, shade V100 the part that you're looking for: P(X < 43) z = -2 Z = 2 Step 3: Use the z-table to look up the z-score you calculated in step 2. z = +2 has a corresponding area of 0.4772 Step 4: Draw a graph and plot the z-score and its corresponding area. Then, shade the part that you're looking for: P(1150 <X < 1250) sha ed part 0.4977 -3 -2 -1 1 2 3 -2.83 0.4772 0.4772 Since we are looking for the probability less than 43 minutes, the shaded part will be on the left part of - 2.83. -3 -2 -1 2 3 Since we are looking for the probability between 1 150 hours and 1 250 hours, the shaded part will be between -2 and 2. Step 5: Subtract your z-score from 0.500. P(X < 43) = 0.500 0.4977 P(X < 43) = 0.0023 Step 5: Add the two z-score values. P(1150 <X < 1250) = 0.4772 + 0.4772 P(1150 <X < 1250) = 0.9544 %3D Step 6: Convert the decimal in Step 5 to a percentage. P(X < 43) = 0.23% %3D Step 6: Convert the decimal in Step 5 to a percentage. P(1150 <X < 1250) = 95.44% %3D . Therefore, the probability that a randomly selected 50 senior high school students will complete the examination in less than 43 minutes is 0.23%. No, it's not reasonable since the probability is less than 1. . Therefore, the probability of randomly selected 100 bulbs to have a sample mean between 1 150 hours and 1 250 hours is 95.44%.
Transcribed Image Text:Solution for #1.a: Step1: Identify the parts of the problem. Given: u= 46.2 minutes; o= 8 minutes; X= 43 minutes; Find: P(X < 43) n = 50 students %3D Step 2: Use the formula to find the z-score. Solution: 43 - 46.2 8 Given: u= 1200 hours; o = 250 hours; Vn V50 X = 1 150 & 1 250 hours n = 100 bulbs Unknown: P(1150 < X < 1250) z = -2.83 Step 2: Use the formula to find the z-score. Step 3: Use the z-table to look up the z-score you calculated in step 2. z = -2.83 has a corresponding area of 0.4977 1150 1200 1250 1200 250 Vn V100 250 Step 4: Draw a graph and plot the z-score and its corresponding area. Then, shade V100 the part that you're looking for: P(X < 43) z = -2 Z = 2 Step 3: Use the z-table to look up the z-score you calculated in step 2. z = +2 has a corresponding area of 0.4772 Step 4: Draw a graph and plot the z-score and its corresponding area. Then, shade the part that you're looking for: P(1150 <X < 1250) sha ed part 0.4977 -3 -2 -1 1 2 3 -2.83 0.4772 0.4772 Since we are looking for the probability less than 43 minutes, the shaded part will be on the left part of - 2.83. -3 -2 -1 2 3 Since we are looking for the probability between 1 150 hours and 1 250 hours, the shaded part will be between -2 and 2. Step 5: Subtract your z-score from 0.500. P(X < 43) = 0.500 0.4977 P(X < 43) = 0.0023 Step 5: Add the two z-score values. P(1150 <X < 1250) = 0.4772 + 0.4772 P(1150 <X < 1250) = 0.9544 %3D Step 6: Convert the decimal in Step 5 to a percentage. P(X < 43) = 0.23% %3D Step 6: Convert the decimal in Step 5 to a percentage. P(1150 <X < 1250) = 95.44% %3D . Therefore, the probability that a randomly selected 50 senior high school students will complete the examination in less than 43 minutes is 0.23%. No, it's not reasonable since the probability is less than 1. . Therefore, the probability of randomly selected 100 bulbs to have a sample mean between 1 150 hours and 1 250 hours is 95.44%.
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