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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 normally distributed. If a sample of 25 canned goods is selected, what is the probability that the mean of the sample will be greater than 220 milligrams?

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 normally distributed. If a sample of 25 canned goods is selected, what is the probability that the mean of the sample will be greater than 220 milligrams?

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.3: Measures Of Spread
Problem 26PFA
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Answer no3 with on point answer and clear solutions the example is on the secod pic

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%.
distributed. If a sample of 25 canned goods is selected, what is the probability the standard deviation is 15 milligrams. Assume that the variable is normally 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 normall distributed. If a sample of 25 canned goods is selected, what is the probability that the mean of the sample will be greater than 220 milligrams? 4. The average public elementary school has 468 students with a standard deviation of 87, If a random samnle of 38 nuhlic elem
Transcribed Image Text:distributed. If a sample of 25 canned goods is selected, what is the probability the standard deviation is 15 milligrams. Assume that the variable is normally 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 normall distributed. If a sample of 25 canned goods is selected, what is the probability that the mean of the sample will be greater than 220 milligrams? 4. The average public elementary school has 468 students with a standard deviation of 87, If a random samnle of 38 nuhlic elem
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