MATH 225N Week 5 Assignment (2020) - Chamberlain College of Nursing | Applications of the Normal Distribution – Excel

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MATH 225N Week 5 Assignment (2020) : Applications of the Normal Distribution – Excel – Chamberlain College of Nursing 33 Questions with answers Question Sugar canes have lengths, X , that are... normally distributed with mean 365.45 centimeters and standard deviation 4.9 centimeters. What is the probability of the length of a randomly selected cane being between 360 and 370 centimeters?  • Round your answer to four decimal places. The mean is μ=365.45 , and the standard deviation is σ=4.9 . As the probability between two values is to be calculated, subtract the probability of the lower value from the higher value. In this case, you have to use the NORMDIST function twice. 1. Open Excel and click on any empty cell. Click Insert function, fx . 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 370 for X , 365.45 for Mean, 4.9 for Standard_dev, and TRUE for Cumulative, all for the higher value of X . Thus, the answer, rounded to four decimal places, is 0.8234 . 5. Click on any other empty cell. Click Insert function, fx . 6. Search for NORMDIST in the search for a function dialog box and click GO. 7. Make sure NORMDIST is on top in select a function. Then click OK. 8. In the function arguments of NORMDIST function, enter 360 for X , 365.45 for Mean, 4.9 for Standard_dev, and TRUE for Cumulative, all for the lower value of X . Thus, the answer, rounded to four decimal places, is 0.1330 . Now subtract, 0.8234−0.1330=0.6904 . Thus, the probability of the length of a randomly selected cane being between 360 and 370 centimeters is 0.6904 . Question The number of miles a motorcycle, X, will travel on one gallon of gasoline is modeled by a normal distribution with mean 44 and standard deviation 5. If Mike starts a journey with one gallon of gasoline in the motorcycle, find the probability that, without refueling, he can travel more than 50 miles.  • Round your answer to four decimal places. The mean is μ=44, and the standard deviation is σ=5. The probability that Mike can travel, without refueling, more than 50 miles is shown below.   A normal curve is over a horizontal axis and is divided into 3 regions. Vertical line segments extend from the horizontal axis to the curve at the mean, 44 and at 50. The right region is shaded.   First find the probability to the left of 50 and subtract from 1. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 50 for X, 44 for Mean, 5 for Standard_dev, and TRUE for Cumulative. This probability, rounded to four decimal places, is 0.8849. Now subtract, 1−0.8849=0.1151. Thus, the desired probability is P(X>50)=0.1151. Question A worn, poorly set-up machine is observed to produce components whose length X follows a normal distribution with mean 14 centimeters and variance 9. Calculate the probability that a component is at least 12 centimeters long. • Round your answer to four decimal places. The mean is μ=14, and the standard deviation is σ=9–√=3. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 12 for X, 14 for Mean, 3 for Standard_dev, and TRUE for Cumulative. The probability, rounded to four decimal places, is P(X<12)≈0.2525.  The desired probability is P(X≥12), so subtract from 1 to get P(X≥12)=1−0.2525=0.7475. 1. An organization has members who possess IQs in the top 4% of the population. If IQs are normally distributed, with a mean of 100 and a standard deviation of 15, what is the minimum IQ required for admission into the organization? Use Excel, and round your answer to the nearest integer: 126 2. The top 5% of applicants on a test will receive a scholarship. If the test scores are normally distributed with a mean of 600 and a standard distribution of 85, how low can an applicant score to still qualify for a scholarship? Use Excel, and round your answer to the nearest integer. 740 -Here, the mean, μ, is 600 and the standard deviation, σ, is 85. Let x be the score on the test. As the top 5% of the applicants will receive a scholarship, the area to the right of x is 5%=0.05. So the area to the left of x is 1−0.05=0.95. Use Excel to find x. -Open Excel. Click on an empty cell. Type =NORM.INV(0.95,600,85) and press ENTER. -The answer rounded to the nearest integer, is x≈740. Thus, an applicant can score a 740 and still be in the top 5% of applicants on a test in order to receive a scholarship. 3. The weights of oranges are normally distributed with a mean of 12.4 pounds and a standard deviation of 3 pounds. Find the minimum value that would be included in the top 5% of orange weights. Use Excel, and round your answer to one decimal place. 17.3 - - - - - - - - - - - - - - - - - - - - - - - - - - - 28. The times to complete an obstacle course is normally distributed with mean 73 seconds and standard deviation 9 seconds. What is the probability using the Empirical Rule that a randomly selected finishing time is less than 100 seconds? 99.85% 29. After collecting the data, Douglas finds that the finishing times for cyclists in a race is normally distributed with mean 149 minutes and standard deviation 16 minutes. What is the probability that a randomly selected race participant had a finishing time of less than 165 minutes? Use the empirical rule 84% 30. Charles has collected data to find that the total snowfall per year in Reamstown has a normal distribution. Using the Empirical Rule, what is the probability that in a randomly selected year, the snowfall was less than 87 inches if the mean is 72 inches and the standard deviation is 15 inches? 84% 31. Christopher has collected data to find that the total snowfall per year in Laytonville has a normal distribution. What is the probability that in a randomly selected year, the snowfall was greater than 53 inches if the mean is 92 inches and the standard deviation is 13 inches? Use the empirical rule Notice that 53 inches is three standard deviations less than the mean. Based on the Empirical Rule, 99.7% of the yearly snowfalls are within three standard deviations of the mean. Since the normal distribution is symmetric, this implies that 0.15% of the yearly snowfalls are less than three standard deviations below the mean. Alternatively, 99.85% of the yearly snowfalls are greater than three standard deviations below the mean. 32. The times to complete an obstacle course is normally distributed with mean 87 seconds and standard deviation 7 seconds. What is the probability that a randomly selected finishing time is greater than 80 seconds? Use the empirical rule Alternatively, 84% of the finishing times are greater than one standard deviation below the mean. 33. 1. Sugar canes have lengths, X, that are normally distributed with mean 365.45 centimeters and standard deviation 4.9centimeters. What is the probability of the length of a randomly selected cane being between 360 and 370 centimeters?  • Round your answer to four decimal places. .6904 2. On average, 28 percent of 18 to 34 year olds check their social media profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a random variable X, which has a standard deviation of five percent. Find the probability that the percent of 18 to 34 year olds who check social media before getting out of bed in the morning is, at most, 32.  • Round your answer to four decimal places. .7881 • The mean is μ=28, and the standard deviation is σ=5. • 1. Open Excel and click on any empty cell. Click Insert function, fx. • 2. Search for NORMDIST in the search for a function dialog box and click GO. • 3. Make sure NORMDIST is on top in select a function. Then click OK. • 4. In the function arguments of the NORMDIST function, enter 32 for X, 28 for Mean, 5 for Standard_dev, and TRUE for Cumulative. • Thus, the answer, rounded to four decimal places, is P(X<32)≈0.7881. 3. In a survey of men aged 20-29 in a country, the mean height was 73.4 inches with a standard deviation of 2.7 inches. Find the minimum height in the top 10% of heights.  • Use Excel, and round your answer to one decimal place. 76.9 4. Two thousand students took an exam. The scores on the exam have an approximate normal distribution with a mean of μ=81 points and a standard deviation of σ=4 points. The middle 50% of the exam scores are between what two values? • Use Excel, and round your answers to the nearest integer. 78-84 [Show More]

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