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4. The average public elementary school has 468 students with a standard deviation of 87. If a random sample of 38 public elementary schools is selected, what is the probability that the number of students enrolled is between 445 and 485?

4. The average public elementary school has 468 students with a standard deviation of 87. If a random sample of 38 public elementary schools is selected, what is the probability that the number of students enrolled is between 445 and 485?

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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Answer no.4 with on point answer and clear solutions the example is on the second pic tysm

0 milligrams? 4. The average public elementary school has 468 students with a standard deviation of 87. If a random sample of 38 public elementary schools is selected, what is the probability that the number of students enrolled is between 445 and 485?
Transcribed Image Text:0 milligrams? 4. The average public elementary school has 468 students with a standard deviation of 87. If a random sample of 38 public elementary schools is selected, what is the probability that the number of students enrolled is between 445 and 485?
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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